Edward M. Purcell David J. Morin - Electricity and Magnetism Solutions Manual (2013) PDF
Edward M. Purcell David J. Morin - Electricity and Magnetism Solutions Manual (2013) PDF
Edward M. Purcell David J. Morin - Electricity and Magnetism Solutions Manual (2013) PDF
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David Morin
morin@physics.harvard.edu
⃝
c D. Morin, D. Purcell, and F. Purcell 2013
Chapter 1
Electrostatics
Solutions manual for Electricity and Magnetism, 3rd edition, E. Purcell, D. Morin.
morin@physics.harvard.edu (Version 1, January 2013)
1.93 · 10−2 g
6.02 · 1023 · = 5.9 · 1019 . (1)
197 g
Each atom has a positive charge of 1 e = 1.6 · 10−19 C, so the total charge in the cube
is (5.9 · 1019 )(1.6 · 10−19 C) = 9.4 C. The repulsive force between two such cubes 1 m
apart is therefore
( )
q2 kg m3 (9.4 C)2
F = k 2 = 9 · 109 2 2 = 8 · 1011 N. (2)
r s C (1 m)2
The weight of an aircraft carrier is mg = (108 kg)(9.8 m/s2 ) ≈ 109 N. The above F is
therefore equal to the weight of 800 aircraft carriers. This is just another example of
the fact that the electrostatic force is enormously larger than the gravitational force.
1.35. Balancing the weight
Let the desired distance be d. We want the upward electric force e2 /4πϵ0 d2 to equal
the downward gravitational force mg. Hence,
1
2 CHAPTER 1. ELECTROSTATICS
Ty Fg mg
= 10 =⇒ = 10 =⇒ = 10. (4)
Tx Fe q 2 /4πϵ 0r
2
Therefore,
( )
1 s2 C 2
q2 = (4πϵ0 )mgr2 = (0.4)π 8.85 · 10−12 (0.3 kg)(9.8 m/s2 )(0.5 m)2
10 kg m3
= 8.17 · 10−12 C2 =⇒ q = 2.9 · 10−6 C. (5)
(a) Consider a charge q at √ a particular corner. If the square has side length ℓ, then
√
one of the other q’s is 2 ℓ away, two of them are ℓ away, and the −Q is ℓ/ 2
away. The net force on the given q, which is directed along the diagonal touching
it, is (ignoring the factors of 1/4πϵ0 since they will cancel)
q2 q2 Qq
F = √ + 2 cos 45◦ 2 − √ . (6)
( 2 ℓ) 2 ℓ (ℓ/ 2)2
(b) To find the potential energy of the system, we must sum over all pairs of charges.
Four pairs involve the charge −Q, four involve the edges of the square, and two
involve the diagonals. The total potential energy is therefore
( ) √ ( )
1 (−Q)q q2 q2 4 2q q q
U= 4· √ +4· +2· √ = −Q + √ + = 0, (8)
4πϵ0 ℓ/ 2 ℓ 2ℓ 4πϵ0 ℓ 2 4
in view of Eq. (7). The result in Problem 1.6 was “The total potential energy
of any system of charges in equilibrium is zero.” With Q given by Eq. (7), the
system is in equilibrium (because along with all the q’s, the force on the −Q
charge is also zero, by symmetry). And consistent with Problem 1.6, the total
potential energy is zero.
order x2 )
( )
Qq 1 1
F (x) = − −
4πϵ0 (ℓ − x)2 (ℓ + x)2
( )
Qq 1 1
≈ − −
4πϵ0 ℓ2 1 − 2x/ℓ 1 + 2x/ℓ
Qq ( )
≈ − (1 + 2x/ℓ) − (1 − 2x/ℓ)
4πϵ0 ℓ2
Qqx
= − . (9)
πϵ0 ℓ3
Alternatively: We can find the potential energy of the charge q at position (x, 0),
and then take the (negative) derivative to find the force. The energy is a scalar, so we
don’t have to worry about directions. We have
( )
Qq 1 1
U (x) = + . (11)
4πϵ0 ℓ − x ℓ + x
We’ll need to expand things to order x2 because the order x terms will cancel:
( )
Qq 1 1
U (x) = +
4πϵ0 ℓ 1 − x/ℓ 1 + x/ℓ
(( ) ( ))
Qq x x2 x x2
≈ 1+ + 2 + 1− + 2
4πϵ0 ℓ ℓ ℓ ℓ ℓ
( 2
)
Qq 2x
= 2+ 2 . (12)
4πϵ0 ℓ ℓ
The constant term isn’t important here, because only changes in the potential energy
matter. Equivalently, the force is the negative derivative of the potential energy, and
the derivative of a constant is zero. The force on the charge q is therefore
dU Qqx
F (x) = − =− , (13)
dx πϵ0 ℓ3
forces is equal and opposite to the sum of the vertical components of the tensions.
This gives
( )
qQ q2 q2 qQ
2 sin θ+ = 2T sin θ =⇒ 3 = 2T ℓ2 − . (14)
4πϵ0 ℓ2 4πϵ0 (2ℓ sin θ)2 16πϵ0 sin θ 2πϵ 0
One obvious location satisfying√this requirement has the proton on the y axis with
r1 = r2 = 2, that is, with y = 15/2 ≈ 1.94. In general, Eq. (17) defines a curve in
the xy plane, and a surface of revolution around the x axis in space. This surface is
the set of all points where the proton can be placed to give U = 0. The surface looks
something like a prolate ellipsoid, but it isn’t.
Let’s now consider the case where all three charges lie on the x axis. Assume that
the proton lies to the right of the right electron. We then have r1 = x − 1/2 and
r2 = x + 1/2, so Eq. (17) becomes
√
1 1 2± 5
+ = 1 =⇒ x − 2x − 1/4 = 0 =⇒ x =
2
. (18)
x − 1/2 x + 1/2 2
The negative root must be thrown out because it violates our assumption that x > 1/2.
(With x < 1/2, the distance r1 isn’t represented by x − 1/2).√ So we find x = 2.118.
The distance from the right electron at x = 1/2 equals (1 + 5)/2. The ratio of this
5
distance to the distance between the electrons (which is just 1) is therefore the golden
ratio. If we assume x < −1/2, then the mirror image at x = −2.118 works equally
well. You can quickly check that there is no solution for x between the electrons, that
is, in the region −1/2 < x < 1/2. There are therefore two solutions with all three
charges on the same line.
1.41. Work for an octahedron
Consider an edge that has two protons at its ends (you can quickly show that at least a
2a
one such edge must exist). There are two options for where the third proton is. It
can be at one of the two vertices such that the triangle formed by the three protons
is a face of the octahedron. Or it can be at one of the other two vertices. These two
possibilities are shown in Fig. 2.
There are 15 pairs of charges, namely the 12 edges and the 3 internal diagonals.
Summing over these pairs gives the potential energy. By examining the two cases √
shown, you can show that for the first configuration the sum is (the term with the 2
comes from the internal diagonals)
( )
e2 1 1 1 e2
U= 6· −6· −3· √ = −(2.121) . (19) a
4πϵ0 a a 2a 4πϵ0 a
2a
And for the second configuration:
( )
e2 1 1 1 1 e2
U= 4· −8· +2· √ −1· √ = −(3.293) . (20)
4πϵ0 a a 2a 2a 4πϵ0 a
Both of these results are negative. This means that energy is released as the octahedron
is assembled. Equivalently, it takes work to separate the charges out to infinity. You Figure 2
should think about why the energy is more negative in the second case. (Hint: the
two cases differ only in the locations of the leftmost two charges.)
1.42. Potential energy in a 1-D crystal
Suppose the array has been built inward from the left (that is, from negative infinity)
as far as a particular negative ion. To add the next positive ion on the right, the
amount of external work required is
( 2 ) ( )
1 e e2 e2 1 e2 1 1 1
− + − + ··· = − 1 − + − + ··· . (21)
4πϵ0 a 2a 3a 4πϵ0 a 2 3 4
An alternative solution is to compute the potential energy of a given ion due to the full
infinite (in both directions) chain. This is essentially the same calculation as above,
except with a factor of 2 due to the ions on each side of the given one. If we then sum
over all ions (or a very large number N ) to find the total energy of the chain, we have
counted each pair twice. So in finding the potential energy per ion, we must divide by
2 (along with N ). The factors of 2 and N cancel, and we arrive at the above result.
1.43. Potential energy in a 3-D crystal
The solution is the same as the solution to Problem 1.7, except that we have an
additional term. We now also need to consider the “half-space” on top of the ion, in
addition to the half-plane above it and the half-line to the right of it. In Fig. 12.4 the
half-space of ions is on top of the plane of the paper (from where you are viewing the
page).
If we index the ions by the coordinates (m, n, p), then the potential energy of the ion
at (0, 0, 0) due to the half-line, half-plane, and half-space is
( ∞ ∞ ∞ ∞ ∑ ∞ ∞
)
e2 ∑ (−1)m ∑ ∑ (−1)m+n ∑ ∑ (−1)m+n+p
U= + √ + √ .
4πϵ0 a m=1 m n=1 m=−∞
m2 + n2 p=1 n=−∞ m=−∞ m2 + n2 + p2
(22)
The triple sum takes more computer time than the other two sums. Taking the limits
to be 300 instead of ∞ in the triple sum, and 1000 in the other two, we obtain decent
enough results via Mathematica. We find
e2 (0.874)e2
U= (−0.693 − 0.115 − 0.066) = − , (23)
4πϵ0 a 4πϵ0 a
which agrees with Eq. (1.18) to three digits. This result is negative, which means that
it requires energy to move the ions away from each other. This makes sense, because
the six nearest neighbors are of the opposite sign.
1.44. Chessboard
W is probably going to be positive, because the four nearest neighbors are all of the
opposite sign. Fig. 3 shows a quarter (or actually slightly more than a quarter) of a
7 × 7 chessboard. Three different groups of charges are circled. The full chessboard
consists of four of the horizontal group, four of the diagonal group, and eight of the
Figure 3 triangular group. Adding up the work associated with each group, the total work
required to move the central charge to a position far away is (in units of e2 /4πϵ0 s)
( ) ( ) ( )
1 1 1 1 1 1 1 1 1
W =4 − + + 4 −√ − √ − √ +8 √ − √ + √ ≈ 1.4146,
1 2 3 2 2 2 3 2 5 10 13
(24)
which is positive, as we guessed.
For larger arrays we can use a Mathematica program to calculate W . If we have
an N × N chessboard, and if we define H by 2H + 1 = N (for example, H = 50
corresponds to N = 101), then the following program gives the work W required to
remove the central charge from a 101 × 101 chessboard.
H=50;
4*Sum[(-1)^(n+1)/n, {n,1,H}] +
4*Sum[Sum[(-1)^(n+m+1)/(n^2+m^2)^(.5), {n,1,H}], {m,1,H}]
H
H
7
This program involves dividing the chessboard into the regions shown in Fig. 4; the
sub-squares have side length H. (If you want, you can reduce the computing time by
about a factor of 2 by dividing the chessboard as we did in Fig. 3.) The results for
various N × N chessboards are (in units of e2 /4πϵ0 s):
Figure 4
N 3 7 101 1001 10,001 100,001
W 1.1716 1.4146 1.6015 1.6141 1.6154 1.6155
where the second factor on each side of the equation comes from the act of taking the
horizontal component. Simplifying this gives
1 3
= 2 =⇒ y 2 + 9a2 = 32/3 (y 2 + a2 )
(y 2 + a2 )3/2 (y + 9a2 )3/2
√
9 − 32/3
=⇒ y=a ≈ (2.53)a. (26)
32/3 − 1
In retrospect, we know that there must exist a point on the y axis with Ex = 0,
by a continuity argument. For small y, the field points leftward, because the two
middle charges dominate. But for large y, the field points rightward, because the two
outer charges dominate. (This is true because for large y, the distances to the four
charges are all essentially the same, but the slope of the lines to the outer charges is
smaller than the slope of the lines to the middle charges (it is 1/3 as large). So the
x component of the field due to the outer charges is 3 times as large, all other things Q
being equal.) Therefore, by continuity, there must exist a point on the y axis where
Ex equals zero.
θ θ R cos θ
1.46. Charges on a circular track
Let’s work with the general angle θ shown in Fig. 6. In the problem at hand, 4θ = 90◦ ,
so θ = 22.5◦ . The tangential electric field at one of the q’s due to Q is 4θ
θ R θ
Q
sin θ, (27)
4πϵ0 (2R cos θ)2 q q
and the tangential field (in the opposite direction) at one q due to the other q is 2θ
q
cos 2θ. (28) Figure 6
4πϵ0 (2R sin 2θ)2
Equating these fields gives
Q q cos2 θ cos 2θ cos 2θ
sin θ = cos 2θ =⇒ Q = q =q , (29)
(cos θ)2 (sin 2θ) 2
sin θ sin2 2θ 4 sin3 θ
8 CHAPTER 1. ELECTROSTATICS
where we have used sin 2θ = 2 sin θ cos θ. Letting θ = 22.5◦ gives Q = (3.154)q.
Some limits: If all three charges are equally spaced (with θ = 30◦ ) then Q = q, as
expected. If θ → 0 then Q ≈ q/(4θ3 ). (Two of these powers of θ come from the r2 in
Coulomb’s law, and one comes from the act of taking the tangential component of Q’s
field.) If the q’s are diametrically opposite (with θ = 45◦ ) then Q = 0, as expected.
where the minus sign indicates that the field points downward (if Q is positive). This
result can be written as −λ/2πϵ0 R, where λ is the linear charge density. Interestingly,
it can also be written as −Q/4πϵ0 A, where A = πR2 /2 is the area of the semicircle.
where the constant a is the value of r at θ = 0, that is, the height of the curve
along the y axis. We therefore have a family of curves indexed by a.
9
(b) Assume that the material has been shaped and positioned so that the electric
field at the origin takes on the maximum possible value. Assume that the field
points in the y direction. Then all the small elements of charge dq on the surface
of the material must give equal contributions to the y component of the field at
the origin. This is true because if it weren’t the case, then we could simply move
a tiny piece of the material from one point on the surface to another, thereby
increasing the field at the origin, in contradiction to our assumption that the field
at the origin is maximum. From part (a), any vertical cross section (formed by
the intersection of the
√ surface with a plane containing the y axis) must therefore
look like the r = a cos θ curve we found. Equivalently, the desired shape √ of
the material is obtained by forming the surface of revolution of the r = a cos θ
curve, around the y axis.
Let’s try to get a sense of what the surface looks like.
√ We know that the height is
a. To find the width, note that x = r sin θ = a sin θ cos θ. Taking the derivative
of this function
√ of θ, you can show that the maximum value of x is achieved when
tan θ = 2; the maximum value is (4/27)1/4 a. The width is twice this value, or
2(4/27)1/4 a ≈ 1.24a. So the ratio of width to height is about 5 to 4.
Let’s compare our shape with a sphere of the same volume. The volume of our
shape can be obtained√ by slicing it into horizontal disks. The radius of a disk is
x = r sin θ = a sin θ cos θ.√ And since y = −r cos θ = −a(cos θ)3/2 , the height of
a disk is dy = (3/2)a sin θ cos θ dθ. So the volume is
∫ 0 ∫ π/2 √ √
( )2 3
V = (πx2 ) dy = π a sin θ cos θ · a sin θ cos θ dθ
−a 0 2
∫ π/2
3 3
= πa sin3 θ cos3/2 θ dθ. (34)
2 0
Writing sin2 θ as 1 − cos2 θ yields integrals of sin θ cos3/2 θ and − sin θ cos7/2 θ,
which give 2/5 and −2/9, respectively. The sum of these is 8/45, so the volume is
V = (4/15)πa3 . Since the diameter of a sphere with volume V is (6V /π)1/3 , we see 1.06 d
that a sphere with the same volume would have a diameter of (8/5)1/3 a ≈ 1.17a.
Compared with a sphere of the same volume, our shape is therefore stretched by
a factor of (1.24)a/(1.17)a ≈ 1.06 in the x direction, and squashed by a factor of
a/(1.17)a ≈ 0.85 along the y direction. Cross sections of our shape and a sphere 0.85 d
d
with the same volume are shown in Fig. 8.
1.50. Field from a hemisphere
(a) Consider the ring shown in Fig. 9, defined by the angle θ and subtending an angle
dθ. Its area is 2π(R cos θ)(R dθ), so its charge is σ(2πR2 cos θ dθ). The horizontal d
component of the field at the center of the hemisphere is zero, by symmetry. So Figure 8
we need only worry about the vertical component from each piece of the ring,
which brings in a factor of sin θ. Adding up these components from all the pieces
of the ring gives the magnitude of the field at the center of the hemisphere, due
to the given ring, as R
θ
σ(2πR2 cos θ dθ) σ sin θ cos θ dθ
dE = 2
sin θ = . (35)
4πϵ0 R 2ϵ0 Figure 9
The field points downward if σ is positive. Integrating over all the rings (θ runs
from 0 to π/2) gives the total field at the center as
∫ π/2 π/2
σ sin θ cos θ dθ σ sin2 θ σ
E= = = . (36)
0 2ϵ0 2ϵ0 2 0 4ϵ0
10 CHAPTER 1. ELECTROSTATICS
This is half as large as the field at the center of the end face of a half-infinite hollow
cylinder (see Problem 1.11). You should convince yourself why the cylinder’s field
must indeed be larger (although the factor of 2 is by no means obvious). Hint:
consider the field contributions from corresponding bits of the surfaces subtending
the same solid angle. The cylinder’s surface is tilted with respect to the line to
the center of the end face.
In terms of the total charge Q = 2πR2 σ on the hemisphere, the result in Eq. (36)
can be written as Q/8πϵ0 R2 . If you solved Exercise 1.47 you will note that
the present field is smaller (by a factor of π/4) than the field at the center of
a semicircle with the same charge Q. This is because the semicircle’s charge is
generally higher up, so the act of taking the vertical component doesn’t reduce
the field as much as for the hemisphere. Equivalently, building a hemispherical
cage out of a large number of semicircular pieces of wire would effectively yield a
hemisphere with a larger surface charge density near the top than near the base.
(b) The field due to a solid hemisphere with radius R and uniform volume charge
density ρ can be found by slicing up the solid hemisphere into concentric hemi-
spherical shells with thickness dR. The effective surface charge density of each
shell is ρ dR, so the result from part (a) tells us that the field from each shell is
(ρ dR)/4ϵ0 . Integrating over R simply turns the dR into an R, so the total field is
Q0 ρR/4ϵ0 . Again, this is half as large as the field from a half-infinite solid cylinder
(see Problem 1.11).
θ/2
1.51. N charges on a circle
θ/2 Let Q0 be the point charge at the top of the circle, and consider the nth point charge
θ/2 away from it (call it Qn ), as shown in Fig. 10 for n = 4 and N = 12. The angle θ
equals n(2π/N ), so the distance from Q0 to Qn is rn = 2R sin(θ/2) = 2R sin(nπ/N ).
N =12 The horizontal components of all the fields at Q0 cancel in pairs, so we’re concerned
Q4
only with the vertical component, which brings in a factor of sin(θ/2). The vertical
component of the field at Q0 due to Qn is therefore
Figure 10 Q/N
En = ( )2 sin(nπ/N ). (37)
4πϵ0 2R sin(nπ/N )
∑
N −1
Q 1
E= ( ). (38)
16πϵ0 R2 n=1 N sin(nπ/N )
If you compute this sum numerically for various values of N , you will find that it grows
with N , although very slowly (like a log). The sum does in fact diverge as N → ∞, due
to the behavior of the terms with small n (and likewise for n close to N , because the
terms are symmetric around n = N/2). If n ≪ N , (we can write ) sin(nπ/N ) ≈ nπ/N ,
so for small n the field from Qn behaves like 1/ N (nπ/N ) = 1/nπ. And since
∑M
the sum n=1 1/n diverges (like ln M ), we see that the total field diverges. We are
assuming that N is large enough so that n can become very large and still have the
approximation sin(nπ/N ) ≈ nπ/N be valid. This is of course true in the N → ∞
limit.
Note that the only possible cause for the divergence of the total field is the behavior
of the fields from nearby Qn . There is a finite total charge on the ring, so the field
from the non-infinitesimally-close charges must be finite, because those charges don’t
11
involve any infinitesimal distances that would make the fields diverge. Equivalently,
once the sin(nπ/N ) term in the denominator of the field becomes non-infinitesimal,
the fields go like 1/N , so the sum (which involves fewer than N terms) is bounded.
We saw above that the vertical field contribution from each of the nearby charges
equals 1/nπ, which is finite. In short, in Eq. (37) the small charge Q/N and the small
factor of sin(nπ/N ) from the vertical component cancel the square of the small distance
2R sin(nπ/N ) in the denominator. But the total field ends up diverging because there
are so many charges that are very close to Q0 .
Even though the field at Q0 diverges in the N → ∞ limit, the actual force on Q0
goes to zero. The force equals the field times the charge Q/N , and since the field only
diverges like ln N , the force behaves like (ln N )/N , which goes to zero for large N .
A continuous circle of charge is equivalent to the N → ∞ limit. So if an additional
point charge with finite (non-infinitesimal) charge q were placed exactly on the cir-
cumference of the (infinitesimally thin) circle, the force on it would be infinite, due to
the infinite field. However, in reality there are no true point charges or infinitesimally
thin distributions of charge.
(a) Let F0 be the force between two charges of q = 10−6 C each, at a distance of
3q
a = 0.2 m. Then F0 = q 2 /4πϵ0 a2 = 0.225 N, as you can verify. The force
A
between B and C has magnitude (2)(2)F0 = 4F0 , and the force between A and
either B or C has magnitude (3)(2)F0 = 6F0 . From Fig. 11, the magnitude of a
the force on A is
FA = 2 cos 30◦ · 6F0 = 2.34 N. (39) B C 4F0
The magnitude of the force on C is (squaring and adding the horizontal and 2q 2q
vertical components) 6F0
Figure 11
[ ]1/2
FC = (4 + 6 cos 60◦ )2 + (6 sin 60◦ )2 F0 = (8.72)F0 = 1.96 N. (40)
dq λR dθ
dE = = . (42)
4πϵ0 ℓ2 4πϵ0 (R2 + z 2 )
Let the x axis split the semicircle in half. Then the net Ey field is zero, by symmetry.
The (magnitudes of the) z and x components of the dE field in Eq. (42) are obtained
by multiplying it by z/ℓ and x/ℓ. (The latter of these can be obtained in two steps:
12 CHAPTER 1. ELECTROSTATICS
multiply by R/ℓ to get the component in the x-y plane, then multiply by x/R to get
the x component.) So we have
λRz λR2
Ez = , and Ex = . (44)
4ϵ0 (R2 + z 2 )3/2 2πϵ0 (R2 + z 2 )3/2
Hence Ez /Ex = πz/2R, which can be written a little more informatively as Ez /Ex =
z/(2R/π). This is the slope of the E vector at a point at height z on the z axis.
The slope covers a horizontal distance 2R/π while covering a vertical distance z. The
straight line that points in the direction of the electric field at the point (0, 0, z)
therefore passes through the point (2R/π, 0, 0) in the plane of the semicircle. This
point is independent of z, as desired.
This point also happens to be the “center of charge” of the semicircle, or equivalently
the center of mass of a semicircle made out of a piece of wire (we’ll leave it to you to
verify this). So the result of this exercise is consistent with the following fact (which
you may want to try to prove): Far away from a distribution of charges, the electric
field points approximately toward the center of charge of the distribution. For nearby
points it generally doesn’t, although it happens to (exactly) point in that direction for
points on the axis of the present setup.
(b) Imagine cutting the two-dimensional setup into thin strips defined by a large
number of vertical planes, rotated at small angles with respect to each other, all
passing through the axis of the cylinder. Each thin strip is similar to the one-
dimensional setup in part (a), except for the fact that the strip gets narrower as
it approaches the top of the hemisphere (it is somewhat like a curved pie piece).
The linear charge density is therefore effectively smaller at the top. We know
from part (a) that a uniform linear charge density leads to the downward field
from the circular part of a strip canceling the upward field from the straight part.
A smaller density on the circular part therefore means that its downward field
can’t fully cancel the upward field from the straight part. The net field therefore
points upward.
This result is consistent with the results from Problem 1.11 (which says that
the upward field at C due to the cylinder is σ/2ϵ0 ) and either Problem 1.12 or a l l
A
Exercise 1.50 (which say that the downward field at C due to the hemisphere is
σ/4ϵ0 ). b
1.55. Field from a finite rod B
In Fig. 13, define the distances: ℓ = 0.05 m, a = 0.03 m, and b = 0.05 m. The linear Figure 13
charge density of the rod is λ = (8 · 10−9 C)/(0.1 m) = 8 · 10−8 C/m. At point A the
field points leftward and has magnitude
∫ a+2ℓ ( ) ( )
1 λ dx λ 1 1 λ 2ℓ
EA = = − =
4πϵ0 a x2 4πϵ0 a a + 2ℓ 4πϵ0 a(a + 2ℓ)
( 3
)( )( )
9 kg m −8 C 2(.05 m)
= 9 · 10 2 2 8 · 10
s C m (.03 m)(.13 m)
N
= 1.85 · 104 . (47)
C
As a check, if a ≫ ℓ this result approaches (1/4πϵ0 )(2ℓλ/a2 ), which is correctly the
field from a point charge 2ℓλ at a distance a.
At point B, only the vertical component of the field survives, by symmetry. So the
field points downward and has magnitude
∫ ℓ
1 λ dx b
EB = ·√ , (48)
4πϵ0 −ℓ b2 + x2 b2 + x2
where the second factor gives the vertical component. This integral can be evaluated
with a trig substitution, x = b tan θ =⇒ dx = b dθ/ cos2 θ (or you can just look it up),
which yields
∫ tan−1 (ℓ/b) ∫ −1
1 λb2 dθ/ cos2 θ 1 λ tan (ℓ/b)
EB = = cos θ dθ
4πϵ0 − tan−1 (ℓ/b) b3 (1 + tan2 θ)3/2 4πϵ0 b − tan−1 (ℓ/b)
tan−1 (ℓ/b)
1 λ 1 λ 2ℓ
= sin θ = √
4πϵ0 b − tan−1 (ℓ/b) 4πϵ0 b ℓ + b2
2
( )
kg m3 (8 · 10−8 C/m) 2(.05 m)
= 9 · 109 2 2 √
s C (.05 m) (.05 m)2 + (.05 m)2
N
= 2.04 · 104 . (49)
C
∫
The cos θ dθ integral here is just what you would obtain if you parameterized the
rod in terms of θ; see Eq. (1.38).
14 CHAPTER 1. ELECTROSTATICS
(a) The total flux through the cube is q/ϵ0 , by Gauss’s law. The flux through every
face of the cube
∫ is the same, by symmetry. Therefore, over any one of the six
faces we have E · da = q/6ϵ0 .
(b) Because the field due to q is parallel to the surface of each of the three faces that
touch q, the flux through these faces is zero. The total flux through the other
three faces must therefore add up to q/8ϵ0 , because our cube is one of eight such
cubes surrounding q. Since the three faces are symmetrically located with respect
to q, the flux through each must be (1/3)(q/8ϵ0 ) = q/24ϵ0 .
Note: if the charge were a true point charge, and if it were located just inside or
just outside the cube, then the field would not be parallel to each of the three
faces that touch the given corner. The flux would depend critically on the exact
location of the point charge. Replacing the point charge with a small sphere,
whose center lies at the corner, eliminates this ambiguity.
(a) You can quickly show that the desired point with E = 0 must satisfy x > a.
Equating the magnitudes of the fields from the two given charges then gives
2q q
= =⇒ 2(x − a)2 = x2
4πϵ0 x2 4πϵ0 (x − a)2
√
√ 2a √
=⇒ 2(x − a) = x =⇒ x = √ = (2 + 2)a ≈ (3.414)a. (50)
2−1
A few field lines, are shown in Fig. 14.1 Note that the field points in four different
directions near the E = 0 point. This is consistent with the fact that the zero
vector is the only vector that can simultaneously point in different directions.
(b) Consider a field line that emerges from the 2q charge and ends up at the x =
(3.414)a point where the field is zero. (There are actually no field lines that end
up right at this point, but we can pick a line infinitesimally close.) Field lines
that emerge at a smaller angle (with respect to the x axis) end up at the −q
charge, and field lines that emerge at a larger angle end up at infinity. Consider
the Gaussian surface indicated in Fig. 15; the surface is formed by rotating the
black curve around the x axis. This surface follows the field lines except very
close to the 2q charge, where it takes the form of a small spherical cap. The total
charge enclosed within this surface is simply −q, so from Gauss’s law there must
be an electric-field flux of q/ϵ0 pointing in to the surface. By construction, the
only place where there is flux is the spherical cap, so all of the q/ϵ0 flux must
occur there. But the total flux emanating from the charge 2q is 2q/ϵ0 , so the
spherical cap must represent half of the total area of a small sphere surrounding
the charge 2q. (Very close to the 2q charge, that charge dominates the electric
field, so the field is essentially spherically symmetric.) The cap must therefore
be a hemisphere, so the desired angle is 180◦ .
1 This figure technically isn’t a plot of field lines, because you can see that some of the lines begin in
empty space, whereas we know that field lines can begin and end only at charges or at infinity. So the
density of the lines on the page doesn’t indicate the field strength. (Well, it fails to do that even if we
15
0
2q -q E=0
-1
-2
-3
-2 -1 0 1 2 3 4 5 (in units of a)
Figure 14
0 2q
-q E=0
-1
-2
-3
-2 -1 0 1 2 3 4 5 (in units of a)
Figure 15
If the charges take on the more general values of N q and −q, then the spherical
cap represents 1/N of the complete sphere. So the task is to find the angle
subtended by a cap with 1/N of the total area. This requires an integral (whereas
in the above case we could simply say we had a hemisphere). But one nice case
is N = 4, which leads to an angle of 60◦ .
(a) Let the ring lie in the horizontal plane. A small piece of the ring with charge dq
produces a field dq/4πϵ0 R2 at the center. At a small vertical distance z above
the center, the magnitude of the field due to the dq piece is essentially the same
(it differs only at order z 2 /R2 , by the Pythagorean theorem ), so the vertical
component is obtained by simply tacking on a sin θ type factor, which is z/R
here. Integrating over the whole ring turns the dq into Q, so the desired vertical
don’t have any lines that abruptly end, because a 2D picture can’t mimic the actual density in 3D space.)
However, every curve shown is at least part of a field line, so the figure is still helpful in visualizing the flow
of the actual field lines.
16 CHAPTER 1. ELECTROSTATICS
field is Qz/4πϵ0 R3 . Alternatively, you can calculate the field exactly (as in the
solution to Exercise 1.48) and then take the z ≪ R limit.
(b) The solution to Problem 1.8 tells us that in the plane of the ring, the field near
the center, at radius r, points radially inward (assuming Q is positive) with mag-
nitude λr/4ϵ0 R2 . But since λ = Q/2πR, this can be written as Qr/8πϵ0 R3 .
Consider a point near the center, with a nonzero r value, and also a nonzero z
value. To leading order, the horizontal component of the field is still Qr/8πϵ0 R3 ,
and the vertical component is still Qz/4πϵ0 R3 , from part (a). That is, these re-
sults are actually valid for all points in space near the origin, not just in the plane
of the ring or on the axis. You can check this by writing out the exact expressions
for the fields. For example, in part (a) the effective values of R change slightly
if the point is off the axis, but this doesn’t change the field, to leading order.
Alternatively, note that due to symmetry, the horizontal component Er (r, z) is
an even function of z. This means that Er (r, z) has no linear dependence on
z. The variation with z therefore starts only at order z 2 , which is negligible for
small z. So Er is essentially independent of z near the axis. Similar reasoning
works with Ez as a function of r.
For simplicity, let’s define A ≡ Q/8πϵ0 R3 . Then the horizontal and vertical field
components have magnitudes Ar and 2Az, respectively. The top and bottom
faces of the small cylinder have a combined area of 2πr02 . And the vertical cylin-
drical side has an area of (2πr0 )(2z0 ). There is outward flux through the top and
bottom, and inward flux through the side, so the net outward flux equals
(2πr02 )(2Az0 ) − (4πr0 z0 )(Ar0 ) = 0, (51)
as desired. If we work backwards, this exercise actually provides a much quicker
method, compared with the one in Problem 1.8, for finding the horizontal com-
ponent of the field near the center of the ring, assuming that we know the vertical
component.
It is interesting to ask what happens to the mass if the charge density is kept constant
but the radius is doubled. You might think that since there is 23 = 8 times as much
18 CHAPTER 1. ELECTROSTATICS
stuff (charge) present, the mass should be 8 times as large. However, the charge e is
squared in the above formula for the potential energy, so this yields a factor of 82 = 64.
But there is also one power of r0 in the denominator, so this cuts the result down to
32. Equivalently, the result for the energy in Eq. (53) in the solution to Exercise 1.61
is proportional to R5 , and 25 = 32.
(a) There is no change in speed inside the shell, because the electric field is zero
inside. So we just need to find the speed of the particle when it reaches the
surface. The charge on the shell is 4πR2 σ, so the potential energy of the particle
at the surface of the shell is
(4πR2 σ)(−q) Rσq
V (R) = =− . (55)
4πϵ0 R ϵ0
The initial potential energy was zero, so this loss √
in potential energy shows up as
kinetic energy. Hence, mv 2 /2 = Rσq/ϵ0 =⇒ v = 2Rσq/ϵ0 m.
(b) Let’s find the potential energy U of the particle, due to one of the cones, when it
is located at the tip of the cones. We’ll slice the cone into rings and then integrate
over the rings. Consider a thin ring around the cone, located at a slant distance
x away from the tip. The charge in this ring is dQ = σ2πr dx, where the radius
r is given by r/x = R/L =⇒ r = xR/L. Every point in the ring is the same
distance x from the tip, so
( )
dQ(−q) σ2π(xR/L) dx q Rσq dx
dU = =− =− . (56)
4πϵ0 x 4πϵ0 x 2ϵ0 L
Integrating from x = 0 to x = L simply turns the dx into an L, so we have
U = −Rσq/2ϵ0 . We need to double this because there are two cones, so we end
√ energy of −Rσq/ϵ0 as in part (a). We therefore obtain
up with the same potential
the same speed of v = 2Rσq/ϵ0 m, independent of L.
λ ( )
15, 000 N/C = 2 =⇒ λ = 15, 000 N/C πϵ0 r
2πϵ0 r
( )
( ) s2 C2
= 15, 000 N/C (3.14) 8.85 · 10−12 (1.5 m)
kg m3
= 6.3 · 10−7 C/m. (57)
P
The amount of excess charge on 1 km of the positive wire is then (1000 m)λ = 6.3 ·
y θ
r 10−4 C.
You should verify the algebra leading to the last line. The force per unit height
is therefore Fh /h = σ 2 b(ln 2)/πϵ0 , which is finite. The field diverges as x → 0,
rod (into but only like a log. This isn’t large enough to outweigh the fact that there is only
P and out a small range of x that is very close to the strip. Basically, the area under the
of page) ln x curve near x = 0 is finite.
Interestingly, we see that for a given angular width of a rod, all rods yield the
same contribution to the vertical electric field at P .
(b) As noted in the statement of the exercise, it is no surprise that the above result
is incorrect, because the same calculation would supposedly yield the field just
inside the cylinder too, where it is zero instead of σ/ϵ0 . The calculation does,
correct however, give the next best thing, namely the average of these two values. We’ll
E direction correct see why shortly.
P
distance The reason why the calculation is invalid is that it doesn’t correctly describe
the field due to rods on the cylinder very close to the given point, that is, for
incorrect
E direction rods characterized by θ ≈ 0. It is incorrect for two reasons. The closeup view
in Fig. 22 shows that the distance from a rod to the given point is not equal to
2R sin(θ/2). Additionally, it shows that the field does not point along the line
2Rsin(θ/2) from the rod to the top of the cylinder. It points more vertically, so the extra
incorrect intersection factor of sin(θ/2) in Eq. (62) isn’t valid.
distance of rod with
page What is true is that if we remove a thin strip from the top of the cylinder (so we
now have a gap in the circle representing the cross sectional view), then the above
Figure 22 integral is valid for
∫ the remaining part of the cylinder. The thin strip contributes
negligibly to the dθ integral, so we can say that the field due to the remaining
part of the cylinder is equal to the above result of σ/2π. By superposition, the
total field due to the entire cylinder is this field of σ/2π plus the field due to the
thin strip. But if the point in question is infinitesimally close to the cylinder, then
the thin strip looks like an infinite plane, the field of which we know is σ/2ϵ0 .
The desired total field is then
σ σ σ
Eoutside = Ecylinder minus strip + Estrip = + = . (63)
2ϵ0 2ϵ0 ϵ0
By superposition we also obtain the correct field just inside the shell:
σ σ
Einside = Ecylinder minus strip − Estrip = − = 0. (64)
2ϵ0 2ϵ0
The relative minus sign arises because the field from the cylinder-minus-strip is
continuous across the gap, but the field from the strip is not; it points in different
directions on either side of the strip.
21
for any value of r up to the radius∫of the sphere. We quickly see that this proportion-
ality holds if ρ(x) ∝ 1/x, because (x2 /x) dx = x2 /2. So this is the desired form of ρ;
it is inversely proportional to the radius. Although ρ diverges at the origin, the charge
there is still finite (the amount of charge within a sphere with radius r is proportional
to r2 ) because of the 4πx2 in the volume element in the above integral. Note that the
field right at the center isn’t well defined.
Alternatively (or rather, equivalently), since Gauss’s law tells us that we want Qr =
Br2 , for some constant B, we have dQr /dr = 2Br. But a general expression for
dQr /dr is 4πr2 ρ (if we imagine adding on a thin shell). Equating these two expressions
for dQr /dr gives ρ = B/2πr ∝ 1/r, as desired.
The solution for the case of a cylinder is similar. Again let Qr,ℓ be the charge inside
radius r, for a length ℓ of the cylinder. Since Gauss’s law tell us that Er ∝ Qr,ℓ /rℓ,
we therefore want Qr,ℓ ∝ rℓ if Er is to be constant. That is, we want
∫ r
2πxℓρ(x) dx ∝ rℓ, (66)
0
Qb (4/3)πa3 ρ aρ
E0 ≡ Eb,B = 2
= 2
= , (67)
4πϵ0 a 4πϵ0 a 3ϵ0
and the field is directed downward. The field at A due to the big sphere is Eb,A = 0.
The field at A due to the small (negative) sphere has magnitude
Qs (4/3)π(a/2)3 ρ aρ E0
Es,A = 2
= = = , (68)
4πϵ0 (a/2) 4πϵ0 (a/2)2 6ϵ0 2
and is directed upward. The field at B due to the small sphere has magnitude
Qs (4/3)π(a/2)3 ρ aρ E0
Es,B = 2
= 2
= = , (69)
4πϵ0 (3a/2) 4πϵ0 (3a/2) 54ϵ0 18
σ0 5σ0 σ0 The field from an infinite sheet with charge density σ has magnitude σ/2ϵ0 . It is
___ ___ ___
2ε0 2ε0 2ε0 directed away from the sheet if σ is positive, and toward it if σ is negative. The total
field in the given setup equals the superposition of the fields from each sheet; the result
is shown in Fig. 23. The field has magnitude (3σ0 + 2σ0 )/2ϵ0 = 5σ0 /2ϵ0 inside the
sheets and (3σ0 − 2σ0 )/2ϵ0 = σ0 /2ϵ0 outside the sheets. In all regions it is directed
away from the 3σ0 sheet.
If the sheets intersect at right angles, the field is again obtained by superposition, but
Figure 23 now the two individual fields are orthogonal. Fig.√24 shows the results in the four
regions. The magnitude of the field everywhere is 32 + 22 (σ0 /2ϵ0 ) ≈ (1.8)σ0 /ϵ0 . In
-2σ0 3σ1
___ all regions it is directed at least partially away from the 3σ0 sheet and partially toward
2ε0 the −2σ0 sheet.
2σ1
___ 1.71. Intersecting sheets
2ε0
(a) The electric field from a given sheet points away from the sheet and has uniform
3σ0 magnitude σ/2ϵ0 . The three fields at a given point therefore take the form shown
in Fig. 25. At the location shown, the net field is directed exactly rightward and
has magnitude
σ σ σ
E = E2 + (E1 + E3 ) cos 60◦ = +2 cos 60◦ = . (70)
2ϵ0 2ϵ0 ϵ0
The field has the same magnitude and direction (to the right) everywhere in the
rightmost “pie piece,” because the field due to a sheet doesn’t depend on the
Figure 24
distance from the sheet. Similarly, in the other five pie pieces the magnitude is
σ/2ϵ0 , and the direction is parallel to the line that bisects the angle of the pie
σ piece.
σ σ
E3
60
E2
1 3
E1
2
Figure 25
σ
23 σ σ
r
(b) Fig. 26 shows the field at points on a symmetrically-located hexagon. Let the r r
“radius” of the hexagon be r, and consider a hexagonal tube with length ℓ perpen-
dicular to the page. The surface area of this tube is 6rℓ, and the charge enclosed is
6rℓσ. Since the electric field is everywhere perpendicular to the surface, Gauss’s
law gives
∫
Q 6rℓσ σ Figure 26
E · da = =⇒ E · 6rℓ = =⇒ E = , (71)
ϵ0 ϵ0 ϵ0
in agreement with the result in part (a). Again, note that E is independent of
r. While Gauss’s law is always valid, it was actually useful in the present setup
because we were able to find a simple surface that is everywhere perpendicular
to the electric field (because the electric field is uniform in each pie piece).
(c) For general N , the electric field is everywhere
( perpendicular
) to a regular 2N -gon.
The surface area of this 2N -gon is (2N ) 2 sin(π/2N ) rℓ, and the charge enclosed
is (2N )rℓσ. So Gauss’s law gives
( ) (2N )rℓσ σ
E · (2N ) 2 sin(π/2N ) rℓ = =⇒ E = . (72)
ϵ0 2ϵ0 sin(π/2N )
As expected, this is independent of r. And it agrees with the above result when
N = 3. For large N , we have sin(π/2N ) ≈ π/2N , so E ≈ N σ/πϵ0 . In the case of
large N , the sheets are very close to each other, so we effectively have a continuous
volume charge distribution that depends on r. The separation between adjacent
sheets grows linearly with r, so we have ρ(r) ∝ 1/r. More precisely, you can show
that ρ(r) = N σ/πr. This is consistent with the result from Exercise 1.68, where
we found that a cylinder with a density of the form ρ(r) ∝ 1/r produces a field
whose magnitude is independent of r (inside the cylinder).
σ + ρx (d − x)ρ σ − ρd + 2ρx
− = . (73)
2ϵ0 2ϵ0 2ϵ0 ρ
x
The plot of E as a function of x is shown in Fig. 27. E is continuous at x = d but not
at x = 0. If the plane had a nonzero thickness, then the field would be continuous at
x = 0. The case shown in the plot has ρd > σ. If we instead had σ > ρd, then at the x=d
discontinuity at x = 0, E would jump to a positive value.
Ex
1.73. Sphere in a cylinder
σ+ρd
_____
From the reasoning in the solution to Problem 1.27, the electric field inside a uniform 2ε0
cylinder is E = ρr/2ϵ0 , where r points away from the axis. And the electric field inside ρd
__
d ε0
a uniform sphere is E = ρr/3ϵ0 , where r points away from the center. x
The given setup may be considered to be the superposition of a uniform cylinder with σ
__
σ+ρd
density ρ and a uniform sphere with density −3ρ/2. This produces the desired net - _____ ε0
2ε0
density of −ρ/2 within the sphere.
Figure 27
24 CHAPTER 1. ELECTROSTATICS
Consider a point inside the sphere. Let its position with respect to the axis of the
cylinder be rc , and let its position with respect to the center of the sphere be rs . Then
y using the above forms of the fields, along with superposition, we find that the field at
this point is
ρrc (−3ρ/2)rs ρ
E = Ec + Es = + = (rc − rs ). (74)
2ϵ0 3ϵ0 2ϵ0
Let’s look at this vector rc − rs . If the given point lies in the x-y plane, then we have
rc the situation shown in Fig. 28; the axis of the cylinder (the z axis) points out of the
rs page. The difference rc − rs is simply the vector ax̂. If the given point does not lie
x in the x-y plane, then rc will still be parallel to the x-y plane, but rs will now have
rc - rs = ax
a z component. However, this z component doesn’t affect the x-y component of the
difference rc − rs , so the x-y component still equals ax̂. In short, a “top” view of the
setup (looking along the z axis) makes it clear that the projection of rc − rs onto the
x-y plane always equals ax̂.
From Eq. (74), the x-y component of the field inside the sphere therefore equals
(ρa/2ϵ0 )x̂, which is uniform, as desired. In the special case where the sphere is centered
Figure 28 on the z axis (so that a = 0), there is additionally zero x component, so the field points
only in the z direction inside the sphere, with a magnitude dependent only on z. This
dependence is linear, so a charge will undergo simple harmonic motion if its initial
velocity is in the z direction.
(πr2 ℓ)ρ ρr
E · 2πrℓ = =⇒ E = . (76)
ϵ0 2ϵ0
The full vector form of this field is E = r̂ρr/2ϵ0 , where the r̂ vector here represents
the direction away from the z axis. So rr̂ is the vector (x, y, 0). Hence Ecyl =
(ρ/2ϵ0 )(x, y, 0).
Finally, if we draw a slab Gaussian surface with area A and thickness 2z, inside the
slab and centered in the slab, Gauss’s law gives
(2zA)ρ ρz
E · 2A = =⇒ E = . (77)
ϵ0 ϵ0
This field points in the z direction, so it is just Eslab = (ρ/ϵ0 )(0, 0, z).
If we now give the three objects the densities ρ1 , ρ2 , ρ3 , we find that the total field at
a given point (x, y, z) inside all three objects (which means inside the sphere) equals
1 ( ρ1 ρ2 )
Etotal = (x, y, z) + (x, y, 0) + ρ3 (0, 0, z) . (78)
ϵ0 3 2
25
(a) At a radius r inside a sphere with charge density ρ, the electric field is effectively
due to the charge inside radius r. So the field is
(4πr3 /3)ρ ρr
E= 2
= . (81)
4πϵ0 r 3ϵ0
The field points radially outward (for positive ρ), so we can write the E vector
compactly as E = ρr/3ϵ0 .
The force on the smaller ball due to the larger sphere is found by integrating the
effect of the E field on all the pieces of the ball. This integral is easy to do if
we group the pieces in pairs that are symmetrically located on either side of the
center of the ball (at position a). Consider two pieces located at positions a + r′
and a − r′ . Then because the above E field is linear in r, the r′ parts of the sum
cancel, and we end up with two pieces effectively located at the center of the ball.
We can build up the entire ball from such pairs, so we see that the ball can be
effectively treated like a point charge at its center, as far as the force from the
larger sphere goes. The force is therefore the same as if it were a point charge of
the same charge q.
It’s no surprise that the ball acts effectively like a point charge at its center,
because what we just did is basically find the average value of r over the ball,
which is the same thing we do when we find the center of mass of a uniform
object. And the CM of a sphere is at its center. So more generally, if we have an
oddly-shaped charged object instead of a nice ball (and even if it isn’t uniform),
and if it is located entirely within a uniform sphere of charge, then the force on
it is the same as if all of its charge were located at its “center of charge.”
(b) Now consider the slightly different setup where we remove the charge in the larger
sphere where the ball is. This is a more realistic scenario, because it corresponds
to hollowing out a cavity and then filling it up with another material. It turns
out that the force on the ball doesn’t change. This follows from the fact that the
larger-sphere-with-cavity is the superposition of the complete larger sphere plus
a sphere of negative charge density located where the ball is. So the total force
on the ball is the sum of the forces from the original complete sphere plus the
26 CHAPTER 1. ELECTROSTATICS
new sphere of negative charge. But the latter provides no force on the ball, by
symmetry, because it is located in exactly the same place as the ball. The total
force is therefore the same as before.
The result of this exercise is consistent with the result in Problem 1.27, which
tells us that the electric field is uniform inside a spherical cavity within a uniform
sphere.
where we have made the change of variables x ≡ 2r/a0 . Note that we don’t need to
know C, because it cancels. You can verify the integral:
∫ x1 x1
x2 e−x dx = −(x2 + 2x + 2)e−x = 2 − (x21 + 2x1 + 2)e−x1 . (83)
0 0
For x1 = ∞ (that is, r1 = ∞) the integral is 2. And for x1 = 2 (that is, r1 = a0 ) the
integral is 2 − 10e−2 . The fraction of the full electron charge −e that lies inside radius
a0 is therefore (2 − 10e−2 )/2 = 1 − 5/(2.718)2 = 0.323. The net positive charge inside
r = a0 is then q = (1 − 0.323)(1.6 · 10−19 C) = 1.08 · 10−19 C. This field at this radius
is ( )
1 q 9 kg m
3
1.08 · 10−19 C
= 9 · 10 2 2 ≈ 3.5 · 1011 N/C, (84)
2
4πϵ0 a0 s C (0.53 · 10−10 m)2
which is huge!
1.77. Electron jelly
The force on a proton, at radius r, from the electron jelly is effectively due to the jelly
that is inside radius r. The force points toward the center of the sphere. If the net
force on a proton is zero, the force from the other proton must also point along the
line (away) from the center. The two protons must therefore lie on the same diameter.
They also must be the same distance r from the center; this is true because they feel
the same force (in magnitude) from each other, so they must also feel the same force
from the jelly, which implies that they must have the same value of r.
Since volume is proportional to r3 , the negative charge inside radius r equals −2e(r3 /a3 ).
The field at radius r due to the jelly is therefore
2e(r3 /a3 ) er
− =− . (85)
4πϵ0 r2 2πϵ0 a3
( )
The field at one proton due to the other is e/ 4πϵ0 (2r)2 . So the total field at one of
the protons will equal zero if
er e a3 a
3
= =⇒ r3 = =⇒ r = . (86)
2πϵ0 a 4πϵ0 (2r)2 8 2
This factor of 1/2 is reasonably clear in retrospect. If all of the −2e electron charge
were located in a point charge at the center, it would provide a force on one of the
protons that is 8 times the force due to the other proton (because the other proton is
twice as far away and half as big). So the forces will balance if we reduce the effective
electron charge by a factor of 8. This is accomplished by reducing the effective radius
of the jelly by a factor of 2.
27
hole
E1 + E2 0 + σ/ϵ0 σ
Aσ = Aσ = Aσ · . (88)
2 2 2ϵ0
But the force on the disk equals the charge on the disk times the field at the location
of the disk, due to all the other charge in the system (that is, the shell with the disk
removed). Equation (88) therefore tells us that the (radial) field of the shell-minus-disk
must be σ/2ϵ0 , as desired.
Figure 31
28 CHAPTER 1. ELECTROSTATICS
are therefore
( )
1 1 4σ0 8σ 2 N
FA = (E1 + E2 )σ = 0+ (−4σ0 ) = − 0 = −90.4 2 ,
2 2 ϵ0 ϵ0 m
( )
1 1 4σ0 3σ0 7σ 2 N
FB = (E1 + E2 )σ = − (7σ0 ) = 0 = 39.5 2 ,
2 2 ϵ0 ϵ0 2ϵ0 m
( )
1 1 3σ0 9σ02 N
FC = (E1 + E2 )σ = 0− (−3σ0 ) = = 50.8 2 . (89)
2 2 ϵ0 2ϵ0 m
The sum of these forces per area is (σ02 /ϵ0 )(−8 + 7/2 + 9/2). This equals zero as it
must, because a system can’t exert a net force on itself (otherwise momentum wouldn’t
be conserved).
Alternatively, we can find the forces by calculating the field at the location of a given
plate due to the other two plates. For example, the bottom two plates produce a field
of (7 − 3)σ0 /2ϵ0 = 2σ0 /ϵ0 at the location of the top plate, which therefore feels a force
per unit area equal to (2σ0 /ϵ0 )(−4σ0 ) = −8σ02 /ϵ0 , as above. The (E1 +E2 )/2 averages
in Eq. (89) are simply a way of finding the field at the location of one sheet due to the
others.
The field just outside the balloon is E = Q/4πϵ0 R2 , and the field inside is zero. So
the average field at the surface is E/2. The charge density is σ = Q/4πR2 . From
Eq. (1.49), the force per unit area (that is, the pressure) is therefore
F E Q 1 Q Q2
P ≡ =σ = 2
· 2
= . (90)
A 2 4πR 2 4πϵ0 R 32π 2 ϵ0 R4
Consider the two hemispheres defined by the horizontal great circle. The horizontal
components of the forces on the different parts of the upper hemisphere cancel by
symmetry, so we care only about the vertical components. The vertical component of
the force on a little patch with area A is P A cos θ, where θ is the angle that the plane
of the patch makes with the horizontal. If we write this as P (A cos θ), we see that the
vertical force on a patch equals P times the projection of the area of the patch onto
the horizontal plane. Since P is constant, and since the sum of all the projections of
the patches in a hemisphere is simply the great-circle area πR2 , we find that the total
upward force on the hemisphere is
Q2
P · πR2 = . (91)
32πϵ0 R2
By comparison with Coulomb’s force law, this has the correct units of (charge)2 /[ϵ0 ·
(length)2 ]. It makes sense that it grows with Q and decreases with R.
If you don’t want to use the above reasoning involving the projection, you can do
an integral. Slice the sphere into rings, with θ being the angle of a ring down from
the top of the sphere. The area of a ring is da = 2π(R sin θ)(R dθ), and the vertical
component of the force on the ring is (P da) cos θ. Integrating from θ = 0 to θ = π/2
29
(a) The field is nonzero only in the region a < r < b, where it equals E = Q/4πϵ0 r2 .
The total energy is therefore
∫ b ( )2 ∫ b ( )
ϵ0 E 2 ϵ0 Q 1 Q2 1 1
U= dv = 4
2
4πr dr = − . (94)
a 2 2 4πϵ0 a r 8πϵ0 a b
If b = a then U = 0, of course, because the shells are right on top of each other,
so the charges cancel and we effectively have no charge anywhere in the system.
If b → ∞ then U = Q2 /8πϵ0 a, which is correctly the energy of a single shell with
charge Q (see Problem 1.32). If a → 0 then U correctly goes to infinity, because
the field diverges (sufficiently quickly) near a point charge. Equivalently, it takes
an infinite amount of energy to compress a given amount of charge down to a
point.
The result in Eq. (94) can be interpreted as follows. As mentioned above, the
energy stored in a system consisting of one spherical shell of radius r is Q2 /8πϵ0 r.
Given this result, consider building up the present two-sphere system from scratch
(that is, by bringing charges in from infinity) in two steps. It takes an energy
Q2 /8πϵ0 a to construct the shell of radius a. Then, with that shell in place, it
takes an energy Q2 /8πϵ0 b − Q2 /4πϵ0 b to construct the outer shell of radius b.
The first term comes from the self energy of this outer shell. The second term
comes from the potential energy of the negative outer shell due to the positive
inner shell already in place (which acts like a point charge at its center). The
sum of the energies of these two steps yields the result in Eq. (94).
(b) Now let’s imagine starting with two neutral shells and then gradually transferring
positive charge from the outer shell to the inner shell. At the start, there is no
electric field between the shells, so it takes no work to transfer an initial bit of
charge dq. But as more charge is piled onto the inner shell, the field grows, and
it takes more work to bring in the successive bits dq.
30 CHAPTER 1. ELECTROSTATICS
At a moment when there is charge q on the inner shell, the field between the
shells is q/4πϵ0 r2 , so the force on a little charge dq is q dq/4πϵ0 r2 . The work you
must do on this dq is the integral of your force times the displacement, or
∫ a ( )
−q dq q dq 1 1
dW = 2
dr = − , (95)
b 4πϵ0 r 4πϵ0 a b
where we have included the minus sign in the force because your force points
inward (it is opposite to the electric force). However, you can always put the
sign in by hand at the end; you certainly have to do positive work to move the
positive charge dq toward the positively charged inner shell.
We must now integrate the above work dW over all the bits dq that we bring in.
This gives ( ) ( )
∫ Q
q dq 1 1 Q2 1 1
W = − = − , (96)
0 4πϵ0 a b 8πϵ0 a b
in agreement with the result in part (a). It may seem mysterious that the po-
tential energy of a system can be found by integrating ϵ0 E 2 /2 over the volume.
But the agreement of the two above methods, applied to our setup involving two
shells, should help convince you that the ϵ0 E 2 /2 method does indeed give the
energy.
The field at radius r inside the cylinder is due only to the charge inside radius r. This
charge has linear density πr2 ρ, so the field equals (πr2 ρ)/2πϵ0 r = ρr/2ϵ0 . The energy
stored in the internal field, within a length ℓ, is then
∫ a ( )2 ∫ a
ϵ0 ρr πρ2 ℓ πρ2 a4 ℓ
Uint = (2πr dr)ℓ = r3 dr = . (98)
2 0 2ϵ0 4ϵ0 0 16ϵ0
Finding the energy per unit length simply involves erasing the ℓ. Using
( ρ = λ/πa)2 ,
2
we can write the sum of Uext and Uint , per unit length, as (λ /4πϵ0 ) ln(R/a) + 1/4 ,
as desired. As mentioned in the statement of the exercise, this diverges as R → ∞. It
also diverges as a → 0.
Chapter 2
These two results are equal, as desired. The electric potential ϕ, if taken∫to be zero
at (0, 0), is just the negative of our result, because we define ϕ by ϕ = − E · ds, or
equivalently E = −∇ϕ. Hence ϕ(x, y) = y 3 − 3x2 y. The negative gradient of this is
( )
∂ϕ ∂ϕ ∂ϕ
−∇ϕ = − , , = (6xy, 3x2 − 3y 2 , 0), (101)
∂x ∂y ∂z
31
l
B C
-3q 32 CHAPTER 2. THE ELECTRIC POTENTIAL
z where we have ignored the units (which are C/m) in the second expression. The plot
-5
-5
5 10 15
of 4πϵ0 ϕ is shown in Fig. 33. ϕ goes to +∞ at z = 0, and −∞ at z = 3. You can
-10
quickly verify that ϕ = 0 at z = 2 and z = 6. √ You can also show that ϕ achieves a
local maximum of ϕ ≈ (0.34)/4πϵ0 at z = 6 + 3 2 ≈ 10.24. Since Ez = −∂ϕ/∂z, the
Figure 33 field is zero at z = 10.24, so a charge placed there will be in equilibrium (unstable with
regard to motion in the z direction if the charge is positive, stable if it is negative).
2.34. Extremum of ϕ
By symmetry, the E field at points on the y axis has no x or z component. And we
know that Ey equals −∂ϕ/∂y. So if ϕ has a local maximum or minimum at some point
on the y axis, then ∂ϕ/∂y, and hence Ey , equals zero. The full vector E therefore also
equals zero.
At the point (0, y, 0) with y > 1, the Ey component equals (ignoring the factor of
1/4πϵ0 , along with the units of the various quantities)
2 1 y−1
Ey = −2· ·√ , (107)
y 2 (y − 1) + 1
2 2
(y − 1)2 + 12
33
where the last factor gives the y component of the field from the two negative charges.
Setting Ey = 0, moving one of the terms to the other side of the equation, and
squaring, we find
(y 2 − 2y + 2)3
y4 = . (108)
(y − 1)2
Another way of obtaining this relation is to (as you can check) write down the potential
(again ignoring the factor of 1/4πϵ0 and units),
2 1 φ
ϕ(0, y, 0) = −2· √ , (109)
y (y − 1)2 + 12 2.5
2.0
and then set ∂ϕ/∂y = 0. The result is Eq. (108), of course, because Ey = −∂ϕ/∂y. We 1.5
can solve Eq. (108) numerically; Mathematica gives the numerical result of y = 1.621. 1.0
0.5
Plots of ϕ(y) and Ey (y) (times 4πϵ0 ) for points on the y axis are shown in Fig. 34. For y
- 0.5 1 2 3 4
large y, we know that both ϕ and Ey must be negative, because for large y we have
a charge 2 C at a distance y, and two −1 C charges at a distance essentially equal to
y − 1. So the negative charges win out.
Ey
You can show that Ey reaches a maximum negative value at y = 2.153; you will again 5
need to solve an equation numerically. The existence of such a point between y = 1.621 4
and y = ∞ follows from a continuity argument similar to the one involving ϕ: Since 3
Ey = 0 at these two points, Ey must have a local maximum or minimum somewhere 2
1
between.
0 y
1 2 3 4
2.35. Center vs. corner of a square
Dimensional analysis tells us that for a given charge density σ, the potential at the Figure 34
center of a square of side s must be proportional to Q/s, where Q is the total charge,
σs2 . (This is true because the potential has the units of q/4πϵ0 r, and Q is the only
charge in the setup, and s is the only length scale.) Hence ϕ is proportional to
σs2 /s = σs. So for fixed σ it is proportional to s.
Equivalently, if we imagine∫ increasing the size of the square by a factor f in each
direction, the integral ϕ ∝ (σ da)/r picks up a factor of f 2 in the da and a factor of
f in the r, yielding a net factor of f in the numerator.
Said in yet another (equivalent) way, if we imagine increasing the size of the square
by a factor f in each direction, and if we look at corresponding pieces of the small
and large versions, then the piece in the large version has f 2 times as much charge,
but it is f times as far away from a given point, so its contribution to the potential is
f 2 /f = f times the contribution in the small version.
If we assemble 4 squares of side b, we make a square of side 2b. The potential at the
center is 4ϕ1 (the sum of 4 corner potentials of the side b square). But from the above
reasoning that ϕ ∝ s, this potential of 4ϕ1 must also be 2 times the center potential
of the side-b square. So we have 4ϕ1 = 2ϕ0 , or ϕ0 = 2ϕ1 . Hence the desired ratio is 2.
It therefore takes twice as much work to bring a charge in from infinity to the center,
as it does to a corner. From Eqs. (2.25) and (2.30), the analogous statement for a
disk of charge is that it takes π/2 as much work to bring a charge in from infinity to
the center, as it does to the edge. But that problem can’t be solved via the above
scaling/superposition argument.
The above result for the square actually holds more generally for any rectangle with
uniform charge density. All of the steps in the above logic are still valid, so the
potential at the center is twice the potential at a corner.
34 CHAPTER 2. THE ELECTRIC POTENTIAL
4.55 √ √
As a double check, this does equal 8/ 3 when x = 0. Plugging in x = 2 gives the
4.50 potential at the midpoint of an edge as 4.4555. Although this is smaller than the 4.6188
x potential at the center, the plot in Fig. 35 shows that ϕ achieves a maximum value
0.2 0.4 0.6 0.8 1.0 1.2 1.4
about halfway out to the edge. The maximum happens to be located at x = 0.761,
Figure 35 and the value is ϕmax = 4.6242. Since this is larger than the value of ϕ at the center,
the proton will not escape if it heads directly toward the midpoint of an edge.
2.37. Field on the earth
The radius of the earth is r = 6.4 · 106 m, so we have
1 Q ( kg m3 ) 1C V
E = = 9 · 109 2 2 = 2.2 · 10−4 ,
4πϵ0 r 2 s C (6.4 · 106 m)2 m
1 Q ( kg m3 ) 1C
ϕ = = 9 · 109 2 2 = 1400 V. (111)
4πϵ0 r s C (6.4 · 106 m)
This value of 1400 V is larger than you might think it should be, given the large radius
of the earth. The point is that a coulomb is a large quantity of charge, on an everyday
scale. Or equivalently ϵ0 has a small value in the system of units we use.
2.38. Interstellar dust
The potential is
( 2 2
)
Q −12 s C
= −0.15 V =⇒ Q = (−0.15 V)4π 8.85 · 10 (3 · 10−7 m)
4πϵ0 R kg m3
= −5 · 10−18 C. (112)
(114)
This is larger than the radius of the silver nucleus, which is about 5 · 10−15 m, so it
was reasonable to consider coulomb repulsion only. The strength of the electric field
at this position is
1 Qr3 /a3 1 Qr
E= 2
= . (117)
4πϵ0 r 4πϵ0 a3
Outside the sphere the field is simply Q/4πϵ0 r2 . The potential at the surface of the
sphere relative to infinity is
∫ a ∫ a
Q dr 1 Q
ϕ(a) − ϕ(∞) = − E dr = − 2
= , (118)
∞ ∞ 4πϵ0 r 4πϵ 0 a
and the potential at the center of the sphere relative to the surface is
∫ 0 ∫ 0
1 Qr dr 1 Q
ϕ(0) − ϕ(a) = − E dr = − = . (119)
a a 4πϵ0 a3 4πϵ0 2a
Adding the two preceding equations gives ϕ(0) − ϕ(∞) = (1/4πϵ0 )(3Q/2a). For the
problem at hand, this yields (assuming ϕ(∞) = 0 as usual)
or 28.4 megavolts.
36 CHAPTER 2. THE ELECTRIC POTENTIAL
At point P2 with a general x value, we have (using the integral table in Appendix K) Figure 36
∫ d (√ )
1 λ dz λ (√ )d λ x2 + d2 + d
ϕ2 = √ = ln x2 + z 2 + z = ln √ .
4πϵ0 −d x2 + z 2 4πϵ0 −d 4πϵ0 x2 + d2 − d
(125)
These two potentials are equal when
√
x 2 + d2 + d √ √
√ = 3 =⇒ 4d = 2 x2 + d2 =⇒ x = 3 d. (126)
x +d −d
2 2
1
∫ d
λ dz ′ λ (√ )d
ϕ= √ = ln x + (z − z) + (z − z)
2 ′ 2 ′
4πϵ0 −d x2 + (z ′ − z)2 4πϵ0 −d
(√ )
λ x + (d − z) + d − z
2 2
= ln √ . (127)
4πϵ0 x2 + (d + z)2 − d − z
You can verify that if x = 3d/2 and z = d, this equals (λ/4πϵ0 ) ln 3, which is the same
as the value of ϕ in Exercise 2.43. And the sum of the distances from (3d/2, 0, d) to
the ends of the rod at (0, 0, d) and (0, 0, −d) equals 3d/2 + 5d/2 = 4d, which is the
same as the sum of the distances for each of the points in Exercise 2.43. So this is all
consistent with the equipotential curve in the x-z plane being an ellipse.
If x and z lie on the ellipse described by x2 /(a2 − d2 ) + z 2 /a2 = 1, then solving for x2
gives x2 = (a2 − z 2 )(a2 − d2 )/a2 . Using this, you can show that the quantity under
the square root in Eq. (127) can be written as
( )2
a2 ± zd
x2 + (d ± z)2 = . (128)
a
We therefore have
( 2 ) ( ) ( )
λ a − zd + a(d − z) λ (a − z)(a + d) λ a+d
ϕ= ln = ln = ln .
4πϵ0 a2 + zd − a(d + z) 4πϵ0 (a − z)(a − d) 4πϵ0 a−d
(129)
As desired, this result is independent of z and x, so ϕ is constant on the ellipse
described by x2 /(a2 − d2 ) + z 2 /a2 = 1. The ellipse containing the point (3d/2, 0, d),
along with the two points from Exercise 2.43, has its a value (which is the length of
the major axis, which lies along the z axis) equal to 2d.
38 CHAPTER 2. THE ELECTRIC POTENTIAL
Between the two objects, the field from the stick points rightward, and the field
from the point charge points leftward. We want these two fields to cancel. The
distance from the point x = a to the point charge is ℓ − a, so we want
Q Q
= =⇒ (ℓ − a)2 = a(a + ℓ)
4πϵ0 a(a + ℓ) 4πϵ0 (ℓ − a)2
=⇒ ℓ2 − 2aℓ + a2 = a2 + aℓ
=⇒ a = ℓ/3. (131)
(b) There are no other points. Ignoring the inside of the stick, the field can’t be zero
anywhere else on the x axis because to the right of the point charge, both objects
produce rightward-pointing fields, so they can’t cancel. And to the left of the
stick, both objects produce leftward-pointing fields, so again they can’t cancel.
For points not on the x axis, both objects produce a field that has a nonzero
component pointing away from the x axis, so again the sum can’t be zero.
The existence of an E = 0 point inside the stick, mentioned in the statement
of the exercise, follows from a continuity argument: The field points rightward
(and in fact diverges) just to the right of the right end of the stick. And it points
leftward just to the left of the left end of the stick. So somewhere in between it
must be zero.
Figure 37
(c) Some equipotential curves and field lines are shown in Fig. 37 (but see Foot-
note 1). The field lines are everywhere perpendicular to the equipotential curves
(because E = −∇ϕ, and the gradient of a function is perpendicular to the level
39
surfaces). The equipotential curves make the transition from two surfaces (one
around each object) to one surface (essentially a sphere at infinity) by intersecting y
at the E = 0 point at x = ℓ/3 that we found in part (a). The field is indeed zero
wherever equipotential curves intersect, because the field must be perpendicular
to both lines at the crossing, and the only vector that is perpendicular to two
different directions is the zero vector. Zoomed-in views of the equipotentials and
field lines near the E = 0 point are shown in Fig. 38 and Fig. 39, respectively. x
The field lines that start off heading nearly along the x axis toward the point
x = ℓ/3 end up taking nearly a right turn and then head off to infinity. Their
direction is vertical near the x axis due to the symmetric nature of the crossing.
And it is vertical at infinity because the two objects have equal charge; you can
use Gauss’s law to show this. But it isn’t vertical in between.
Figure 38
2.46. Right triangle ϕ
The potential at P equals the area integral, y
∫ ∫ b ∫ ax/b
σ da σ dy
ϕP = = dx √ . (132)
4πϵ0 r 4πϵ0 0 0 x + y2
2
as desired. Alternatively, you can derive this result directly, by performing an integral a
P
over x and θ,∫ instead of x and y (by substituting y = x tan θ). You will need to use b
the integral dθ/ cos θ = ln[(1 + sin θ)/ cos θ].
Remark: It is interesting to compare the potentials at the two vertices of a given very
thin right triangle, or equivalently at the left vertices of the two triangles shown in Fig. 40.
In the first case we can take the a ≪ b limit of Eq. (134), which you can show (with the a
help of ln(1 + ϵ) ≈ ϵ) leads to ϕ ≈ σa/4πϵ0 . (You can verify that this is consistent with
the result you would obtain for a thin pie piece. You should also think about why it is
P
b
Figure 40
40 CHAPTER 2. THE ELECTRIC POTENTIAL
independent of b.) In the second case we have the reverse situation with b ≪ a, which
leads to ϕ ≈ (σb/4πϵ0 ) ln(2a/b). So if we take a = ℓ and b = 100ℓ in the first case, we
obtain ϕ ≈ σℓ/4πϵ0 . And if we take b = ℓ and a = 100ℓ in the second case, we obtain
ϕ ≈ (ln 200)σℓ/4πϵ0 . It makes sense that the second case has the larger ϕ, because the
charge is generally closer to that vertex P in that case. But the log behavior isn’t obvious.
For the disk, we can slice it into concentric rings, which gives the potential at the
center as ∫ ∫ d/2
1 dq 1 2πrσ dr πσd
ϕdisk = = = . (137)
4πϵ0 r 4πϵ0 0 r 4πϵ0
Setting ϕdisk = ϕsquare yields d/s = (3.525)/π = 1.122. As was to be expected, the disk
is larger than the inscribed
√ circle, for which d = s; but smaller than the circumscribed
circle, for which d = 2 s = (1.414)s.
l As in Problem 2.7, our strategy will be to find the potential at radius r, and then
take the derivative to find the field. The calculation is the same as in Problem 2.7,
R θ except that the limits of integration are modified. If we define θ in the same way
r P
is in Fig. 12.28, it now runs from π/2 to π. Following the steps in the solution to
Problem2.7, the potential at point P in Fig. 41 is (we’ll keep things in terms of the
density σ)
∫ π
2πR2 σ sin θ dθ
Figure 41 ϕ(r) = √
π/2 4πϵ0 R + r − 2rR cos θ
2 2
√ π
σR
= R + r − 2rR cos θ
2 2
2ϵ0 r π/2
σR ( √ )
= (R + r) − R2 + r2 . (138)
2ϵ0 r
We are concerned with√small r, because
√ we want to know the field at the center. For
small r we can write R2 + r2 = R 1 + r2 /R2 ≈ R(1 + r2 /2R2 ). So the potential
near the center is
σR ( ) σR ( )
ϕ(r) = r − r2 /2R = 1 − r/2R (139)
2ϵ0 r 2ϵ0
The field at the center is then
dϕ σ
E(r) = − = . (140)
dr 4ϵ0
You can check that you arrive at the same result if you take P to be on the left side
of the center. You will need to be careful about the limits of integration and various
signs.
41
∫
(ϕ/2) ρ dv = ϕq/2. So if the spheres of radii R1 and R2 have charge q and Q − q,
respectively, the sum of the two potential energies is
( 2 )
q q Q−q Q−q 1 q (Q − q)2
U= · + · = + . (145)
4πϵ0 R1 2 4πϵ0 R2 2 4πϵ0 2R1 2R2
Minimizing this by setting the derivative with respect to q equal to zero yields
( )
dU 1 q (Q − q)
0= = − . (146)
dq 4πϵ0 R1 R2
Solving for q gives q = QR1 /(R1 + R2 ). So there is charge QR1 /(R1 + R2 ) on the first
sphere and charge QR2 /(R1 + R2 ) on the second sphere.
The two terms in Eq. (146) (without the minus sign in front of the second term)
are simply the potentials of the two spheres. So the condition of minimum energy
is equivalent to the condition of equal potentials. Note that the second derivative,
d2 U/d2 q = 1/R1 + 1/R2 , is positive, so the extremum is indeed a minimum of U , not
a maximum. This is consistent with the special case where R1 = R2 ; equal division
of the charge involves half as much total energy as piling all of Q on one sphere, from
Eq. (145).
1 Q (√ 2 )b/2
= ln a + (x − x0 )2 + (x − x0 )
4πϵ0 b −b/2
( √ )
1 Q a + (b/2 − x0 ) + (b/2 − x0 )
2 2
= ln √ . (147)
4πϵ0 b a2 + (−b/2 − x0 )2 + (−b/2 − x0 )
This is a rather messy answer, so let’s look at how ϕmid and ϕend behave in some
limits, to feel better about the result. Consider the limit b → 0, in which case the
cylinder reduces to a thin ring. To first order in b, we can ignore the b2 terms under
the square roots in the above expressions. We obtain
( )
Q a + b/2 Q b Q b Q
4πϵ0 ϕmid ≈ ln ≈ ln 1 + ≈ · = . (151)
b a − b/2 b a b a a
Likewise, ( )
Q b Q
4πϵ0 ϕend ≈ ln 1 + ≈ . (152)
b a a
Both of these results make sense, because all of the charge on the ring is essentially a
distance a from the midpoint of the axis, which is essentially the same as an endpoint.
The difference in the potentials is therefore zero.
In the limit b → ∞, both ϕmid and ϕend go to zero because of the Q/b coefficient, and
because we are holding the total charge Q constant. If we instead let the charge per
unit length be constant, so that Q = λb, then you can show that 4πϵ0 ϕmid ≈ 2λ ln(b/a)
and 4πϵ0 ϕend ≈ λ ln(2b/a). For large b, the “2” inside the log doesn’t matter much, so
we see that ϕmid is approximately twice as large as ϕmid . The reason for this basically
comes down to the fact that if you’re in the middle of a long cylinder, you see two long
cylinders on either side of you (and also the fact that the potential due to a stick grows
only like the log of the length, so that a long stick yields roughly the same potential
as one that has twice the length).
A rough sketch of the field lines is shown in Fig. 42. The field is zero at the center,
by symmetry. We haven’t worried about drawing the density of field lines correctly;
for a very long cylinder, you can show that the field at the surface is twice the field at Figure 42
the midpoint of an end face.
2.52. Spherical cavity in a slab
(a) The given setup can be considered to be the superposition of the given slab and
a sphere with charge density −ρ. Our goal is to show that the potential at a
point on the surface of the slab at infinity equals the potential at the center of
the cavity. From the standard Gauss’s-law argument with a pillbox extending
from −x to x, the field due to the slab in the interior is x̂ρx/ϵ0 . (Alternatively,
the relevant part of the slab is effectively a sheet with surface charge density
ρ(2x)). And outside the slab the field is x̂ρR/ϵ0 (the whole slab is effectively
a sheet with surface charge density ρ(2R)). Similar Gauss’s-law arguments give
the field due to the (negative) sphere in the interior as −r̂ρr/3ϵ0 , and outside as
−r̂ρR3 /3ϵ0 r2 . Integrating the slab and sphere fields to find the total potential
relative to the center of the cavity, we find that the potential inside the cavity
equals −ρx2 /2ϵ0 + ρr2 /6ϵ0 .
At the rightmost point of the cavity, we have x = r = R, so the potential of
this point (relative to the center) is (ρR2 /ϵ0 )(−1/2 + 1/6) = −ρR2 /3ϵ0 . If we
now march along the surface of the slab to infinity, only the field from the sphere
matters. The change in potential as we march out is
∫ ∞
−ρR3 ρR3 ρR2
∆ϕ = − dr = = . (153)
R 3ϵ0 r2 3ϵ0 R 3ϵ0
(Alternatively, the change in potential from R to ∞ is just −Q/4πϵ0 R, with
Q = −4πR3 ρ/3.) This is correctly positive, because the potential increases as
44 CHAPTER 2. THE ELECTRIC POTENTIAL
we move away from the negative sphere. This ∆ϕ exactly cancels the negative
potential at the rightmost point of the cavity, so we end up with zero potential
on the surface of the slab at infinity, as desired. The algebra here basically boils
down to 1/2 = 1/6 + 1/3. The 1/2 is the (magnitude of the) decrease in potential
due to the slab. The 1/6 + 1/3 is the increase in potential due to the sphere,
inside and outside.
(b) The potential inside the cavity is ϕ = −ρx2 /2ϵ0 + ρr2 /6ϵ0 . In the plane of the
paper, we have r2 = x2 + y 2 , so ϕ becomes√ρ(y 2 − 2x2 )/6ϵ0 . Points on the ϕ = 0
curve therefore satisfy y 2 = 2x2 =⇒ y = ± 2x. This is consistent with Fig. 2.49,
where the slope of the straight line looks to be a little larger than 1.
(c) We found in part (a) that as we move from the rightmost point of the cavity out
to infinity along the surface of the slab, the potential increases by ρR2 /3ϵ0 . If
we want to end up at the same potential as at the rightmost point of the cavity,
we must then move away from the slab by a distance that causes the potential
to decrease by ρR2 /3ϵ0 . Since the field outside the slab equals x̂ρR/ϵ0 (far away,
the sphere can be ignored), the change in potential as we move away from the
slab is −(ρR/ϵ0 )∆x. This equals −ρR2 /3ϵ0 when ∆x = R/3, as desired.
√ ( )2
disk 2a 2a π2 − 4
y 2 + a2 −y = =⇒ y 2 +a2 = y + =⇒ y = a ≈ (0.467)a. (154)
π π 4π
The equipotential surface is (roughly) represented by the curve in Fig.44. The direction
Figure 44 of the curve near the edge of the disk happens to be perpendicular to the plane of the
disk. This follows from the result of Exercise 2.57(b) below; the tangential component
of the field diverges near the edge of the disk, so the field is essentially parallel to the
disk. The equipotential curve is therefore is perpendicular to the disk.
45
(a) Slicing the disk into concentric rings, we find the potential at the center to be
(with ℓ = 1 cm)
∫ ∫ 3ℓ
1 dq 1 2πrσ dr σℓ
ϕ= = = . (155)
4πϵ0 r 4πϵ0 ℓ r ϵ0
Plugging in the various quantities gives
( )
− 10−5 mC2 (0.01 m)
ϕ= = −11, 300 V. (156)
8.85 · 10−12 kg
s2 C2
m3
(b) The electron’s final kinetic energy at infinity equals the loss in potential energy.
This loss has magnitude
Since this is only about 2% of the electron’s rest energy, namely mc2 = 8.2 ·
10−14 J, a nonrelativistic calculation will suffice:
( )1/2
1 2(1.81 · 10−15 J)
mv 2 = 1.81 · 10−15 J =⇒ v = = 6.3 · 107 m/s, (158)
2 9.1 · 10−31 kg
which is about 20% of the speed of light. This answer is very close to the answer
obtained via the correct relativistic calculation: Conservation of energy gives
1.81 · 10−15 J
γmc2 = mc2 + |∆U | =⇒ γ = 1 + = 1.022. (159)
8.2 · 10−14 J
Hence (with β ≡ v/c),
√
β = 1 − 1/γ 2 = 0.206 =⇒ v = βc = 6.2 · 107 m/s. (160)
2 Q2 0.0675 Q2
U= 2
≈ . (162)
3π ϵ0 a ϵ0 a
From Problem 1.32 or Exercise 2.58, the energy required to build up a hollow spherical
shell with radius a and charge Q is
1 Q2 0.0398 Q2
U= ≈ . (163)
8πϵ0 a ϵ0 a
As expected, this is smaller than the result for the disk, because the charges are
generally closer to each other in the case of the disk.
46 CHAPTER 2. THE ELECTRIC POTENTIAL
(a) A little area element within a wedge in Fig. 2.50 has area r dr dθ. So at the given
point P , the magnitude of the field due to this little area is σ(r dr dθ)/4πϵ0 r2 =
σ dr dθ/4πϵ0 r. But only the vertical component (in the plane of the page) sur-
vives, which brings in a factor of cos θ, yielding σ cos θ dr dθ/4πϵ0 r. If we integrate
this from r = 0 up to either the r1 or r2 in Fig. 2.50, we obtain an infinite result.
However, the divergence from one wedge cancels the divergence from the other,
because the field due to the short wedge cancels the field due to the part of the
long wedge out to a distance r1 . The uncanceled part of the long wedge comes
from r values ranging from r1 out to the end at r2 .1 The net vertical component
of the field from the two opposite wedges in Fig. 2.50 is therefore
∫ r2 ( )
σ cos θ dr dθ σ r2
E∥ = = ln cos θ dθ. (164)
r1 r1 4πϵ 0 r 4πϵ 0 r1
θ
P R We must now integrate this result over θ. We can integrate from 0 to π/2 and
r2 θ then double the result; this will end up covering the whole disk.
ηR The task now is to find r1 and r2 in terms of θ. We can find r2 by using the law
R of cosines in the triangle involving r2 in Fig. 45. This gives R2 = (ηR)2 + r22 −
2(ηR)r2 cos θ. Solving this quadratic equation for r2 and choosing the positive
root gives
(√ )
r2 = R 1 − η 2 sin2 θ + η cos θ . (165)
Figure 45
The process involving r1 (with the triangle containing the angle π − θ) is the
same except for the replacement of cos θ with − cos θ. So twice the integration of
the result in Eq. (164) from 0 to π/2 gives
∫ π/2 ( √ )
σ 1 − η 2 sin2 θ + η cos θ
E∥ = ln √ cos θ dθ. (166)
2πϵ0 0 1 − η 2 sin2 θ − η cos θ
(b) Let’s look at small values of η first. We can ignore terms of order η 2 , so the log
term in Eq. (166) becomes
( ) ( )
1 + η cos θ (1 + η cos θ)2 ( )
ln = ln ≈ ln 1 + 2η cos θ ≈ 2η cos θ, (167)
1 − η cos θ 1 − η cos θ
2 2
where we have used ln(1 + x) ≈ x. Substituting this into Eq. (166) gives
∫
ση π/2
ση ( π ) ση
E∥ ≈ cos2 θ dθ = = , (168)
πϵ0 0 πϵ0 4 4ϵ0
ηR where we have used the fact that the average value of cos2 θ equals 1/2. It
P is reasonable that this result is linear in η, because as the given point P is
moved away from the center, the unbalanced part of the disk (the shaded part in
unbalanced Fig. 46) grows linearly with η (for small η). Intuitively, the ση/4ϵ0 field from the
part of disk unbalanced part should take the same form, up to a numerical factor, as the field
disk
from an effective line of charge with small width 2ηR (indicated by the dotted
circle 2ηR 1 If the given point doesn’t lie exactly in the plane of the disk, then there is actually no divergence. But
centered at P it is still worth mentioning the canceling divergences, because E∥ is well defined even if the point does lie
in the plane (whereas E⊥ isn’t, for an infinitesimally thin disk).
Figure 46
47
lines in Fig. 46). And indeed, you can show that the former is π/4 times the
latter.
Now let’s look at values of η very close to 1. With η ≡ 1 − ϵ, the square root
term in Eq. (166) becomes (dropping terms of order ϵ2 )
√
√ √ 2ϵ sin2 θ
1 − (1 − ϵ)2 sin2 θ ≈ cos2 θ + 2ϵ sin2 θ = cos θ 1 +
cos2 θ
( )
ϵ sin2 θ ϵ sin2 θ
≈ cos θ 1 + = cos θ + , (169)
cos2 θ cos θ
√
where we have used 1 + x ≈ 1 + x/2. In the numerator in the log term in
Eq. (166), the leading-order term (which happens to be zeroth order in ϵ, in this
case) is simply 2 cos θ. But in the denominator, the cos θ terms cancel, so the
leading-order term is ϵ sin2 θ/ cos θ + ϵ cos θ = ϵ/ cos θ. The argument of the log
term is therefore (2 cos θ)/(ϵ/ cos θ) = 2 cos2 θ/ϵ. So Eq. (166) becomes (we’ll
drop the ln(2 cos2 θ) term, because it is small compared with the leading-order
ln ϵ term; so all that matters is the 1/ϵ behavior of the argument)
∫ π/2 ( )
σ 2 cos2 θ
E∥ ≈ ln cos θ dθ
2πϵ0 0 ϵ
∫
σ ln ϵ π/2 σ ln ϵ
≈ − cos θ dθ = − . (170)
2πϵ0 0 2πϵ0
Note that this result is positive since ϵ < 1. We see that the field diverges as we
get very close to the edge of the disk, but it diverges slowly, like a log. The log
behavior makes sense, because as we get close to the edge, the short wedge in
Fig. 2.50 cancels only a small part of the long wedge. So we need to integrate
the 1/r field (see Eq. (164)) due to the long wedge almost down to r = 0. And
the integral of 1/r diverges like ln r near r = 0.
You can check, for various values of η near 0 or 1, that Eqs. (168) and (170)
correctly agree with a numerical evaluation of Eq. (166).
Equation (2.32) then gives the energy (relative to the configuration where the charge
is distributed over a cylinder with radius R) in a length ℓ of the cylinder as
∫ ∫ [ 2 ( ) ]
1 1 a ρa R ρ 2
U= ρϕ dv = ρ ln + (a − r2 ) 2πrℓ dr
2 2 0 2ϵ0 a 4ϵ0
2 ∫ a
[ ( ) ]
U πρ R 1
=⇒ = a2 ln + (a2 − r2 ) r dr
ℓ 2ϵ0 0 a 2
2
[ 4 ( ) ( 4 )]
πρ a R 1 a a4
= ln + −
2ϵ0 2 a 2 2 4
2 2 4
[ ( ) ]
ρ π a R 1
= ln +
4πϵ0 a 4
2
[ ( ) ]
λ R 1
= ln + , (173)
4πϵ0 a 4
as desired. If R = a, so that all of the charge is initially distributed over the surface
of the cylinder, then it takes an amount of work per unit length equal to λ2 /16πϵ0
to move the charge inward and distribute it uniformly throughout the volume of the
cylinder.
This field points in the positive r̂ direction, so it agrees with the result in Eq. (2.36),
q
kqℓ ( ) l/2 β
3
2 cos θ r̂ + sin θ θ̂ , (177) r
r
l/2
when θ = 0. -q
In the transverse direction, we have the situation shown in Fig. 47. The magnitudes of
Figure 47
the two fields are equal. The horizontal components cancel, but the downward com-
ponents add. The distances from the given point to the two charges are essentially
equal to r, so the magnitudes of the fields are kq/r2 . The (negative) vertical compo-
nents are obtained by multiplying by sin β, which is approximately equal to (ℓ/2)/r in
the small-angle approximation. The vertical field is therefore directed downward with
magnitude
( )
kq ℓ/2 kqℓ
E≈2 = 3 . (178)
r2 r r
This agrees with the result in Eq. (2.36) when θ = π/2, because the θ̂ vector points
downward at the given point (in the direction of increasing θ, which is measured down
from the vertical). This field is half as large as the field on the vertical axis, for a
given value of r.
kq kq 2kq
Er = + − 2 cos θ
(r − d) 2 (r + d) 2 r1 d
kq kq 2kq r
= + − 2 √
(r − d) 2 (r + d) 2 r +d 2
r + d2
2
[ ]
kq 1 1 2 d
= + − l l
r2 (1 − ϵ)2 (1 + ϵ)2 (1 + ϵ2 )3/2
[
kq ( ) ( ) ( )]
≈ 1 + 2ϵ + 3ϵ 2
+ 1 − 2ϵ + 3ϵ 2
− 2 1 − 3ϵ2
/2
r2 Figure 48
kq [ 2 ] 9kqℓ2
= 9ϵ = . (179)
r2 2r4
There are various ways of obtaining the above Taylor series, but perhaps the easiest
is to note that, for example, 1/(1 − ϵ)2 equals the derivative of 1/(1 − ϵ), which itself
is just the sum of the geometric series 1 + ϵ + ϵ2 + · · · .
Our result for Er is positive, so the field points away from the square. Along the other
diagonal, it points toward the square. This implies that if we traverse a large circle
around the quadrupole, there are four locations where the radial component of the P
field is zero. This should be contrasted with the field of a dipole, which has only two
such locations where the radial component is zero.
l/2
-λ
Figure 49
50 CHAPTER 2. THE ELECTRIC POTENTIAL
the potential at P (due to the positive wire) relative to the point midway between the
wires is ∫ r1 ∫ r1 ( )
λ dr λ ℓ/2
ϕ=− Er dr = − = ln . (180)
ℓ/2 ℓ/2 2πϵ0 r 2πϵ0 r1
As a double check on the sign, if r1 is very small (although we’re not concerned
with this case), then ϕ is a large positive quantity,( as it should
) be. Likewise, the
potential due to the negative wire is −(λ/2πϵ0 ) ln (ℓ/2)/r2 . When we add these
two potentials, the
( ℓ dependence
) drops out, and we end up with a total potential of
ϕ = (λ/2πϵ0 ) ln r2 /r1 . Using the same approximate forms of r1 and r2 that we used
in Eq. (2.35) in the r ≫ ℓ limit, we find
where we have used 1/(1 − ϵ) ≈ 1 + ϵ. We can now use the Taylor approximation
ln(1 + ϵ) ≈ ϵ to write
( ) ( )
λ r2 λ ℓ cos θ λℓ cos θ
ϕ(r, θ) = ln ≈ = . (182)
2πϵ0 r1 2πϵ0 r 2πϵ0 r
The 1/r dependence in ϕ is the same as the 1/r dependence in the individual fields
from the wires. This is analogous to the fact that in the 3D dipole case in Section 2.7,
the 1/r2 dependence in ϕ was the same as the 1/r2 dependence in the individual fields
from the point charges.
We can now find the electric field via E = −∇ϕ. In polar coordinates the gradient
operator is given by ∇ = r̂(∂/∂r) + θ̂(1/r)(∂/∂θ). So the electric field equals
( ) ( )
∂ λℓ cos θ 1 ∂ λℓ cos θ λℓ ( )
E = −r̂ − θ̂ = cos θ r̂ + sin θ θ̂ . (183)
∂r 2πϵ0 r r ∂θ 2πϵ0 r 2πϵ0 r2
The calculation of the shapes of the field-line curves and the constant-potential curves
is nearly the same as in Section 2.7. The equation for the constant-ϕ curves is imme-
diately obtained from Eq. (182). The set of points for which ϕ takes on the constant
value ϕ0 is given by
( )
λℓ cos θ λℓ
= ϕ0 =⇒ r = cos θ =⇒ r = r0 cos θ, (184)
2πϵ0 r 2πϵ0 ϕ0
where r0 ≡ λℓ/2πϵ0 ϕ0 is the radius associated with the angle θ = 0. In the lower half
plane, both ϕ0 and cos θ are negative, so r is still positive, as it should be. As an
exercise, you can show that r = r0 cos θ describes a circle with diameter r0 . So the
equipotential curves are circles; see the solid lines in Fig. 50.
As explained in Section 2.7, the slope of a given curve at a given point, relative to the
r̂ and θ̂ basis vectors at that point, is dr/rdθ. So the slope of the r = r0 cos θ curve is
1 dr 1 sin θ
= (−r0 sin θ) = − . (185)
r dθ r0 cos θ cos θ
Remember that this is the slope with respect to the local r̂-θ̂ basis (which varies with
position), and not the fixed x̂-ŷ basis.
51
θ=0
1
θ = π/2
Field lines and constant-potential curves for a dipole. The two sets of curves are orthogonal at all
intersections. The solid lines show constant-ϕ curves (r = r0 cos θ), and the dashed lines show E
field lines (r = r0 sin θ).
Figure 50
Now consider the E field. As in Section 2.7, we’ll do things in reverse order, first
finding the slope of the tangent, and then using that to find the equation of the field-
line curves. The slope of the tangent is immediately obtained from the Er and Eθ
components given in Eq. (183). We have
Er cos θ
= . (186)
Eθ sin θ
This slope is the negative reciprocal of the slope of the tangent to the constant-ϕ
curves, given in Eq. (185), as it should be. To find the equation for the field-line
curves, we can use the fact that the slope in Eq. (186) must be equal to dr/rdθ. We
can then separate variables and integrate to obtain
∫ ∫
1 dr cos θ dr cos θ dθ
= =⇒ = =⇒ ln r = ln sin θ + C. (187)
r dθ sin θ r sin θ
where r0 ≡ eC is the radius associated with the angle θ = π/2. As an exercise, you
can show that r = r0 sin θ describes a circle with diameter r0 . So the field-line curves
are also circles; see the dashed lines in Fig. 50.
(a) Setting a = 1 and ignoring the factor of q/4πϵ0 , the potential due to the two
charges, at locations in the xy plane, is
4 1
ϕ(x, y) = √ −√ . (189)
2
(x + 2) + y 2 (x + 1)2 + y 2
52 CHAPTER 2. THE ELECTRIC POTENTIAL
√
Using the Taylor expansion 1/ 1 + ϵ ≈ 1 − ϵ/2 + 3ϵ2 /8, and keeping terms up
to second order in x and y, we have
4 1
ϕ(x, y) = √ −√
4 + (4x + x2 + y 2 ) 1 + (2x + x2 + y 2 )
2 1
=√ −√
2 2
1 + (x + x /4 + y /4) 1 + (2x + x2 + y 2 )
( )
1( ) 3( )2
≈ 2 1 − x + x /4 + y /4 + x + · · ·
2 2
2 8
( )
1( ) 3 ( )2
− 1 − 2x + x + y + 2x + · · ·
2 2
2 8
= 1 + (1/4)(y 2 − 2x2 ). (190)
(Alternatively, you can obtain this from the Series operation in Mathematica.)
If we had included the z dependence, then a z 2 term would appear in the same
manner as the y 2 term. That is, the term in parentheses would be y 2 + z 2 − 2x2 .
y
In terms of all the given parameters, you can show that
( )
q y 2 − 2x2
ϕ(x, y) ≈ 1+ . (191)
4πϵ0 a 4a2
x Some level surfaces of the function y 2 − 2x2 are shown in Fig. 51. The origin is a
saddle point; it is a maximum with respect to variations in the x direction, and
a minimum with respect to variations in the y direction.√The constant-ϕ lines
passing through the equilibrium point are given by y = ± 2x (near the origin).
If we zoom in closer to the origin, the curves keep the same general shape; the
picture looks the same, with the only change being the ϕ value associated with
Figure 51 each curve.
(b) The electric field is the negative gradient of the potential, so we have
q
E = −∇ϕ = (2x, −y). (192)
8πϵ0 a3
The field lines are the curves whose tangents are the E field vectors, by defini-
tion. Equating the slope of a curve with the slope of the tangent E vector, and
separating variables and integrating, gives
∫ ∫
dy Ey dy y dy 1 dx
= =⇒ =− =⇒ =−
dx Ex dx 2x y 2 x
1 B
=⇒ ln y = − ln x + A =⇒ y = √ , (193)
2 x
y
where A is a constant of integration, and B √≡ eA . Different values of B give
different field lines. Technically, this y = B/ x result is valid only in the first
quadrant. But since the setup is symmetric with respect to the yz plane (at least
near the origin), and also with respect to rotations around the x axis, the general
x form of the field lines in the xy plane is shown in Fig. 52. If we zoom in closer to
the origin, the lines keep the same general shape.
If we include the z dependence, then the correct expression for the field near the
origin has the (2x, −y) vector in Eq. (192) replaced with (2x, −y, −z). As a check
on this, the divergence of this vector is zero, which is correct because ∇·E = ρ/ϵ0
Figure 52
53
and because there are no charges at the origin. Although the abbreviated vector
in Eq. (192) is sufficient for making a picture of what the field lines look like, it
has nonzero divergence, so its utility goes only so far. See Exercise 2.65 for an
extension of this exercise.
(a) First note that the equilibrium point must lie on the line containing the two
charges, because otherwise the fields from the two charges won’t point along the
same line, which means that there is no way for the fields to exactly cancel.
There are various cases to consider. (1) If both charges are nonzero and have the
same sign, then the equilibrium point must lie between them, and it is easy to
see that there is only one such point. (2) If both charges are nonzero and have
opposite sign, then the equilibrium point must lie outside them, closer to the
smaller charge. The one exception to this case is when the charges are equal and
opposite, in which case there is no equilibrium point (technically, it is located at
infinity). (3) If one of the charges is zero, then there is no equilibrium point. (4)
If both charges are zero (a trivial case which probably isn’t worth considering),
then every point is an equilibrium point.
(b) If we choose the origin of our coordinate system to be the equilibrium point, with
the two charges lying on the x axis, and if we Taylor-expand the potential around
the origin, then it can’t have any terms that are linear in the coordinates. This
is true because if it did, the electric field (which is the negative gradient of the
potential) would have constant nonzero terms, violating the fact that E = 0 at the
equilibrium point. So the potential must take the form of ϕ = A + ax2 + by 2 + cz 2
(plus higher order terms, which we can ignore close to the origin). But b = c
because the system is symmetric under rotations around the x axis. So ϕ =
A + ax2 + b(y 2 + z 2 ). The relation ∇2 ϕ = −ρ/ϵ0 tells us that ∇2 ϕ = 0, because
ρ = 0 at the equilibrium point (and everywhere else, except at the locations of
the charges). Therefore, a = −2b. So the potential must take the form of
ϕ = A + b(−2x2 + y 2 + z 2 ). (194)
In the√xy plane (that is, for z = 0), the equipotential lines are therefore given by
y = ± 2x, as desired.
Note that all of the above reasoning is still valid even if we replace one (or both)
of the point charges with a stick (as in Exercise 2.45, for example), as long as all
of the charge in the setup lies on a single line.
Remark: The above reasoning can also be applied to the equipotential lines that cross
at the origin in Fig. 12.42 in√the solution to Problem 2.19. We claim that the slopes of
these lines are equal to ±1/ 2. This follows from Gauss’s law and symmetry around
the z axis (instead of the x axis in the above setup with two points); the potential must
look like
ϕ = A + b(−2z 2 + x2 + y 2 ) (195)
near the origin. In the xz plane (that is, for y = 0), the equipotential lines are therefore
√
given by z = ±x/ 2. All of the above results can be summarized by saying that if
a system is symmetric under rotations around a given axis, then at an equilibrium
√
point the equipotential lines make an angle of tan−1 ( 2) ≈ 54.7◦ with respect to the
symmetry axis.
54 CHAPTER 2. THE ELECTRIC POTENTIAL
(a) From the same reasoning we used in Problem 2.19, the field on the z axis equals
2Qz
E(x) = . (196)
4πϵ0 (z 2 + R2 )3/2
3 z
2 The dependence on z is the same√as in the case of the ring in Problem 2.19, so the
maximum still occurs at z = R/ 2. (The form of the answer is actually exactly
1
the same; the total charge appears in the numerator, with the total charge being
0 x 2Q in the present setup, whereas it was just Q in Problem 2.19.)
-1 (b) A sketch of some equipotential curves is shown in Fig. 53; we have chosen R = 1.
-2 The full surfaces are obtained by rotating the curves around the x axis. Close to
-3
the two point charges, the curves are circles, which means that the equipotentials
-3 -2 -1 0 1 2 3 are spheres in 3D space. Far away, the curves become large circles (or spheres in
3D) around the whole setup. The transition between the double spheres and the
Figure 53 single sphere occurs where the equipotentials cross at the origin, as shown. From
√
Exercise 2.65, the slopes of the lines at the crossing are ± 2.
(c) From Fig. 53, it appears that the transition from concave up√to concave down
occurs at about z =√3R/2 (you can show that it’s actually 2R). This isn’t
equal to the z = R/ 2 location of the maximum field we found in part (a), so
apparently the result analogous to the one in Problem 2.19 does not hold. Let’s
see why.
Consider the point on the z axis where the transition from concave up to concave
down occurs. From the reasoning in Problem 2.19, we know that ∂Ex /∂x = 0
at this point. In Problem 2.19 we noted that ∂Ey /∂y = 0 was also zero at this
point, due to the symmetry under rotations around the z axis. However, the
present scenario with two point charges is not symmetric around the z axis. It is
symmetric around the x axis instead. So we can’t say that ∂Ey /∂y = 0 is zero.
And indeed, the cross section of the equipotential surface in the y-z plane is a
circle, which is concave down. So ∂Ey /∂y = 0 is positive.
Therefore, since ∂Ex /∂x + ∂Ey /∂y + ∂Ez /∂z = 0 from Gauss’s law,√and since
∂Ex /∂x is zero at the transition point (which happens to be z = 2R), and
since ∂Ey /∂y is positive at any point on the z axis, we see that ∂Ez /∂z must
be negative at the transition point. This means that Ez is decreasing; that is,
the maximum has already occurred.
√ This is consistent with the fact that the
maximum-field
√ z value of R/ 2 from part (a) is smaller than transition-point z
value of 2R.
(a) If we start with the two collections of charges very far apart and then bring
collection 1 toward collection 2, a little piece of charge dq1 = ρ1 dv in collection 1
picks up an energy of dq1 ϕ2 = (ρ1 dv)ϕ2 in the presence of collection 2, where ϕ2 is
evaluated
∫ at the final location of the charge dq1 . So the total energy of the system
is ρ1 ϕ2 dv. This energy ignores the self energies of the two ∫ collections. But
these self energies don’t change throughout the process, so ρ1 ϕ2 dv represents
the increase in energy, that is, the work done.
If we instead bring∫collection 2 toward collection 1, the same reasoning shows that
the work done is ρ2 ϕ1 dv. But ∫the work can’t ∫ depend on how the collections
are brought together, so we have ρ1 ϕ2 dv = ρ2 ϕ1 dv, as desired.
55
And since ∇·E = ρ/ϵ0 , the identity becomes E1 ·E2 = ρ1 ϕ2 /ϵ0 −∇·(E1 ϕ2 ). If we
integrate this over all space, we can use the divergence theorem to write the third
term as an integral over a surface at infinity. But since our distributions have
finite extent, E falls off like 1/r2 (or faster, if the net charge is zero) and ϕ falls
2
off like 1/r (or faster). Since the area of a surface
∫ grows only line r∫ , the product
Eϕ·(area) goes to zero. We therefore arrive at E1 ·E ∫ 2 dv = (1/ϵ0 ) ρ1 ϕ∫2 dv. The
same procedure
∫ with∫ the 1’s and 2’s reversed yields E1 ·E2 dv = (1/ϵ0 ) ρ2 ϕ1 dv.
Hence ρ1 ϕ2 dv = ρ2 ϕ1 dv, as desired.
The constant of proportionality doesn’t matter because the end result will be zero.
The ∂Ex /∂x term in Eq. (2.59) is then
1 x(−3/2)(2x) −2x2 + y 2 + z 2
+ 2 = 2 , (199)
(x2 + y2 2
+z ) 3/2 2 2
(x + y + z ) 5/2 (x + y 2 + z 2 )5/2
with similar expressions for the ∂Ey /∂y and ∂Ez /∂z terms. The sum of all three
terms is zero, because the coefficient of x2 is (−2 + 1 + 1), etc. This is consistent with
div E = ρ/ϵ0 because ρ = 0 outside the sphere.
(a) At position x inside the slab, there is a slab with thickness ℓ − x to the right of
x, which acts effectively like a sheet with surface charge density σR = (ℓ − x)ρ.
Likewise, to the left of x we effectively have a sheet with surface charge density
σL = (ℓ + x)ρ. Since the electric field from a sheet is σ/2ϵ0 , the net field at
position x inside the slab is
(ℓ + x)ρ (ℓ − x)ρ ρx
E= − = , (200)
2ϵ0 2ϵ0 ϵ0
and it is directed away from the center plane (if ρ is positive). You can also
quickly obtain this by using a Gaussian surface that extends a distance x on
either side of the center plane.
Outside the slab, the slab acts effectively like a sheet with surface charge density
ρ(2ℓ), so the field has magnitude (2ρℓ)/2ϵ0 = ρℓ/ϵ0 and is directed away from
the slab. E(x) is continuous at x = ±ℓ, as it should be since there are no surface
charge densities in the setup.
56 CHAPTER 2. THE ELECTRIC POTENTIAL
∫x
(b) The potential relative to x = 0 is ϕ = − 0 E dx. Inside the slab this gives
∫ x
ρx ρx2
ϕin (x) = − =− . (201)
0 ϵ0 2ϵ0
Outside the slab, we must continue the integral past x = ±ℓ. On the right side
of the slab, where x > ℓ, the potential is
∫ ℓ ∫ x ∫ ℓ ∫ x
ρx ρℓ
ϕ(x) = − Ex dx − Ex dx = − dx − dx
0 ℓ 0 ϵ 0 ℓ ϵ0
ρℓ2 ρℓ ρℓ2 ρℓx
= − − (x − ℓ) = − . (202)
2ϵ0 ϵ0 2ϵ0 ϵ0
On the left side of the slab, where x < −ℓ, you can show that the only change in
ϕ is that there is a relative “+” sign between the terms (basically, just change ℓ
E to −ℓ). So the potential outside the slab equals
ρx ρℓ2 ρℓ|x|
__ ρl
__ ϕout (x) = − . (203)
ε0 ε0 2ϵ0 ϵ0
x
From Eqs. (201) and (203) we see that ϕ(x) is continuous at the boundaries at
x = ±ℓ, as it should be. Plots of E(x) and ϕ(x) are shown in Fig. 54.
(c) For a single Cartesian direction, we have ∇ · E = ∂Ex /∂x and ∇2 ϕ = ∂ 2 ϕ/∂x2 .
The following four relations are indeed all true:
And ρ = 0 outside the −a ≤ x ≤ a region. So we have two slabs with opposite charge
densities, with the positive slab on the left.
Since E = −∇ϕ (which in one dimension becomes E = −x̂ ∂ϕ/∂x), we simply need to
integrate E(x) to obtain ϕ(x). We find
E
ρ φ
E0
ε0E0/a E0a/2
a
x x x
-a -a a -a a
Figure 55
region, ϕ is constant, taking on the values it has at the boundaries, namely ±E0 a/2.
The plots of ρ, E, and ϕ are shown in Fig. 55.
A double check: At x = 0, the two slabs act effectively like sheets with charge densities
±σ = ±ρa. They each create a field pointing to the right with magnitude σ/2ϵ0 , so the
total field at x = 0 is 2(ρa)/2ϵ0 = ρa/ϵ0 . And since we found above that ρ = ϵ0 E0 /a,
this field equals E0 , in agreement with the given value.
x x x
-2Bl
Figure 56
The charge density is given by ρ = −ϵ0 ∇2 ϕ, or equivalently by ρ = ϵ0 ∇·E. For |x| > ℓ
we have ρ = 0, and for |x| < ℓ we have
d2 ϕ
ρ = −ϵ0 = 2ϵ0 B. (209)
dx2
So we have a uniform slab of charge between x = −ℓ and x = ℓ. However, we aren’t
quite done, because as mentioned above, there is also a surface charge density on the
planes at x = ±ℓ. This is consistent with ρ = −ϵ0 ∇2 ϕ, because if you tried to calculate
58 CHAPTER 2. THE ELECTRIC POTENTIAL
−ϵ0 ∇2 ϕ or ϵ0 ∇ · E there, you would obtain an infinite result (due to the discontinuity
in Ex ), consistent with the fact that a surface charge occupies zero volume.
To determine the surface charge density σ on the two planes, we can look at the
discontinuity in the field across them. From Fig. 56(b), E has a downward jump of
−2Bℓ at both planes. Gauss’s law tells us that the change in the field at a surface is
equal to σ/ϵ0 . Hence σ = −2ϵ0 Bℓ. You can quickly check that the sign here makes
the discontinuity in E work out properly. The plot of ρ is shown in Fig. 56(c), where
the arrows indicate the negative infinite volume densities associated with the surface
charges.
Our system therefore consists of a thick slab with positive volume charge density
ρ = 2ϵ0 B sandwiched between two sheets with negative surface charge density σ =
−2ϵ0 Bℓ. Note that the total charge per unit area in the system is 2σ + ρ(2ℓ), which
equals zero. This is consistent with the fact that the electric field is zero for |x| > ℓ.
2.72. A spherical charge distribution
The given potential, shown in Fig. 57(a), arises from a spherically symmetric charge
distribution. The potential is more briefly described in spherical coordinates by
ρ0 r2
(for r < a),
ϕ(r) = 4πϵ0 ( ) (210)
3
ρ0 2a
−a2 + (for r > a).
4πϵ0 r
Note that ϕ(r) is continuous at r = a, where it takes on the value ρa2 /4πϵ0 . Also note
that ϕ = −ρa2 /4πϵ0 at r = ∞; it is not necessary to have ϕ = 0 at infinity.
r r r
a a a
ρ0a 3ρ
ρ0a2 - ____ - ___0
- ____ 2πε0 2π
4πε0
Figure 57
The electric field is given by E = −∇ϕ, which reduces to Er = −dϕ/dr for a function
that depends only on r. This gives
−ρ r
0
(for r < a),
2πϵ0
Er (r) = 3 (211)
ρ0 a
(for r > a).
2πϵ0 r2
A plot of Er is shown in Fig. 57(b). You should verify that you obtain the same field
if you work with Cartesian coordinates, where ∇ϕ = (∂ϕ/∂x, ∂ϕ/∂y, ∂ϕ/∂z). Note
59
Again, you should verify that you obtain these same results in Cartesian coordinates,
where ∇2 ϕ = ∂ 2 ϕ/∂x2 + ∂ 2 ϕ/∂y 2 + ∂ 2 ϕ/∂z 2 and ∇ · E = ∂Ex /∂x + ∂Ey /∂y + ∂Ez /∂z.
We therefore have a uniform charge density inside r = a (the total charge there is
4πa3 ρ/3 = −2ρ0 a3 ), and zero charge outside. But as mentioned above, there is also a
surface charge density σ on the sphere with radius a. Equation (212) doesn’t contradict
this fact, because that equation has nothing to say about ρ(r) right at r = a. If you
tried to calculate −ϵ0 ∇2 ϕ or ϵ0 ∇ · E there, you would obtain an infinite result due
to the discontinuity in E, consistent with the fact that a surface charge occupies zero
volume.
To determine σ, we can look at the discontinuity in the field across the sphere at
r = a. From Eq. (211) the field just inside this sphere is −ρ0 a/2πϵ0 , and the field
just outside is ρ0 a/2πϵ0 . Gauss’s law (with a little pillbox) tells us that the change
in the field at the surface, which is ρ0 a/πϵ0 , must equal σ/ϵ0 . Hence σ = ρ0 a/π.
The total charge on the surface is then 4πa2 σ = 4ρ0 a3 , which is twice as large as the
−2ρ0 a3 charge distributed throughout the inside the sphere. The external field of the
entire sphere is therefore the field of a net charge 4ρ0 a3 − 2ρ0 a3 = 2ρ0 a3 , which is
Er (r) = (2ρ0 a3 )/4πϵ0 r2 = ρ0 a3 /2πϵ0 r2 in agreement with Eq. (211). Indeed, working
backward from this external field would be another way of finding the surface density
σ.
A plot of ρ(r) is shown in Fig. 57(c). The spike indicates the infinite volume charge
density associated with the surface charge density. Looking back at the plot of ϕ(r)
Fig. 57(a), note that the first derivative of ϕ (which is related to the field) is not well
defined at r = a, consistent with the discontinuity in E. Also, the second derivative
of ϕ (which is related to the density) is infinite at r = a, consistent with the infinite ρ.
2.73. Satisfying Laplace
In f (x, y) = x2 + y 2 , then
z
2 2
∂ f ∂ f
∇2 f = + 2 = 2 + 2 ̸= 0. (213) y
∂x2 ∂y
x
If g(x, y) = x2 − y 2 , then ∇2 f = 2 − 2 = 0. So g satisfies Laplace’s equation, but f
does not. The plot of g(x, y) looks like the saddle shown in Fig. 58. It is a positive
parabola along the x axis, and a negative parabola along the y axis.
Figure 58
y
60 CHAPTER 2. THE ELECTRIC POTENTIAL
∂ϕ ∂2ϕ
= −kϕ0 e−kz sin kx, = −k 2 ϕ0 e−kz cos kx, (214)
∂x ∂x2
and
∂ϕ ∂2ϕ
= −kϕ0 e−kz cos kx, = k 2 ϕ0 e−kz cos kx. (215)
∂z ∂z 2
Therefore,
Figure 60 ∂2ϕ ∂2ϕ ∂2ϕ
∇2 ϕ = + 2 + 2 = 0. (216)
∂x2 ∂y ∂z
6 To get a sense of what this field looks like, note that Ez /Ex = 1/ tan kx. This is
5 independent of z, so for a given value of x, the slopes of all the field-line curves
4 are the same. This slope is infinite for x = 0, ±π/k, ±2π/k, etc, and it is zero
for x = ±π/2k, ±3π/2k, etc. Fig. 61 shows a few of the curves. The density of
3
the field lines in this figure does not indicate the strength of the field, so we have
2
also drawn Fig. 62, in which the relative field strengths are accurately presented;
1 this figure shows only half of a cycle in x, for the sake of clarity. The constant
0 nature of the slope for a given value of x is clear from this figure. (Equivalently,
x
0 1 2 3 4 5 6 Fig. 61 looks the same at any height.) Since the surface charge density on the
sheet is proportional to the normal component of the field, it is evident from
Figure 61 Fig. 61 that the charge density is high at x = 0, ±π/k, etc., while it is zero at
x = ±π/2k, etc. This is consistent with the result we will find in part (c).
(E, with k =1)
z If you want to calculate the actual shape of the field-line curves, you can do this
as follows. If a curve is described by the function z(x), then the slope dz/dx is
3.0
given by the ratio Ez /Ex that we found above (because the E field is by definition
2.5 tangent to the field-line curve). Therefore, if a particular field line crosses the x
2.0 axis at x = x0 (assume 0 < x < π/2k), we have
1.5 ∫ z ∫ x
dz 1 cos kx dx
1.0 = =⇒ dz =
dx tan kx 0 x0 sin kx
0.5
x ( )
1 1 sin kx
0.0
x =⇒ z = ln(sin kx) = ln . (218)
0.0 0.5 1.0 1.5 2.0 2.5 3.0 k x0 k sin kx0
Figure 62
61
[
] slightly more informative to write this as z = (1/k) ln(sin kx) −
It is actually
ln(sin kx0 ) . This form makes it clear that the different curves have the same
shape but are simply shifted vertically relative to each other by an amount
−(1/k) ln(sin kx0 ). The closer x0 is to zero, the larger this term is, so the higher
the curve goes in the xz plane. This is consistent with Fig. 61. This form also
makes it clear that all the different curves associated with different values of x0
have the same slope for a given value of x.
(c) The component of the field perpendicular to the sheet, very close to the sheet,
satisfies Ez = σ/2ϵ0 . Therefore,
σ = 2ϵ0 Ez = 2ϵ0 kϕ0 e−0 cos kx = 2ϵ0 kϕ0 cos kx. (219)
z=0
Note that since we are told that the only charges in the system are on the sheet,
the system is symmetric with respect to the sheet. So the potential and field
below the sheet are the mirror images of the potential and field above the sheet.
= 0 dx + 2x0 dy + 3y0 dz
0 0 0
= 2x0 y0 + 3y0 z0 . (221)
Since (x0 , y0 , z0 ) is a general point, we can drop the subscripts and write g(x, y, z) =
2xy + 3yz. You can quickly check that the gradient of g is indeed G.
A quicker method of obtaining g is the following. The x component of ∇g = G
tells us that ∂g/∂x = 2y. So g must be a function of the form 2xy + f1 (y, z).
Similarly, the y component tells us that g must take the form 2xy +3yz +f2 (x, z),
and the z component tells us that g must take the form 3yz + f3 (x, y). You can
quickly verify that the only function satisfying all three of these forms is 2xy+3yz
(plus a constant).
(c) If H = (x2 −z 2 , 2, 2xz) then we find ∇×H = (0, −4z, 0) and ∇·H = 2x+0+2x =
4x. Since the curl isn’t zero, there is no associated potential ϕ.
62 CHAPTER 2. THE ELECTRIC POTENTIAL
The first equality is the statement of Stokes’ theorem, and the second is the
statement of Gauss’s theorem (the divergence theorem) applied to the vector
“∇ × A.” The logic of the derivation is as follows. The line integral of A around
the curve C in Fig. 2.52 is zero because the curve backtracks along itself. (We
can make the two “circles” of C be arbitrarily close to each other, and they run
in opposite directions.) Stokes’ theorem then tells us that the surface integral of
∇ × A over S is also zero. The surface S is essentially the same as the closed
surface S ′ consisting of S plus the tiny area enclosed by C. So the surface integral
of ∇ × A over S ′ is zero. But S ′ encloses the volume V , so Gauss’s theorem tells
us that the volume integral of ∇ · (∇ × A) over V is also zero. Since this result
holds for any arbitrary volume V , the integrand ∇ · (∇ × A) must be identically
zero, as we wanted to show.2
This logic here basically boils down to the mathematical fact that the boundary
of a boundary is zero. More precisely, the volume integral of ∇ · (∇ × A) equals
(by Gauss) the surface integral of ∇ × A over the boundary S ′ of the volume V ,
which in turn equals (by Stokes) the line integral of A over the boundary C of
the boundary S ′ of the volume V . But S ′ has no boundary, so C doesn’t exist.
That is, C has zero length. The line integral over C is therefore zero, which
means that the original volume integral of ∇ · (∇ × A) is also zero.
In view of this, there actually wasn’t any need to pick the curve C to be of the
specific stated form. We could have just picked a very tiny circle. The first step
in the above derivation, namely that the line integral of A around the curve C
is zero, still holds (but now simply because C has essentially no length), so the
derivation proceeds in exactly the same way.
2 If ∇ · (∇ × A) were different from zero at some point, then the integral over a small volume containing
this point would be nonzero. This is true because we can pick the volume to be small enough so that
∇ · (∇ × A) is essentially constant, so there is no possibility of cancelation.
64 CHAPTER 2. THE ELECTRIC POTENTIAL
Chapter 3
65
66 CHAPTER 3. ELECTRIC FIELDS AROUND CONDUCTORS
repel a positive-mass particle in the interior, canceling the attraction from the original
positive point mass.
However, there is one aspect in which the gravitational case differs fundamentally from
the electrical case. In the latter, like charges repel; whereas in the former, like masses
attract. This means that in the gravitational case, like masses would attract each
other and collapse down to a point, if they were allowed to move freely. So we would
have to bolt them down in the desired distribution.
(a) The charge distributions and field lines are shown (roughly) in Fig. 65. Let the
four surfaces be labeled 1, 2, 3, 4, starting from the innermost one. There is
charge −q on surface 1. This is true because the field is zero inside the metal of
the conductor, so a spherical Gaussian surface drawn inside the metal of the inner
conductor has no flux, so the net charge enclosed in the sphere must be zero. The
negative surface charge density on surface 1 is higher near the off-center point
charge. Since the inner conductor is neutral, a charge +q must reside on surface
2. This surface charge density is spherically symmetric, because it feels no field
from the charges inside (or outside) of it, due to the zero field inside the metal
of the conductors.
a) b)
4+
+ + E=0 4+
+ + E=0
-
+- 3 - - + 3
+ 2 + + 2 -
+- 1 - + -1 -
+ - - + - wire -
++- + - -+ - -
- ++ - - -
-
- q q - q q
+ +- - + - + - - -
+- + - +
+ +- + + + +
E=0 E=0
Figure 65
3.34. Equipotentials
Assume that the point charge is positive (the general result is the same if it is negative).
Then the near part of the sphere ends up negatively charged, and the far part ends
up positively charged. (The sizes of these two regions depend on the distance from
the point charge to the sphere.) By continuity, there must therefore be a circle on the
sphere where the charge density is zero. But the electric field near the sphere (which
is perpendicular to the conducting surface) is given by σ/ϵ0 . So if σ = 0 on the circle,
then E = 0 also.
The general shapes of the equipotentials are shown in Fig. 66. (The various curves
have been chosen to indicate the general features; their potential values aren’t equally
spaced.) In this specific case, the distance from the point charge to the sphere has
been chosen to be twice the radius of the sphere. The transition from small circles
around the point charge to large circles around the whole system takes place via the
equipotential curve that heads straight into the sphere and then splits in two, encircling
the sphere. At the point where the curve splits, it changes direction abruptly. Since
the electric field must be perpendicular to the equipotential surface at every point,
this means that the electric field must point in two different directions at the splitting
point. The only vector that is perpendicular to two different directions is the zero
vector. So this is a second way of seeing why there must be points on the surface of
the sphere where the electric field is zero.
0
conducting
sphere
-1
-2
-3
-2 -1 0 1 2 3 4
Figure 66
Figure 67
68 CHAPTER 3. ELECTRIC FIELDS AROUND CONDUCTORS
This diverges as x → 0.
In the case of a cone, a slightly different calculation is required, but the field still
diverges. See Problem 1.3.
3.36. Zero flow
The point is that although the field is weaker near the larger sphere, it has an ap-
preciable size over a larger distance than does the field from the smaller sphere. The
fields at the surfaces are proportional to Q/R2 ∝ R/R2 = 1/R. And the fields fall off
on a distance scale of R, because at a radius of 2R, the field has decreased by a factor
1/22 , etc. These two effects cancel.
We can be quantitative. As in Problem 3.10, the thin wire has essentially zero capaci-
tance, so charge can’t pile up on it. But a tiny bit can, so as mentioned in the solution
to Problem 3.10, we effectively have a rigid stick of (a tiny bit of) charge extending
from one sphere to the other. Assuming constant density λ, the repulsive force on
∫ D−R
the stick from the smaller sphere is R1 2 E1 λ dr, where D is the distance between
the centers of the spheres, and where E1 (r) is the field due to the smaller sphere.
∫ D−R
Likewise, the repulsive force from the larger sphere is R2 1 E2 λ dr. If the spheres
are far apart, we can replace the upper limits of these integrations by ∞; the fields
become negligible at large distances. If we set these two forces equal to each other and
cancel the λ’s, we simply have the statement that the potentials of each shell, relative
to infinity, are equal. In a sense, we can think of many weak people forcing the charge y=0 y=s
away from the larger sphere, with only a few strong people forcing it away from the
small sphere. The two total forces exactly cancel, and no charge moves.
3.37. A charge between two plates
Consider the Gaussian surface shown in Fig. 70, which has very large extent in the x Q
and z directions. Two of its faces lie inside the metal of the two plates. The field is
zero inside the metal of the plates; and between the plates the field is negligible at b
points far away from the charge. So there is zero flux through the Gaussian surface.
By Gauss’s law, the total charge inside must be zero, which implies that the total
charge on the inner surfaces of the plates must be −Q.
Dividing the charge Q into many smaller charges, all on the plane y = s, yields the
same total charge on each plate, by superposition. We can take the continuum limit
and smear out the charge Q uniformly onto a sheet with a large area A at y = b. The left right
surface charge density on this sheet is σ = Q/A (this is the sum of the densities on plate plate
its two surfaces). If we can find the total charges on the two plates in this scenario, Figure 70
then we will have found the total charges on the two plates in the original scenario
involving the point charge Q.
Call the charges on (the inner surface of) each plate Q1 and Q2 . The densities are
then σ1 = Q1 /A and σ2 = Q2 /A, so Q1 /Q2 = σ1 /σ2 . The same area A applies to
the two plates because we are assuming A is large. The edge effects will be negligible,
so we can assume that σ1 and σ2 are essentially uniform over the area A. (And σ is
uniform, by construction.)
From the standard argument using Gauss’s law with one face of the Gaussian surface
lying inside the metal of a conductor, the fields in the two different regions are E1 =
σ1 /ϵ0 and E2 = σ2 /ϵ0 . So E1 /E2 = σ1 /σ2 . But the two plates are at the same
potential, so we also know that E1 b = E2 (s − b), because these are the differences in
potential from the middle sheet. Hence E1 /E2 = (s − b)/b. So we have
Q1 σ1 E1 s−b Q1 s−b
= = = =⇒ = . (229)
Q2 σ2 E2 b Q2 b
70 CHAPTER 3. ELECTRIC FIELDS AROUND CONDUCTORS
But we also have Q1 + Q2 = −Q. Solving these two equations for Q1 and Q2 gives
s−b b
Q1 = −Q and Q2 = −Q . (230)
s s
If b ≪ s, then we have Q1 ≈ −Q and Q2 ≈ 0, as expected. In general, the charges are
in the inverse ratio of the distances from the plates to the intermediate sheet (or to
the given point charge Q).
3.38. Two charges and a plane
First note that such a location must exist, due to a continuity argument: If the −Q
charge is placed only slightly below the fixed Q charge, the upward attractive force
from the Q charge will dominate. But if the −Q charge is placed only slightly above
Q the conducting plane, the downward attractive force from the +Q image charge will
dominate. So somewhere in between, the force on the −Q charge must be zero.
Let y be the distance from the −Q charge to the plane. The field above the plane due
l -Q to the two given charges along with the induced charge on the plane is identical to the
y field due to the two given charges along with the two image charges below the plane
shown in Fig. 71. The given −Q charge feels the fields due to the other three charges.
Taking upward to be positive, the force on the given −Q charge is
( )
Q2 1 1 1
Q F = − + . (231)
4πϵ0 (ℓ − y)2 (2y)2 (ℓ + y)2
Setting this equal to zero yields
-Q 1 2(ℓ2 + y 2 )
= =⇒ 7y 4 + 10ℓ2 y 2 − ℓ4 = 0. (232)
Figure 71 4y 2 (ℓ2 − y 2 )2
This is a quadratic equation in y 2 . We are concerned with the positive root (since y 2
is positive), which is
√
(−5 + 4 2)ℓ2
2
y = ≈ (0.0938)ℓ2 =⇒ y = (0.306)ℓ. (233)
7
λ 10−5 C/m V
Esurface = 2 · = ( 2 C2 ) = 7.2 · 104 . (234)
2πϵ0 h π 8.85 · 10 −12 s m
kg m3 (5 m)
The electrical force on the given wire is the force due to the field arising from the image-
charge wire. The total charge on the given wire is q = λL = (10−5 C/m)(200 m) =
2 · 10−3 C. Over nearly the whole length of the wire, the field due to the image-
charge wire is essentially λ/2πϵ0 (2h) = 1.8 · 104 V/m, directed downward. This is a
quarter of the field we found above, which involved two wires and half the distance.
Neglecting the decrease in field near the ends of the wire, the force on the given wire
is qE = (2 · 10−3 C)(1.8 · 104 V/m) = 36 N, directed downward. In terms of the various
parameters, this force is qE = (λL)(λ/2πϵ0 (2h)) = (λ2 /4πϵ0 )(L/h).
71
Remark: The above result is a good approximation in the limit where L ≫ h. Let’s try
to get a handle on the error involved. Since we assumed that the field was uniform over the
length of the wire, what we actually calculated was the force on a finite wire due to an infinite
wire (Fig. 72(a)). This is larger than the force between two finite wires (Fig. 72(b)) because
of the two extra half-infinite wires (Fig. 72(c)) that need to be added to the lower finite wire
to make the infinite wire.
(a)
(b)
(c)
(d)
Figure 72
As as exercise, you can show that the vertical component of the force between the two half-
infinite wires in Fig. 72(d) is λ2 /4πϵ0 , which is independent of the separation (this follows
from a dimensional-analysis argument). To a good approximation, the original finite wire
can be treated as half-infinite for this purpose. (The error is of order (λ2 /4πϵ0 )(h/L).) So
the force between the finite wire and the two half-infinite wires in Fig. 72(c) equals 2λ2 /4πϵ0 .
This is the correction we need to make to our original result. So the actual force between the
two finite wires in Fig. 72(b) is
( )
λ2 L 2λ2 λ2 L 2h
− = 1− , (235)
4πϵ0 h 4πϵ0 4πϵ0 h L
up to corrections of higher order in h/L. In the present setup, 2h/L equals 1/20. So by
ignoring the end effects, we over-estimated the force by about 5%.
Remark: The above reasoning works for any starting angle of the field lines, not just
horizontal. If we measure θ with respect to the upward vertical, then as an exercise you
can
√ shown that a field line that starts out at an angle θ meets the plane at a radius R =
h 3 + 2 cos θ − cos2 θ/(1 − cos θ). (As a sub-problem, you will need to show that the fraction
of the total surface area of a sphere that lies in a spherical cap subtended by the cone with
half-angle θ is (1 − cos θ)/2. The remainder
√ therefore subtends a fraction (1 + cos θ)/2.) For
θ = π/2, this correctly gives R = 3 h. And for θ = 0 and π it gives, respectively, R = ∞
and 0, as it should.
Interestingly, for θ close to π (call it π − ϵ), that is, for field lines that start off pointing
√
nearly straight downward, you can show that R ≈ hϵ/ 2. This is correctly smaller than
hϵ, because that would be where the field line would hit the plane if the line were exactly
straight, whereas we know that it must bend so that it ends up vertical when it meets the
-q r1 r1 q √
plane. But the 1/ 2 factor isn’t obvious.
The factor is parenthesis is approximately equal to 1.2. Note that the total force from
the dipoles is smaller than the force from the closest image charge by a factor of order
(b/ℓ)3 . Two of these powers of b/ℓ come from the fact that the distances from the
given charge to the dipoles are on the order of ℓ instead of b. And the third power
comes from the dipole effect of taking the difference between nearly-canceling forces.
You can also calculate the total force by looking at the forces from the positive and neg-
ative image charges separately. From Fig. 12.53, the forces from the positive charges
cancel, because they are symmetrically located with respect to the given charge. The
force from the closest negative charge is q 2 /4πϵ0 (2b)2 , directed to the right. The forces
from the other negative charges nearly cancel in pairs. The sum of the forces from all
these pairs points to the left, and its magnitude can be written as
∞ ( )
q2 ∑ 1 1
Fneg = −
4πϵ0 n=1 (2nℓ − 2b)2 (2nℓ + 2b)2
∑∞ ( )
q2 1 1 1
= −
16πϵ0 ℓ2 n=1 n2 (1 − b/nℓ)2 (1 + b/nℓ)2
∑∞ (( ) ( ))
q2 1 2b 2b
≈ 1+ − 1−
16πϵ0 ℓ2 n=1 n2 nℓ nℓ
∞
q2 b ∑ 1
= , (243)
4πϵ0 ℓ3 n=1 n3
(a) At a given point P on the right plane in Fig. 12.53, we need to add up the x
components of the fields due to the real charge and all the image charges. The
two nearest charges on either
√ side of the right plane (the real charge and image
charge 1) are a distance b2 + r2 away from P . So the magnitude of the field
from each charge is q/4πϵ0 (b2 + r2 ). Taking the x component brings in a factor
75
√
of b/ b2 + r2 . Both charges produce a positive x component, so that brings in a
factor of 2. Putting it all together gives the first term in Eq. (3.41).
Now consider images charges 2 and 4 in Fig. 12.53. They both are a distance
2ℓ − b from the right plane, so we simply need to replace b with 2ℓ − b in the
above reasoning. And we also need to add on a minus sign since the x component
is negative. So we obtain the first term in the sum in Eq. (3.41) with n = 1.
Similarly, charges 3 and 5 both are a distance 2ℓ + b from the right plane, so they
yield the second term in the sum with n = 1.
In the same manner, charges 6 and 8 with distances 4ℓ − b, and charges 7 and 9
with distances 4ℓ + b, yield the n = 2 terms. And so on.
(b) As in Eq. (3.5), the integral of the first term in Eq. (3.41) (after dividing by −4π
to obtain the density) is
∫ R R
1 2qb qb
− 2πr dr = 2 = −q, (244)
2
4π 0 (b + r ) 2 3/2 2
(b + r ) 1/2
0
where we have set R = ∞, which causes no problem with this term. In the same
manner, the integral arising from the pair of terms with a particular value of n is
( ) R
q(2nℓ − b)
q(2nℓ + b)
−( )1/2 + ( )1/2 . (245)
(2nℓ − b) + r
2 2 2
(2nℓ + b) + r 2
0
As stated in the problem, if we set R = ∞ these two terms equal ±q, so they
cancel. We’ll therefore let R be a large but finite distance. Then when the first
term is evaluated at R it can be approximated as (dropping the b2 term)
( )
q(2nℓ − b) q(2nℓ − b) 1 4nℓb
−( )1/2 ≈ − ( )1/2 1 + 2 4n2 ℓ2 + R2 , (246)
(4n2 ℓ2 + R2 ) − 4nℓb 4n2 ℓ2 + R2
where we have used 1/(1 − ϵ)1/2 ≈ 1 + ϵ/2. The terms that don’t involve b
(or that involve b2 ) will cancel with the corresponding terms in the second term
in Eq. (245), which looks the same except for the overall minus sign and the
replacement b → −b. So we care only about the b terms. You can show that
their sum for a given value of n is 2qbR2 /(4n2 ℓ2 + R2 )3/2 . The factor of 2 out
front comes from the fact that there are two terms in Eq. (245).
We must now sum this over n. For large R, the terms change slowly with n, so
we can approximate the sum by an integral. Let’s relabel n as z. In the original
sum over n, we can multiply each term by dn, because dn simply equals 1 since n
runs over the integers. We can then replace dn by dz. Integrating over z from 0
(although the exact starting point doesn’t matter) to ∞ gives (you should verify
this integral)
∫ ∞ ∞
2qbR2 dz 2qbz
= √ = qb . (247)
0
2 2
(4z ℓ + R ) 2 3/2
4ℓ z + R 0
2 2 2 ℓ
Remembering to include the −q charge in Eq. (244), the total charge on the right
plane is −q + qb/ℓ = −q(ℓ − b)/ℓ. This equals −q if b = 0, and 0 if b = ℓ. These
makes sense, because the left plane or right plane, respectively, is irrelevant in
these two cases.
The charge on the left plane (is obtained)by letting b → ℓ−b throughout the above
calculation, which yields −q ℓ − (ℓ − b) /ℓ = −qb/ℓ. Alternatively, we know that
76 CHAPTER 3. ELECTRIC FIELDS AROUND CONDUCTORS
the total charge on the inner surfaces of the two planes must be −q (by using a
Gaussian surface with two faces lying inside the metal of the conducting planes,
where the field is zero). So −q + qb/ℓ on the right plane implies −qb/ℓ on the left
plane.
R2 R2 R
x= = = ≈ R(1 − h/R) = R − h. (248)
A R+h 1 + h/R
Very close to a spherical shell, the shell looks locally like a plane. So what we essentially
have here is two charges ±Q located on either side of a plane, a distance h from it.
This is exactly what our image-charge setup looked like in the case of a plane. The
same type of reasoning holds if the given charge lies inside the shell. The real and
image charges have now simply switched sides of the plane.
electric field must be perpendicular to at all points. And the only vector that is
perpendicular to two directions is the zero vector.
3.49. Positive or negative density
Let the point charge Q be at radius nR, where n is a numerical factor. (Working with
this factor n makes the solution a little cleaner than it otherwise might be.) From
Problem 3.13 there is an image charge −Q/n located at radius R/n. And then from
Problem 3.16 there is an additional image charge (1 + 1/n)Q located at the center of
the shell, to make the total charge on the shell be Q.
In the cutoff case where the surface charge density σ at the closest point on the shell
changes from negative to positive, σ will be zero. But the field right outside the shell
equals σ/ϵ0 , so σ will be zero if the field is zero. The field equals the sum of the fields
from the three charges (one real and two image). Being careful with the signs of the
three fields, if we set the total field right outside the closest point equal to zero, we
obtain (ignoring the 1/4πϵ0 )
Remark: We can also ask the analogous question where we put a charge Q inside a non-
grounded conducting spherical shell with radius R and net charge Q. A continuity argument
again shows that there must be a cutoff radius, below which the charge density (the sum
of the inner and outer surface densities on the shell) is positive everywhere on the shell.
But the solution is slightly different, due to the fact that charge resides on both the inner
and outer surfaces of the shell. The inner surface has a total charge −Q, nonuniformly
distributed; the density is determined by considering an image charge located outside (see
Problem 3.13). There is also a charge 2Q on the outer surface (to make the total charge on
the shell equal to Q); this charge is uniformly distributed. You can show that the inner and
outer densities cancel at the nearest point on the shell if the given charge Q is located at
√
radius R(5 − 17)/4 ≈ (0.219)R.
(1 + 1/n)Q (−Q/n)
E= + . (250)
(nR) 2 (n − 1/n)2 R2
Setting this equal to zero yields
1 + 1/n 1/n
= . (251)
n2 (n − 1/n)2
78 CHAPTER 3. ELECTRIC FIELDS AROUND CONDUCTORS
Simplifying gives
n4
n+1= =⇒ n5 − 2n3 − 2n2 + n + 1 = 0. (252)
(n2 − 1)2
This equation factors into
(a) From the reasoning in the solution to Problem 1.27, the field due to the upper
sphere, at a point at position r1 with respect to its center, is E = ρr1 /3ϵ0 .
Likewise, the field due to the lower (negative) sphere, at a point at position r2
with respect to its center, is E = −ρr2 /3ϵ0 . The sum of these fields is (with the
subscript “s” standing for “spheres”)
(2πR sin θ)(R dθ)ℓ = (2πR sin θ)(R dθ)s(1 + cos θ). (255)
The charge in this ring is therefore ρ2πR2 s sin θ(1 + cos θ) dθ. Since the ring is
essentially located right on the surface of the bottom sphere (if s is small), each
little piece dq feels a force with magnitude Q dq/4πϵ0 R2 due to the bottom sphere,
where −Q is the charge of the sphere (which equals 4πR3 ρ/3, but we won’t need
this). Only the vertical component survives, and this brings in a factor of cos θ.
So the vertical force due to the bottom sphere on the ring is directed downward
with magnitude
( )
Q ρ2πR2 s sin θ(1 + cos θ) dθ Qρs sin θ cos θ(1 + cos θ) dθ
· cos θ = . (256)
4πϵ0 R2 2ϵ0
79
Integrating this from 0 to π gives the total downward force on the shaded region
as
∫ ( ) π
Qρs π Qρs cos2 θ cos3 θ
sin θ cos θ(1 + cos θ) dθ = − −
2ϵ0 0 2ϵ0 2 3
0
Qρs 2 Qρs
= · = , (257)
2ϵ0 3 3ϵ0
3ϵ0 E
σ = ρ(s cos θ) = ρ cos θ = 3ϵ0 E cos θ, (259)
ρ
as desired. (Note that the relevant thickness here is not the thickness of the
shaded region in Fig. 3.35.) This σ = 3ϵ0 E cos θ result implies that the magnitude
of the field at the poles is three times the uniform field E. The conducting-sphere
limit is obtained in the ρ → ∞ and s → 0 limit, with the product ρs equalling
3ϵ0 E.
Put charges Q and −Q on the two conductors in each of the two given capacitors. In
the bottom capacitor in Fig. 78, one of the conductors consists of the two outer plates,
wire
because they are connected by a wire. The charge distributions on the various surfaces A
are shown. All the factors of 1/2 arise from symmetry. In the bottom capacitor, the s/2 Q/2
-Q/2
potential difference (which is the difference between either of the outside plates and -Q/2
s/2 Q/2
the inner plate) equals the field times the separation. The field is half of what it is in
the top capacitor (because the density σ is half), and the separation is also half. So
Figure 78
80 CHAPTER 3. ELECTRIC FIELDS AROUND CONDUCTORS
the potential difference is (1/2)(1/2) = 1/4 of what it is in the top capacitor. Since
the charge Q on each capacitor is the same, we have
Remark: From similar reasoning involving Gaussian surfaces with one side lying inside a
conductor, it follows that the density on the bottom face of the top plate is −σ1 , and the
density on the top face of the bottom plate is −σ2 . Assuming that there is zero net charge on
the outer two plates, this leaves at total of σ1 + σ2 = σ for the outer surfaces of these plates.
It must get divided evenly, because otherwise these two surfaces would create a nonzero field
between them, which would change the above fields and make the outer plates not be at the
same potential. If any additional charge is dumped on the outer plates, it simply gets divided
evenly between their two outer surfaces.
Solving for q1 and q2 gives q1 = (Q1 + Q2 )/2 and q2 = (Q1 − Q2 )/2. So from top to
bottom, the charges on the four plates are
Q1 + Q2 Q1 − Q2 Q2 − Q1 Q1 + Q2
, , , (262)
2 2 2 2
We can also state which four of the eight surfaces (top and bottom of the each of the
four plates) these charges lie on. None of the charges can border the E = 0 regions,
because otherwise the standard argument involving a Gaussian surface with one face
lying inside the metal of the conductor would imply a nonzero field in these regions.
So the four charges lie on the top of the top plate, the bottom of the next, the top of
the next, and finally the bottom of the bottom plate.
Figure 80
82 CHAPTER 3. ELECTRIC FIELDS AROUND CONDUCTORS
between the positive and negative (sets of N ) plates in the capacitor can then be
written as
( )
σ1 s Qs (2N − 1)Aϵ0 (2N − 1)Aϵ0
ϕ= = =⇒ Q = ϕ =⇒ C = .
ϵ0 (2N − 1)Aϵ0 s s
(264)
As N → ∞, this is essentially equal to 2N Aϵ0 /s. The capacitance in Eq. (264) is
larger than the capacitance we would obtain if we juxtapose the two pairs of N plates
to create two plates with area N A. If we keep the separation s, then the capacitance
of the resulting standard two-plate capacitor would be C = ϵ0 (N A)/s. As mentioned
in the solution to Problem 3.21, the reason why the capacitance in Eq. (264) is larger
than C = ϵ0 (N A)/s is because we effectively have 2N − 1 identical area-A capacitors
of alternating orientation lined up next to each other, instead of N area-A capacitors
(with the same orientation).
So the capacitance becomes C ≈ 2πϵ0 bL/(a − b). But 2πbL is the area A of the inner
cylinder, and a − b is the separation s between the cylinders. So the capacitance can
be written as C = ϵ0 A/s, which agrees with the standard result for the parallel-plate
capacitor.
(a) Let Q1 and Q3 be the final charges on the inner and outer shells, respectively.
The outward-pointing field between the inner and middle shells is due only to
the inner shell, and it equals (ignoring the 1/4πϵ0 since it will cancel) Q1 /r2 . So
the potential difference between the inner and middle shells is Q1 (1/R − 1/2R),
with the inner shell at the higher potential.
83
If the inner and outer shells are at the same potential, then Q1 (1/R − 1/2R)
must also be the potential difference between the outer and middle shells, with
the outer shell at the higher potential. The field between the middle and outer
shells must therefore point inward. This field is due to the inner two shells, so
it points inward with magnitude (Q − Q1 )/r2 , given that −Q is the charge on
the middle shell. Note that Q must be larger than Q1 . The potential difference
between the outer two shells is then (Q − Q1 )(1/2R − 1/3R), with the outer shell
at the higher potential.
Equating the inner-middle and outer-middle potential differences gives
( ) ( )
1 1 1 1 Q1 Q − Q1 Q
Q1 − = (Q − Q1 ) − =⇒ = =⇒ Q1 = .
R 2R 2R 3R 2 6 4
(267)
And then Q3 = 3Q/4, to make the total charge on the inner and outer shells be
equal to Q.
(b) The potential difference between the inner and middle shells, which is the same
as the difference between the outer and middle shells, is (bringing the 1/4πϵ0
back in, and using Q1 = Q/4)
( )
Q1 1 1 Q
ϕ= − = . (268)
4πϵ0 R 2R 32πϵ0 R
Therefore Q = (32πϵ0 R)ϕ, so the capacitance is 32πϵ0 R.
(c) Note that Q3 didn’t appear anywhere in the calculation in part (a). It can
therefore take on any value, and the inner-middle and outer-middle potential
differences will still be equal, provided that Q1 = Q/4. So if we add charge q
to the outer shell, it will simply stay there, uniformly distributed on the outside
surface of the shell. It will raise the potential everywhere inside by q/4πϵ0 (3R),
but since this change is uniform inside, all differences remain the same. If any
charge flowed across the wire from the outer shell to the inner shell, the final
charge Q1 on the inner shell would violate the Q1 = Q/4 result we found above
(because the middle shell is now isolated, so its charge of −Q doesn’t change).
1.0
84 CHAPTER 3. ELECTRIC FIELDS AROUND CONDUCTORS
0.5
0.0 a A plot of C/C0 is shown in Fig. 81. (The expression for the capacitance given in the
0 1 2 3 4
statement of the problem is valid only for a prolate spheroid, that is, one with a > 1.)
Figure 81 As promised, C is larger than C0 .
In the limit where a is very large (so we have a long stick-like object), you can show with
a Taylor series that C/C0 ≈ 2a/ ln(4a3 ) ≈ 2a/(3 ln a), to leading order. This grows
with a, so a long spheroid-shaped stick can have a capacitance much greater than that
of a sphere of equal volume. If you want to write things more generally in terms of
both a and b, the capacitance given in the problem equals C ≈ 4πϵ0 a/ ln(2a/b) in the
a ≫ b limit. As an example, consider a spheroid with a = 1 km and b = 1 mm. Its
volume is that of a sphere with radius (ab2 )1/3 = 0.1 m, but its capacitance is that of a
sphere with radius r = 69 meters, because C ≈ 4πϵ0 (103 m)/ ln(2 · 106 ) = 4πϵ0 (69 m).
a
b
h 3.62. Deriving C for a spheroid
stick a We’ll assume here the validity of the result from Exercise 2.44, namely that the po-
tential due to a stick with uniform charge density is constant over an ellipsoid that
has the ends of the stick as its foci. Such an ellipse is shown in Fig. 82. The axes
have lengths 2a√and 2b, and the stick has length 2h. From the properties of an ellipse,
Figure 82 we have a = h2 + b2 . To find the (constant) potential over this ellipse, we may
conveniently pick points at the ends of either axis. An end of the major axis yields a
slightly simpler integral (you can check that an end of the minor axis yields the same
result). We have
∫ h ( ) ( √ )
1 λ dy λ a+h λ a + a2 − b2
ϕ= = ln = ln √
4πϵ0 −h a − y 4πϵ0 a−h 4πϵ0 a − a2 − b2
( )
λ 1+ϵ
= ln , (271)
4πϵ0 1−ϵ
where √
b2
ϵ≡ 1− . (272)
a2
The charge contained within the ellipsoid is the charge on the stick,
√
Q = (2h)λ = 2λ a2 − b2 = 2λaϵ. (273)
From the reasoning near the end of Section 3.4, in the region of space exterior to the
conducting ellipsoid, the field due to the ellipsoid with charge Q is identical to the field
due to the uniform stick with charge Q. This is true because the latter field satisfies
the boundary conditions for the ellipsoid (the field is perpendicular to the surface, and
the total flux is Q/ϵ0 ), so the uniqueness theorem tells us that this solution must be
the solution for the ellipsoid.
Using Q = 2λaϵ in the Q = Cϕ relation for the ellipsoid gives
Q 2λaϵ 8πϵ aϵ
C= = ( ) = ( 0 ), (274)
ϕ λ 1+ϵ 1+ϵ
ln ln
4πϵ0 1−ϵ 1−ϵ
as desired.
3.63. Capacitance coefficients for shells
We can write in general,
Q1 = C11 ϕ1 + C12 ϕ2 ,
Q2 = C21 ϕ1 + C22 ϕ2 . (275)
85
With charge Q1 on the inner shell, the field between the shells equals Q1 /4πϵ0 r2 , so
the potential difference is
∫ a ∫ a ( )
Q1 dr Q1 1 1
ϕ2 − ϕ1 = − E dr = − 2
= − . (276)
b b 4πϵ0 r 4πϵ0 a b
Hence,
4πϵ0 ab
Q1 = (ϕ1 − ϕ2 ). (277)
a−b
Comparing this with Eq. (275) gives
4πϵ0 ab 4πϵ0 ab
C11 = and C12 = − . (278)
a−b a−b
In the region external to the outer shell of radius a, both shells look like point charges.
So the potential at radius a is simply ϕ2 = (Q1 + Q2 )/4πϵ0 a, which yields Q2 =
4πϵ0 aϕ2 − Q1 . Using the Q1 from Eq. (277) allows us to write Q2 in terms of ϕ1 and
ϕ2 :
4πϵ0 ab 4πϵ0 ab 4πϵ0 a2
Q2 = 4πϵ0 aϕ2 − (ϕ1 − ϕ2 ) = − ϕ1 + ϕ2 . (279)
a−b a−b a−b
Comparing this with Eq. (275) gives
4πϵ0 ab 4πϵ0 a2
C21 = − and C22 = . (280)
a−b a−b
As expected, C12 = C21 . In the event that we have a standard capacitor with charge
Q on the inner shell and −Q on the outer, the field is zero outside the outer shell. So
ϕ2 = 0, and ϕ1 equals the ∆ϕ between the shells. Both of the equations in Eq. (275)
then reduce to Q = [4πϵ0 ab/(a − b)]∆ϕ, which agrees with the result in Eq. (3.18) for
a two-sphere capacitor.
Q1 = C11 ϕ1 + C12 ϕ2 ,
Q2 = C21 ϕ1 + C22 ϕ2 . (281)
which is the statement that capacitances in parallel simply add. Equating the right-
hand sides of Eqs. (287) and (288) gives
70
C1 (100 V) = (C1 + C2 )(30 V) =⇒ C2 = C1 · = 233 pF. (289)
30
87
Solving for λ and plugging the result into Eq. (292) gives
( )2
2πϵ0 ϕ ℓ πϵ0 ℓϕ2
U = ln(r2 /r1 ) =
ln(r2 /r1 ) 4πϵ0 ln(r2 /r1 )
( −12 s2 C2
)
π 8.85 · 10 kg m3 (0.3 m)(45 V)
2
= = 5.9 · 10−8 J. (294)
ln(4/3)
Alternatively: We can solve the problem using capacitance. Using the value of ϕ
we found above, the capacitance of the tubes is given by
Q λℓ 2πϵ0 ℓ
C= = = , (295)
ϕ (λ/2πϵ0 ) ln(r2 /r1 ) ln(r2 /r1 )
which is independent of λ, as it should be. (The capacitance depends only on the
geometry of the system, and not on the charge that is placed on the conductors.) The
energy stored is then
( )
1 2 1 2πϵ0 ℓ πϵ0 ℓϕ2
Cϕ = ϕ2 = , (296)
2 2 ln(r2 /r1 ) ln(r2 /r1 )
in agreement with the first solution.
88 CHAPTER 3. ELECTRIC FIELDS AROUND CONDUCTORS
(a) The initial energy stored in the capacitor is ϵ0 E 2 /2 times the volume As. The
field is E = σ/ϵ0 , so the initial energy is
( )2
ϵ0 σ σ 2 As
Ui = (As) = . (309)
2 ϵ0 2ϵ0
This is larger than the initial energy, in contrast with the situation in part (a).
So if there weren’t anything else going on, we would have a violation of energy
conservation. But there is indeed something else going on; the battery is doing
work. It must dump more charge onto the plates to increase the density from σ
to σ ′ = 2σ. It does this by moving charges from the negative plate to the positive
plate, through the constant potential difference ϕ = σs/ϵ0 . The amount of charge
moved is q = (σ ′ − σ)A = σA, so the work done is W = qϕ = (σA)(σs/ϵ0 ) =
91
σ 2 As/ϵ0 . Conservation of energy therefore gives the final kinetic energy of the
slab as
σ 2 As σ 2 As σ 2 As σ 2 As
W + Ui = Uf + K =⇒ + = + K =⇒ K = . (312)
ϵ0 2ϵ0 ϵ0 2ϵ0
Basically, of the σ 2 As/ϵ0 work done, half goes into increasing U , and half goes
into K. It makes sense that the K here is larger than the K in part (a), because
the present case ends up with more charge on the plates, so the forces involved
are larger.
The voltage difference between the sheets is V = Es = Qs/2ϵ0 yb, so the capacitance
is C = Q/V = 2ϵ0 yb/s. (This is just the standard parallel-plate expression C = ϵ0 A/s
with area A = 2yb.) Equation 3.32 then gives1
( ) ( )
Q2 d 1 Q2 d s Q2 s
F = = =− . (314)
2 dy C 2 dy 2ϵ0 yb 4ϵ0 by 2
The negative sign indicates that the energy of the capacitor decreases as y increases.
So the direction of the force on A is downward, because the decrease in energy will
show up as kinetic energy, or work done on some other object. (Equivalently, the F in
Eq. (3.32) was defined as the force that some other object must apply to A to keep it
at rest. This force is upward, in the direction of decreasing y; hence the negative sign.)
We can write y in terms of V via the above relation, V = Qs/2ϵ0 yb =⇒ y = Qs/2ϵ0 V b.
Hence, the magnitude of the force is
( )2 √
Q2 s 2ϵ0 V b ϵ0 V 2 b Fs
F = = =⇒ V = . (315)
4ϵ0 b Qs s ϵ0 b
Note that the force F is independent of y. This is consistent with the discussion at
the end of the solution to Problem 3.26. Although we (justifiably) ignored the edge
effects in computing the capacitance, it is in fact these edge effects that produce the
force. These edge effects depend on the charge density on the plates, and for a given
voltage V , this density is independent of y.
Alternatively: We can find the force by directly calculating how the stored energy
changes with y. The stored energy is
1 Q2 s
U= QV = . (316)
2 4ϵ0 yb
1 The parameter y need not represent the distance between the plates, for Eq. (3.32) to hold. That
equation is valid for any y describing the relative position of the conductors in a capacitor; it gives the force
in the direction corresponding to the parameter y.
92 CHAPTER 3. ELECTRIC FIELDS AROUND CONDUCTORS
Q2 s dy
dU = − . (318)
4ϵ0 b y 2
If A is attached to some other object D, this decrease in energy equals the increase
in the energy of D, due to the work F dy that sheet A does on D. Therefore, F =
Q2 s/4ϵ0 by 2 . (We’re now defining F to be the force that A applies to D, instead of the
other way around, as it is defined in Eq. (3.32).)
Since the sheets are isolated, Q remains constant as y increases, whereas V does not.
On the other hand, if the sheets are connected to a constant voltage source, Q will
increase with increasing y, and U will also increase. But in this case the voltage source
will supply energy for both the increase in U and the external work. The expressions
for the force F (given in Eqs. (314) and (315)) will be exactly the same; so F is now
constant. See the solution to Problem 3.26 for a more complete discussion of this
point.
3.73. Force on a coaxial capacitor
We’ll need to know the capacitance of the cylinders. Let ℓ be the distance of overlap.
And let ±λ = ±Q/ℓ be the charge densities per unit length on the overlap region of
the cylinders. The (magnitude of the) potential difference ϕ between the cylinders is
∫ r2 ∫ r2 ( )
λ dr λ r2
ϕ= E dr = = ln . (319)
r1 r1 2πϵ 0 r 2πϵ0 r1
If r1 < r2 (so the point is closer to the positive line) then ϕ is positive, which is
correct. We see that the potential is constant on a curve for which r2 /r1 = k, for
some constant k. So our goal is to show that the curve defined by r2 /r1 = k is a
circle. Since r12 = (x − x0 )2 + y 2 and r22 = (x + x0 )2 + y 2 , we can rewrite the relation
r22 = k 2 r12 as
[ ]
(x + x0 )2 + y 2 = k 2 (x − x0 )2 + y 2
=⇒ (k 2 − 1)(x2 + y 2 + x20 ) − 2(k 2 + 1)x0 x = 0
( )
k2 + 1
=⇒ x −2 2
2
x0 x + y 2 = −x20 . (325)
k −1
The fact that the coefficients of x2 and y 2 here are the same means that the curve is
a circle. But let’s finish the calculation anyway. Letting b ≡ (k 2 + 1)x0 /(k 2 − 1) and
completing the square yields
its center on the y axis.) There are various ways to show this. One is to take the
differential of each circle equation to find the slopes of the curves, and to then show
that these slopes are the negative reciprocals of each other, by using the difference of
the circle equations, namely bx − hy = x20 .
3.75. Average of six points
The Taylor expansions are
∂ϕ δ 2 ∂ 2 ϕ δ 3 ∂ 3 ϕ
ϕ(x0 + δ, y0 , z0 ) = ϕ(x0 , y0 , z0 ) + δ + + + ··· ,
∂x 2! ∂x2 3! ∂x3
2 2 3 3
∂ϕ δ ∂ ϕ δ ∂ ϕ
ϕ(x0 − δ, y0 , z0 ) = ϕ(x0 , y0 , z0 ) − δ + − + ··· , (327)
∂x 2! ∂x2 3! ∂x3
and likewise for the y0 ± δ and z0 ± δ points. When we add up all six terms and divide
by 6 to take the average, the terms with odd powers of δ cancel in pairs, and we are
left with
[ ( 2 ) ]
1 ∂ ϕ ∂2ϕ ∂2ϕ
ϕavg = 6ϕ(x0 , y0 , z0 ) + δ 2 + + + O(δ 4
) . (328)
6 ∂x2 ∂y 2 ∂z 2
φ = 100
50
25
φ=0
Figure 84
The results for the “triangle” of entries analogous to those in Exercise 3.76 are:
The bold entries correspond to the seven entries in Exercise 3.76. The agreement is
reasonably good. If the middle box is 48 × 48 instead of the above 6 × 6 or the 3 × 3
we had in Exercise 3.76, then, for example, the 36.091 entry becomes 35.2961. By
96 CHAPTER 3. ELECTRIC FIELDS AROUND CONDUCTORS
looking at how this number changes with the size of the box, it appears to converge
to approximately 35.2 in the continuum limit involving an infinite number of lattice
points.
Interestingly, the computing time doesn’t appear to be helped much by our choice
of initial equipotentials that varied linearly from the outer boundary to the central
square. If we had instead picked ϕ = 0 on the outer boundary and ϕ = 100 at every
other point (all the other points, not just the ones in the middle square), then the
computing time would be only slightly longer. The computing time increases by a
factor of 16 for every doubling of the array’s width, because there are 4 times as many
points that each iteration needs to run through, and also it turns out that we need to
do 4 times as many iterations to achieve a given accuracy.
Chapter 4
Electric currents
Solutions manual for Electricity and Magnetism, 3rd edition, E. Purcell, D. Morin.
morin@physics.harvard.edu (Version 1, January 2013)
(1.25 · 106 s−1 )N e = (1.25 · 106 s−1 )(1011 )(1.6 · 10−19 C) = 0.02 A. (330)
(a) Let x be the distance from the left plate, and let Qr be the charge on the right
plate. From Exercise 3.37 we have Qr = −(2e)x/ℓ, where ℓ = 2 mm is the distance
between the plates. The current flowing out of the right plate is therefore
97
I (10-10 A)
98 CHAPTER 4. ELECTRIC CURRENTS
2
This current lasts for a time t = ℓ/v = (0.002 m)/(106 m/s) = 2 · 10−9 s, which
1 is 2 nanoseconds. The current is constant during this time, so we have the bold
line shown in Fig. 86. The total charge that flows during this time is It, which
equals 2e as expected.
1 2 3 If the path slopes upward at 45◦ , then dx/dt = v cos 45 ◦
√ . From above, the current
t (10-9 s) pulse is therefore reduced in amplitude by a factor 1/ 2 and stretched out in time
√
by a factor 2; see the dotted line in Fig. 86. Again the total charge transferred
Figure 86 is It = 2e.
(b) Following the strategy of the solution to Exercise 3.37, we know that if Q1 and Q2
are the charges on the inner and outer electrodes (with radii a and b, respectively),
then Q1 + Q2 = −2e. How is the charge of −2e distributed between Q1 and Q2
when the alpha particle is at radius r? As in Exercise 3.37, the key points are
that (1) we can smear out the alpha particle into a cylinder of charge, and (2) the
potentials of the two electrodes are the same, which means that the line integrals
of the electric field from radius r to the two electrodes must be equal. The field
inside radius r is proportional to Q1 /r (this points inward since Q1 is negative),
and the field outside radius r is proportional to (2e + Q1 )/r = −Q2 /r (this points
outward since Q2 is negative). Equating the two line integrals gives (note that
both sides of the following equation are positive since dr is negative in the left
integral)
∫ ∫
a
Q1 b
−Q2
dr = dr =⇒ Q1 ln(a/r) = −Q2 ln(b/r)
r r r r
=⇒ Q1 ln(r/a) = Q2 ln(b/r). (334)
Combining this equation with Q1 + Q2 = −2e and solving for Q1 and Q2 gives
where we have used r = a + vt. We see that I(t) is not constant. A plot of
the general shape of I(t) is shown in Fig. 87 (with b chosen to equal 4a). For a
t given value of b, if a is very small then the current starts out very large, because
at t = 0 the smallness of a in the denominator in Eq. (336) wins out over the
Figure 87 largeness of ln(b/a).
In the case of a 45◦ angle of the path, the same modifications that applied in
part
√ (a) also apply here. That is, the curve is stretched
√ horizontally by a factor
of 2, and squashed vertically by a factor of 1/ 2.
(b) Our goal is to obtain a rough upper bound on the resistance of the ocean path,
and then see if this is smaller than the answer to part (a). As stated, we’ll take
the electrodes to be spheres with a radius of 0.1 m. (Any factors of order 1 will
be irrelevant.) Let’s imagine the path of the current to be roughly a hemisphere
expanding out from one sphere, then a tube with a large cross section, and then
a hemisphere tapering down to the other sphere. (Even if the end parts are more
conical than hemispherical, we’ll still get the same order of magnitude.) For the
tubular middle part, it doesn’t matter much what cross section we pick, if our
only goal is to obtain an upper bound on the resistance. Even if the cross section
is just a kilometer in diameter, the resistance is (using a length of 3000 km)
ρL (0.25 Ω m)(3 · 106 m)
R= = ≈ 1 Ω, (338)
A π(103 m)2 /4
which is negligible compared with the resistance of the cable.
What is the resistance of the hemispheres? If d ≈ 0.1 m is their radius, then
from dimensional analysis we expect the resistance to be roughly ρ/d, in order
of magnitude. And indeed, from Problem 4.4 the resistance is proportional to
ρ/d, assuming that the other radius involved is large. In the present case, ρ/d
equals (0.25 Ω m)/(0.1 m) = 2.5 Ω. This is negligible compared with the cable’s
resistance. Even if the electrode’s size was on the order of a millimeter, the
hemisphere’s resistance would still only be 250 ohms, which is again negligible
compared with the cable’s resistance.
(a) If we divide the time t into a very large number, N , of equal small intervals, then
the length of each interval is dt = t/N . The probability that an event happens
in a particular one of these small intervals is therefore p dt = p(t/N ). So the
probability that an event happens in none of these N intervals is (1 − pt/N )N .
Multiplying this by the probability p dt that an event does happen in the next dt
interval tells us that the probability of the next event happening between t and
t + dt is (in the N → ∞ limit)
(1 − pt/N )N p dt = e−pt p dt, (339)
as desired. The integral of this probability must be 1, because the next event
must happen at some time. And indeed,
∫ ∞ ∞
e−pt p dt = −e−pt = 1. (340)
0 0
We are free to pick the t = 0 point as the time of an event, so the probability
in Eq. (339) is also the probability that an interval between events has length
between t and t + dt. That is, if we look at a million successive intervals, then
approximately (106 )(e−pt p dt) of them will have length between t and t + dt.
(b) To find the average value (or expectation value) of a quantity, we must multiply
the probability of a value occurring times the value itself, and then integrate over
all the values. Equivalently, we can look at a million waiting times and calculate
their average by adding up all the times and dividing by a million. So the average
waiting time (starting at any given time, not necessarily the time of an event)
until the next event is
∫ ∞ ∞
twait = t · e−pt p dt = −e−pt (t + 1/p) = 1/p. (341)
0 0
100 CHAPTER 4. ELECTRIC CURRENTS
(You should check this integral by differentiating it.) It makes sense that this
time decreases with p; if p is large, then the events happen more frequently, so
the waiting time is shorter.
Since this 1/p result holds for any arbitrary starting time, we are free to choose the
starting time as the time of an event. A special case of this result is therefore the
statement that the average waiting time between events is 1/p. This is consistent
with the fact that pt is the average number of events that occur during a (not
necessarily infinitesimal) time t.
(c) If we pick a random point in time, then the average waiting time until the next
event is 1/p, from part (b). And the average time since the previous event is
also 1/p, because we can use the same reasoning that we used in part (a), going
backward in time, to calculate the probability that the most recent event occurred
at a time between t and t + dt earlier. The direction of time is irrelevant; the
process is completely described by saying that p dt is the probability of an event
happening in an infinitesimal time dt, and this makes no reference to a direction
of time. The average length of the interval surrounding a randomly chosen point
in time is therefore 1/p + 1/p = 2/p.
(d) From part (a), an event-to-event interval with length between t and t + dt occurs
with probability e−pt p dt (in the sense that out of a million successive intervals,
(106 )(e−pt p dt) of them will have this length). But if you pick a random point
in time, e−pt p dt is not the probability that you will end up in an interval with
length between t and t + dt, because you are more likely to end up in an interval
that is longer.
Consider the simple case where there are only two possible lengths of intervals,
1 and 100, and these occur with equal probabilities of 1/2. If you look at 1000
successive intervals, then about 500 will have length 100. But if you pick a random
point in time, you are of course 100 times more likely to end up in one of the
large intervals. The probability of landing in each type of interval is not 1/2. The
probability of landing in an interval of a given length (1/101 and 100/101 in the
present example) does not equal the probability of that given length occurring
in a list of the lengths (1/2 and 1/2 here). In this example, the average time
between events is 50.5, while the average time surrounding a randomly chosen
point in time is, as you can show, 99.02. (These results don’t have anything to do
with the above results involving p, because the present example isn’t a random
process described by a given probability per unit time. But it illustrates the basic
point.)
In short, the probability of landing in an interval with length between t and t + dt
is proportional both to e−pt p dt (because the more intervals there are of a certain
length, the more likely you are to land in one of them), and to the length t of the
intervals (because the longer they are, the more likely you are to land in one of
them).
(e) Consider a large number N of intervals. The number of intervals with length
between t and t + dt is N (e−pt p dt). The total length of these intervals with
length between t and t + dt is therefore N (e−pt p dt)t. The total length of all of
the N intervals is the integral of this, which you can quickly show equals N/p,
as it should.
The probability of picking a point in time that lands in one of the intervals
with length between t and t + dt equals the total length associated with these
intervals, divided by the total length of all of the N intervals, which gives
(N e−pt pt dt)/(N/p) = e−pt p2 t dt. As mentioned in part (d), this probability
101
is proportional to both e−pt p dt and t. The expectation value of the length of the
interval that the given point lands in is obtained by multiplying this probability
by the interval length t and integrating. This gives
∫ ∞
e−pt ( )∞ 2
e−pt p2 t2 dt = − 2 + 2pt + p2 t2 = , (342)
0 p 0 p
as desired. (Again, you should check this integral by differentiating it.) To sum
up, there are two different probabilities in this problem: (1) the probability that
a randomly chosen interval has length between t and t + dt (this equals e−pt p dt),
and (2) the probability that a randomly chosen point in time falls in an interval
with length between t and t + dt (this equals e−pt p2 t dt). In the first case, by
“randomly” we mean that we label each interval with a number and then pick a
random number. The length of each interval is irrelevant in this case, whereas it
is quite relevant in the second case.
I V /R V 1 V
J= = = = . (345)
A A A ρL/A ρL
The drift velocity is therefore (the factor of 2 in the N here comes from the fact that
there are both Na+ and Cl− ions)
J V 12 V
v = = =
Ne ρLN e (0.25 Ω m)(2 m)(2 · 3 · 1026 m−3 )(1.6 · 10−19 C)
= 2.5 · 10−7 m/s. (346)
to n-type material, and the diffusion of electrons from the n-type to p-type material.
But for the purposes of this exercise, it isn’t critical which is which.
Poisson’s equation, ∇2 ϕ = −ρ/ϵ0 , tells us that
d2 ϕ ρ ρ 2
=− =⇒ ϕ = A + Bx − x . (347)
dx2 ϵ0 2ϵ0
This can also be written in the alternative general form, ϕ = −(ρ/2ϵ0 )(x + C)2 + D.
We see that ϕ varies quadratically with x if ρ is constant. The curvature is positive
in the region where ρ is negative, and negative in the region where ρ is positive.
We are told that the slope of ϕ is zero outside the charge layers. It must therefore
also be zero just inside the boundaries. If there were a discontinuity in the slope of ϕ,
then the second derivative would be infinite. Poisson’s equation would then imply an
infinite ρ, such as that arising from a surface charge. But there are no surface charges
in this setup.
The zero slope at the boundaries implies that ϕ must look like the curve shown in
Fig. 88. Let us define ϕ = 0 at the left boundary of the left slab, and let x = 0 be the
location of the midplane. Let ℓ ≡ 10−4 m. The density is −ρ in the left (p-type) region,
and +ρ in the right (n-type) region. In the left region we have half of a rightside-up
parabola centered at x = −ℓ, and in the right region we have half of an upside-down
parabola centered at x = ℓ. So if ∓ρ are the charge densities in the two regions, we
quickly see that the potential ϕ in Eq. (347) (or rather the alternate form listed right
after) must take the forms,
0.3 V
ρ
-l
x
l
-ρ
(p-type material) (n-type material)
Figure 88
Note that this is twice the average field in the slabs, which is −(0.3 V)/(2ℓ).
E − (I1 − I2 )R1 = 0,
−(I2 − I1 )R1 − I2 R2 = 0. (352)
Adding these two equations quickly gives I2 = E/R2 (which corresponds to the loop
around the whole circuit). Either equation then gives
E(R1 + R2 ) E R1 R2
I1 = ≡ , where Reff ≡ . (353)
R1 R2 Reff R1 + R2
The current through the battery is I1 , so E = I1 Reff tells us that the effective resistance
is Reff .
resistor in parallel with the series combination of the other two resistors. So the the
resistances in the second box are
34(170 + 85)
Rab = = 30,
289
85(34 + 170)
Rac = = 60,
289
170(85 + 34)
Rbc = = 70, (355)
b 289
as desired.
10
For the two configurations pictured, the above resistances are the only possibilities that
lead to the given resistances between the terminals. (Each configuration yields three
equations in the three resistances, so they are uniquely determined.) However, there
30 30 are other configurations that also work, for example, the one pictured in Fig. 89 (as
0 40 you can check).
30 The potential assumed by a free terminal when the potentials at the other two termi-
a c nals are fixed is the same for the two boxes. For example, if the potentials at b and c
in the first box are fixed at ϕb and ϕc , then the potential at a divides the difference
Figure 89
ϕb − ϕc in the ratio of 20 to 50. And in the second box the ratio is 34 to 85, which
is the same. The two boxes are therefore indistinguishable by external measurements
(using direct currents). You can show that the configuration in Fig.89 also yields the
same ratio of 2 to 5.
(b-a)(x/l)
4.32. Tapered rod
(b-a)
Let the length of the rods be ℓ. Then the resistance of the cylindrical rod is ρL/A =
x ρℓ/πa2 .
Now consider the tapered cone. The resistance of a cross-sectional disk is ρL/A =
2a dx 2b ρ dx/πr2 , where r = a + (b − a)(x/ℓ) from similar triangles in Fig. 90. So dx =
dr ℓ/(b − a), and we have
∫ ℓ ∫ b ( )
ρ dx ρ ℓ dr ρ ℓ 1 1 ρℓ
R= = = − = . (356)
0 πr
2 πb−a a r 2 πb−a a b πab
l This is a/b times the resistance of the cylindrical rod, as desired. If b < a (or b > a)
then this fraction is larger (or smaller) than 1, which makes sense. If b → 0 then
Figure 90
R → ∞. Note √ that the conical rod has the same resistance as a cylindrical rod
with radius ab. See Problem 4.6 for a discussion of the approximate nature of this
solution.
As a check on this result, consider three objects, all with the same length: a cylinder
with radius a, a tapered cone with radii a and b, and another cylinder with radius
b. For concreteness, assume a < b, as in Fig. 90. Then Eq. (356) tells us that the
resistance of the second of these objects is a/b times that of the first, and also that the
resistance of the third is a/b times that of the second. So the first and third resistances
are in the ratio of b2 to a2 , or equivalently 1/a2 to 1/b2 . This is correct, because the
resistances of the two cylinders are inversely proportional to their cross-sectional areas,
which are proportional to length squared.
4.33. Laminated conductor extremum
In the solution to Problem 4.5, if we replace the number 7.2 with n, and the fraction
1/3 with f , then we can repeat the same reasoning and derive the general result,
σ⊥ 1
=[ ][ ]. (357)
σ∥ f + n(1 − f ) f + (1/n)(1 − f )
105
(Or you can just look at Eqs. (12.204) and (12.206) and make the above replacements.)
When the denominator is multiplied out, it equals
(a) For a given n, taking the derivative of the denominator given in Eq. (358), with
respect to f , shows that it achieves a maximum when f = 1/2, which means
that σ⊥ /σ∥ achieves a minimum when f = 1/2, that is, when the two thicknesses
are equal. This result is independent of n. So no matter what the ratio of
conductivities is, σ⊥ /σ∥ is minimum when the materials have the same thickness.
The minimum value of σ⊥ /σ∥ is quickly found to be 4/(2 + n + 1/n).
It makes sense that there should be an extremum of σ⊥ /σ∥ for an intermediate
value of f (between 0 and 1), because if f = 0 or f = 1, the material consists of
only one substance, so σ⊥ /σ∥ = 1. Therefore, unless σ⊥ /σ∥ is a flat curve (which
it undoubtedly isn’t), it must reach a maximum or minimum for some f between
0 and 1.
(b) For a given f (between 0 and 1), taking the derivative of the denominator given in
Eq. (358), with respect to n, shows that it achieves a minimum when n = 1, which
means that σ⊥ /σ∥ achieves a maximum when n = 1. This result is independent
of f . So no matter what the ratio of thicknesses is, σ⊥ /σ∥ is maximum when the
materials have the same σ. And the maximum value of σ⊥ /σ∥ is 1, of course,
because the two materials are the same.
It makes sense that there should be an extremum of σ⊥ /σ∥ for an intermediate
value of n (between 0 and ∞), because setting one of the σ’s equal to zero makes
σ⊥ (but not σ∥ ) equal to zero; and setting one of the σ’s equal to infinity makes
σ∥ (but not σ⊥ ) equal to infinity. So σ⊥ /σ∥ is zero at both extremes. Therefore,
unless σ⊥ /σ∥ is a flat curve (which, again, it undoubtedly isn’t), it must reach a
maximum or minimum for some n between 0 and ∞.
b B
a
b
a b
A a
3 a's 3 b's
A B
(d) The 1-D lattice has n = 2, so Reff = 2/2 = 1 Ω. This makes sense, because
there is only one path between two adjacent nodes on a line, namely across the
1 Ω resistor connecting them. All the other resistors outside the two nodes are
irrelevant.
(a) In Fig. 91 the three vertices adjacent to A (which are labeled as “a”) are all
at the same potential (by symmetry under rotations around the AB diagonal),
so we can collapse them to one point. (Equivalently, if we connect them with
resistance-less wires, no current will flow in these wires.) Likewise for the three
vertices adjacent to B (which are labeled as “b”). So the circuit is equivalent to
the second setup shown in Fig. 91 (the number of lines is still 12), which can be
simplified as indicated. The equivalent resistance is therefore 5R/6.
Alternatively, we can work in terms of currents. The input current I0 gets divided
evenly, by symmetry, into three I0 /3 currents. It then divides into six I0 /6
currents, and then converges to three I0 /3 currents. The total potential drop
across any of the possible paths from A to B is given by V = (I0 /3)R +(I0 /6)R +
(I0 /3)R = (5/6)I0 R. The effective resistance is then V /I0 = 5R/6.
(b) In Fig. 92 there are four vertices (labeled as “c”) that lie in the plane that is
equidistant from A and B. These vertices are all at the same potential (halfway
between VA and VB ), so we can collapse them to a point. (In the second setup
shown, there are only 10 lines because 2 of the original 12 lines were collapsed).
The circuit can then be simplified as shown, and the equivalent resistance is 3R/4.
(c) From symmetry, the two points marked as a in Fig. 93 are at the same potential,
so we can collapse them to a point. Likewise for the two b’s. The circuit can then
be simplified as shown, and the equivalent resistance is 7R/12. As expected, this
is smaller than the answer to part (b), which in turn is smaller than the answer
to part (a).
107
c B
A 4 c's B A B
A 1/2 1/2
c
1 1
c b 1/2 1/2
a b
a c
A B
1/2 1/2 3/8 3/8
A B
3/2 3/2
Figure 92
1
A B A B
A B
1/2 1/2
a
b 1/2
2 a's 2 b's
a b 1/2 1/2
c d
c d
1
1 1 1 7/12
A B A B A B A B
7/5
1/2 1/2 1/2 1/2
1/2 2/5
2
Figure 93
108 CHAPTER 4. ELECTRIC CURRENTS
Note that the sum of the effective resistances across all 12 resistors is 12(7R/12) =
R1 7R = (8 − 1)R, where the 8 here is the number of corners in the cube. This is a
special case of the general result in Problem 4.9.
From Eqs. (359) and (360), we see that if we want V /V ′ ≈ 1 (that is, the voltage
hardly decreases), then we need R ≪ R2 , which implies R1 ≪ R2 . On the other hand,
if we want V /V ′ ≪ 1 (that is, the voltage decreases quickly), then we need R ≫ R2 ,
which implies R1 ≫ R2 . These results make intuitive sense.
To terminate the ladder after any section, without changing its resistance from that
of the infinite chain, we can simply connect a single resistor R given by Eq. (359) in
parallel with the last R2 , because this R mimics the rest of the infinite chain. For
example, after one step of the ladder, we would have the setup in Fig. 94.
109 (a)
A C
4.37. Some golden ratios
(a) For simplicity, let’s set R = 1. Let the desired resistance be r. Following the
strategy from Exercise 4.36, the circuit to the right of C and D in Fig. 96(a) is
identical to the original infinite chain. So the infinite chain consists of a resistance B D
of 1 connected in series with the parallel combination of 1 and r. The net result
is r, so
(b)
1·r 1 + 2r
1+ =r =⇒ = r =⇒ r2 − r − 1 = 0 A C
1+r 1+r
√
1+ 5
=⇒ r= = 1.618 ≡ r1 . (362)
2
This number is the golden ratio. B D
(b) Again set R = 1, and let the desired resistance be r. The circuit to the right of C
and D in Fig. 96(b) is identical to the original infinite chain. So the infinite chain Figure 96
consists of a resistance of 1 connected in parallel with the series combination of
1 and r. The net result is r, so
1 · (1 + r)
=r =⇒ r2 + r − 1 = 0
1 + (1 + r)
√
−1 + 5
=⇒ r= = 0.618 ≡ r2 . (363)
2
This number is the inverse of the golden ratio. You should convince yourself why
this setup is actually the same as the setup in Problem 4.7(b).
As a double check, the only difference between the two given circuits is the extra
vertical resistor connecting A to B in the second circuit. So r2 should equal the
parallel combination of 1 and r1 . And indeed,
1 · r1
= r2 , (364)
1 + r1
as you can verify. This works out due to the various properties of the golden
ratio.
(a) The power dissipated takes the form of V 2 /R. Both bulbs have the same voltage
drop V , so if Bulb 1 is twice as bright as Bulb 2, it must have half the R. Bulb
2’s resistance is therefore larger by a factor of 2. (The larger resistor is dimmer.)
(b) The power dissipated also takes the form of I 2 R. Both bulbs now have the same
current I, so if Bulb 2 has twice the resistance, as we found in part (a), then it
is twice as bright – the opposite of the case in part (a). (The larger resistor is
brighter.) Note that in part (a) we used the expression P = V 2 /R because both
bulbs (in parallel) had the same V , whereas now we are using the expression
P = I 2 R because both bulbs (in series) have the same I.
We can also compare the total power dissipated in each case. If the resistances are
R and 2R, then in part (a) the total power dissipated is V 2 /R+V 2 /2R = 3V 2 /2R.
In part (b) the total power is I 2 R + I 2 (2R) = 3I 2 R, where I = V /3R. So the
power is V 2 /3R. This is 2/9 of the power in part (a). In units of V 2 /R, the
powers in part (a) are 1 and 1/2, while in part (b) they are 1/9 and 2/9.
110 CHAPTER 4. ELECTRIC CURRENTS
(R + Ri )2 · 1 − R · 2(R + Ri ) Ri − R
0= 4
= =⇒ R = Ri . (365)
(R + Ri ) (R + Ri )3
This is indeed a maximum, because dP/dR > 0 for R < Ri , and dP/dR < 0 for
R > Ri . Equivalently, the second derivative is negative at R = Ri , as you can check.
It makes sense that a maximum exists for some finite value of R, because P = 0
both at R = 0 (because P = I 2 R, with I finite and R zero) and at R = ∞ (because
P = V 2 /R, with V finite and R infinite).
Consider a different question, “Given a fixed external resistance R, what value of the
internal resistance Ri yields the maximum power delivered to the external resistor
R?” In view of the above expression for the power, the answer is simply Ri = 0. This
makes sense; we want the largest possible current passing through the given external
resistor.
dP I0 R2
0= = 2I1 (R1 + R2 ) − 2I0 R2 =⇒ I1 = , (367)
dI1 R1 + R2
which is also what we obtain from Ohm’s law (that is, equating the voltage drops
I1 R1 and I2 R2 , and using I1 + I2 = I0 ). Note that P is indeed a minimum, and not a
maximum, because dP/dI1 is less than (or greater than) 0 if I1 is less than (or greater
than) I0 R2 /(R1 + R2 ). Equivalently, the second derivative of P equals 2(R1 + R2 ),
which is positive.
Alternatively, we can set the differential of P = I12 R1 + I22 R2 equal to zero, which
gives dP = 2R1 I1 dI1 + 2R2 I2 dI2 = 0. But I1 + I2 = I0 tells us that dI2 = −dI1 , so
we obtain R1 I1 = R2 I2 . This is simply the statement of equal voltage drops, as given
by Ohm’s law. Combining this with I1 + I2 = I0 yields the above value of I1 .
4.41. D-cell
(a) The total charge produced by 0.1 A flowing for 30 hours is (0.1 C/s)(30 · 3600 s) =
10, 800 C. This charge passes through a potential difference of 1.5 V, so the
total energy output is E = (10, 800 C)(1.5 J/C) = 16, 200 J. Since the mass
of the battery is 0.09 kg, the energy storage in J/kg is (16, 200 J)/(0.09 kg) =
1.8 · 105 J/kg.
In the example in Section 4.9, the 10 kg battery had an energy output of 8.6·105 J,
which implies 8.6 · 104 J/kg. This is about half as much as the D cell.
(b) Lifting a 70 kg person 1 m requires an energy of mgh = (70 kg)(9.8 m/s2 )(1 m) ≈
700 J. The D cell with 50% efficiency can supply 8,100 J. This corresponds to
about 11.5 m.
111
R1 R2 R3 + R1 R2 R4 + R1 R3 R4 + R2 R3 R4 + R5 (R1 R3 + R2 R3 + R1 R4 + R2 R4 )
Req = .
R1 R2 + R1 R4 + R2 R3 + R3 R4 + R5 (R1 + R2 + R3 + R4 )
(368)
Figure 97
(a) If R5 = 0 we equivalently have the circuit shown in Fig. 98(a), where we have
collapsed R5 to a point since it is short circuited. The parallel combination of R1
and R3 is in series with the parallel combination of R2 and R4 . So the resistance
is
R1 R3 R2 R4 R1 R2 R3 + R1 R2 R4 + R1 R3 R4 + R2 R3 R4
+ = , (369)
R1 + R3 R2 + R4 R1 R2 + R1 R4 + R2 R3 + R3 R4
112 CHAPTER 4. ELECTRIC CURRENTS
R3 R1 R3 R1 R4 R2 R R
R4 R2 R4 R2 R R
Figure 98
according to Q(t) = Q0 e−t/t0 , where t0 = RC is the time constant. Since the charge
of an electron is 1.6·10−19 C, we will have roughly one electron left when 1.6·10−19 C =
(10−4 C)e−t/t0 =⇒ t = −t0 ln(1.6 · 10−15 ) = 34t0 . So if the time constant were, say, 1
second (which would mean R = 106 Ω here), we would be down to roughly one electron
in a little over half a minute. For a 1 kΩ resistor, the time would be 0.034 s.
4.48. Charging a capacitor
The total work done by the battery is Qf E, where Qf is the final charge on the
capacitor. This is true because the battery transfers a charge Qf through a constant
potential difference of E.
The final energy of the capacitor is Qf ϕ/2 = Qf E/2, because the final potential ϕ
across the capacitor equals the voltage E across the battery. (There is no current
flowing after a long time, so there is no voltage drop across the resistor.)
∫
The energy dissipated in the resistor is the integral of the power, that is, RI 2 dt.
From the solution to Problem 4.17, we have I(t) = (E/R)e−t/RC . Therefore,
∫ ∞ ∫ ∞
E 2 ∞ −2t/RC E 2 RC −2t/RC E 2 RC CE 2 Qf E
2
RI dt = R 2 e dt = − e = = = ,
0 R 0 R 2 0 R 2 2 2
(375)
where we have used Qf = CE. The conservation-of-energy statement is then
Qf E Qf E
Wbattery = Ucapacitor + Eresistor =⇒ Qf E = + , (376)
2 2
which is indeed true.
Remark: It is also possible to use the general formulas for Q(t) and I(t) from Problem 4.17
to show that energy is conserved at all times (not just t → ∞), as we know it must be.
But we can show this in a quicker manner by demonstrating that the conservation-of-energy
statement is equivalent to the Kirchhoff loop equation, E − Q/C − RI = 0. We can do this
either by differentiating the former to obtain the latter, or by integrating the latter to obtain
the former. Let’s take the first of these routes. The conservation-of-energy statement at any
intermediate time is ∫ t
Q(t)2
EQ(t) = + RI 2 dt. (377)
2C 0
Differentiating with respect to t gives (using dQ/dt = I and canceling a factor of I)
dQ Q dQ Q
E = + RI 2 =⇒ E = + RI, (378)
dt C dt C
which is the Kirchhoff loop equation, as desired. If you want to go in the reverse direction,
just multiply by I and then integrate with respect to t (using the fact that Q = 0 at t = 0).
(a) 0.05 µA equals 5 · 10−8 C/s, so the number of electrons passing a given point
per second is n = (5 · 10−8 C/s)/(1.6 · 10−19 C) = 3.1 · 1011 s−1 . They move
at essentially speed c, so the average distance between them is d = c/n = (3 ·
108 m/s)/(3.1 · 1011 s−1 ) ≈ 0.001 m = 1 mm. The linear charge density is then
λ = e/d = (1.6 · 10−19 C)/(0.001 m) = 1.6 · 10−16 C/m. The electric field 1 cm
from the beam is therefore
λ 1.6 · 10−16 C/m
E= = ( ) = 2.88 · 10−4 V/m. (381)
2πϵ0 r 2π 8.85 · 10−12 kg
s2 C2
m3 (0.01 m)
Note that the distance between the electrons (1 mm) is small compared with
the distance from the beam (1 cm), so the beam looks roughly like a uniform
distribution of charge.
115
116 CHAPTER 5. THE FIELDS OF MOVING CHARGES
(b) In the electron rest frame, the electrons are “uncontracted” compared with the lab
frame, so their average separation is d′ = γd = (20)(0.001 m) = 0.02 m = 2 cm.
The linear charge density, and hence electric field, is therefore decreased by a
factor 1/γ. So the field 1 cm from the beam is
This is the average (along a line parallel to the beam) of the radial component
of the field. Because the electrons are relatively far apart (2 cm, compared with
the 1 cm distance from the beam), there is a large variation in the field as the
position varies along the line parallel to the beam.
The factor here correctly equals 1.1043 if γ = 5/4 (that is, if v = 3c/5).
√ √
The same reasoning also gives the magnitude of the field in F ′ as E ′ = (E/ 2) 1 + γ 2 .
To find the normal component, En′ , we must multiply this by cos(2θ − 90◦ ), where
tan θ = γ. This trig factor can alternatively be written as sin 2θ, which in turn can be
written as 2 sin θ cos θ. So we have
E √
En′ = E ′ cos(2θ − 90◦ ) = √ 1 + γ 2 · 2 sin θ cos θ
2
√
E √ γ 1 2γ
= √ 1 + γ · 2√
2 √ =√ E. (384)
2 1+γ 2 1+γ 2 1 + γ2
This is the same factor that relates σ ′ to σ. So if E = σ/2ϵ0 is true (which it is), then
En′ = σ ′ /2ϵ0 is also true. That is, Gauss’s law holds in F ′ .
′ ′
Limits: If γ = 1 (that is, v = 0), we obtain
√ σ = σ and En√= E, as expected. And if
γ → ∞ (that is, v → c), we obtain σ ′ = 2σ and En′ = 2E; the sheet√is vertical,
so a given amount of charge in F ′ is now located within a span that is 1/ 2 times as
z large as it was in F .
because if β = 0 then the muon’s field dominates and the net Ex is negative, whereas
if β → 1 then the x component of the muon’s field is zero, so the net Ex (due only to
the proton) is positive.
5.14. Forgetting relativity
Assuming β ̸= 0, the factor (1 − β 2 )/(1 − β 2 sin2 θ)3/2 in Eq. (5.15) is smaller than 1
when θ = 0 and larger than 1 when θ = 90◦ . So there must be an angle in between
where the factor equals 1. This occurs when
1 − (1 − β 2 )2/3
(1 − β 2 sin2 θ) = (1 − β 2 )2/3 =⇒ sin2 θ = . (385)
β2
as desired.
Note: having shown that Gauss’s law holds for a sphere, the same reasoning as in
Section 1.10 can be used to show that it holds for any shape. The critical property of
the field is the 1/r2 dependence on r.
5.16. Cosmic rays
′ ◦
The maximum value of the √field in Eq. (5.15) is achieved when θ = 90 , in which case
′2 ′2
the value is (Q/4πϵ0 r )/ 1 − β 2 = γQ/4πϵ0 r . Therefore,
1 γe
Emax =
4πϵ0 r′2
1 γe ( kg m3 ) (1010 )(1.6 · 10−19 C)
=⇒ r′2 = = 9 · 109 2 2 = 14.4 m2 , (388)
4πϵ0 E s C 1 V/m
(a) The electron had been traveling in the positive direction along the negative x
axis toward the origin, where it rather suddenly stopped. Since the speed of light
is c = 3 · 1010 cm/s, and since the “transition” sphere surrounding the static E
field has a radius of about r = 15 cm, the stopping must have taken place at
t = −r/c = −5 · 10−10 s. (From the figure, it looks like the stopping commenced
at this time, and then lasted for perhaps 0.5 · 10−10 s.) If the electron hadn’t
stopped, it would be located at x = 12 cm, because that is where the external
field lines point. So its speed was (12/15)c = (0.8)c.
(b) At t = −7.5 · 10−10 s, which was ∆t = 2.5 · 10−10 s before the electron stopped,
it was located at x = −v ∆t = −(0.8)(3 · 1010 m/s)(2.5 · 10−10 s) = −6 cm, with
respect to the origin.
(c) To find the field strength at the origin when the electron was at x = −6 cm, we
can use Eq. (5.15) with θ′ = 0. This gives
now effectively only one dimension that is expanding (at least with regard to the
spacing between the lines), namely the dimension lying on the surface of the sphere
and perpendicular to the lines. (We have used the fact that the thickness of the shell
remains constant.) Since the density of field lines is proportional to the field strength,
we arrive at the desired result that the tangential field falls off like 1/r.
cos ϕ0
cos θ0 = . (393)
(1 − β 2 sin2 ϕ0 )1/2
Now let’s show that this is equivalent to tan ϕ0 = γ tan θ0 . If tan θ0 = tan ϕ0 /γ, then
1 1
cos θ0 = ( )1/2 = ( )
tan2 ϕ0 1/2
(394)
1 + tan2 θ0 1+ γ2
1 cos ϕ0
= ( )
sin2 ϕ0 1/2
=( )1/2 ,
1 + (1 − β ) cos2 ϕ0
2 (cos ϕ0 + sin ϕ0 ) − β 2 sin2 ϕ0
2 2
If this flux equals Q/4ϵ0 , then the mirror-image cap will also contain a flux of Q/4ϵ0 ,
which will leave Q/2ϵ0 for the region between the cones, which is half of the total flux
Q/ϵ0 . So we want
sin δ 1
= =⇒ 4 sin2 δ = 1 − β 2 (1 − sin2 δ)
(1 − β 2 cos2 δ)1/2 2
1 − β2
=⇒ sin2 δ = . (397)
4 − β2
For β = 0 we have sin δ = 1/2 =⇒ δ = 30◦ , which is the correct result for a spherically
symmetric field, as you can verify. (A cap with a half-angle of 60◦ has half the
area of a hemisphere.)
√ For β → 1 (and γ → ∞) we have sin2 δ ≈ (1√− β 2 )/3, so
δ ≈ sin δ ≈ 1/( 3 γ). The angle between the two cones is then 2δ ≈ 2/( 3 γ).
5.22. Electron in an oscilloscope
(a) • The kinetic energy of the electron is 250 keV. The total energy, including
the rest energy, mc2 = 500 keV, is therefore 750 keV. But the total energy
2
is also given by
√ γmc = γ(500 √ keV). So γ = 750/500 = 1.5. The β factor is
given by β = 1 − 1/γ 2 = 5/3 = 0.745.
• The momentum is px = γm(βc) = (1.5)(0.745)mc = (1.12)mc.
• The time spent between the plates is t = (0.04 m)/(βc) = 1.79 · 10−10 s.
• The transverse force, which has the constant value of Ee, equals the rate of
change of the transverse momentum. So py = (Ee)t. But the electric field is
E = V /s, where s is the separation between the plates. So py = V et/s. We
could plug in the numbers, but since we want to write the result in units of mc
anyway, it is easier to take the following route. We have py /mc = V et/smc.
Multiply by c/c to obtain
py eV tc 6 keV (1.79 · 10−10 s)(3 · 108 m/s)
= = · = 0.0805. (398)
mc mc2 s 500 keV 0.008 m
(Remember that an eV is the change in energy of an electron moving through
a potential difference of 1 volt. So to be precise, “eV” should actually be
written as “eV.” That is, it is the charge of one electron, e, multiplied by 1
volt. And the V in Eq. (398) is 6 kV.)
• The transverse velocity at exit is given by
py (0.0805)mc
py = γmvy =⇒ vy = = = 1.61 · 107 m/s. (399)
γm γm
• Since the transverse force (and hence transverse acceleration) is constant,
the average transverse velocity is half of the vy we just found, that is,
v y = 8.05 · 106 m/s. (vy is small enough so that nonrelativistic kinemat-
ics works fine here.) The transverse distance traveled is then y = v y t =
(8.05 · 106 m/s)(1.79 · 10−10 s) = 1.44 · 10−3 m, which is 1.44 mm.
• The angle of the trajectory at exit is (using tan θ ≈ θ for small θ)
vy py 0.0805
θ= = = = 0.072 rad = 4.1◦ . (400)
vx px 1.12
121
(b) In the frame in which the electron is initially at rest, the plates are moving to
the left with speed βc = (0.745)c, and their length is (0.04 m)/γ = 0.0267 m.
(The other two dimensions are unchanged.) The plates are above and below
the electron for a time t′ = (0.0267 m)/(0.745c) = 1.19 · 10−10 s (which can also
be obtained from time dilation), during which time the electron is accelerated
upward in the field E ′ = γE = γV /s. The upward momentum acquired is
E ′ et′ = (γE)e(t/γ) = Eet = (0.0805)mc. This is the same as in the lab frame,
which is consistent with the fact that transverse momenta are unaffected by a
Lorentz transformation.
In this frame the electron is non-relativistic, to a good approximation, so the
final vy′ is vy′ ≈ (0.0805)c = 2.42 · 107 m/s. (If you want, you can show that
taking relativity into account would decrease the speed by only about 0.3%.)
The average transverse velocity is then v ′y = vy′ /2 = 1.21 · 107 m/s. The total
transverse distance is therefore y ′ = v ′y t′ = (1.21 · 107 m/s)(1.19 · 10−10 s) = 1.44 ·
10−3 m, in agreement with the result in the lab frame, which is consistent with
the fact that transverse distances are unaffected by a Lorentz transformation. In
short, the transverse distances are the same in the two frames because in the
electron’s frame vy′ is larger by a factor γ, but t′ is smaller by a factor γ.
The y components of the momenta in the two frames are equal, as dictated by the
Lorentz transformations. In short, this is due to the fact that in the electron-neutron
frame the transverse field E ′ (and hence transverse force Fy′ ) is larger by a factor γ,
but the time t′ is smaller by a factor γ. So the product Fy′ t′ is the same as in the lab
frame. E1y E1
This is the desired condition on m. Note, interestingly, that the transverse dis-
tance (P/γm)(b/v) is independent of b. This is due to the facts that the force
decreases like 1/b2 , but the distance (and hence time) of the interaction increases
like b. And the latter matters twice in the nonrelativistic expression for distance,
y = ay t2 /2.
To do the same calculation for q1 , we must remember that since the x speed of
q2 is relativistic, we need to take into account the fact that q2 ’s field lines are
squashed into a “pancake.” The result from Problem 5.6 or Exercise 5.21 shows
that the angular width of the pancake is on the order of 1/γ. So the time of
the main part of the interaction (as far as the force on q1 is concerned) is only
on the order of b/γv. But since q1 is moving slowly, we now have vy ≈ Py /m,
with no need for a γ factor in the denominator. We therefore end up with the
condition (Py /m)(b/γm) ≪ b =⇒ q1 q2 /2πϵ0 γv 2 b ≪ m, in agreement with the
above result.
Note that this condition can be written as q1 q2 /4πϵ0 b ≪ γmv 2 /2. For nonrela-
tivistic motion, this says that the kinetic energy of q2 must be much larger than
the electrical potential energy of the particles at closest approach. It is reasonable
123
that the condition takes this form, because these are the only two energy scales
in the problem.
β0 − β 0.8 − 0.553
β0′ = = = 0.443. (402)
1 − β0 β 1 − (0.8)(0.553)
(a) The force is always largest in the rest frame of the particle. It is smaller in any
other frame by the γ factor associated with the speed v of the particle. So the
force in the new frame (the charge’s frame) is larger; it is γqλ/2πrϵ0 . This force
is repulsive, assuming q and λ have the same sign.
(b) In the new frame the charge q isn’t moving, so even though the magnetic field is
nonzero, the magnetic force is zero, due to the v in FB = qvB.
We therefore need only worry about the electric force. In the new frame the
linear charge density on the rod is increased to γλ, due to length contraction.
So the electric field is γλ/2πrϵ0 . This field produces a repulsive electric force of
FE = γqλ/2πrϵ0 , which agrees with the result in part (a).
1 1 1 + β2 1 + β2
γ2 = √ =√ ( )2 = √ = . (404)
1 − β22 (1 + β 2 )2 − 4β 2 1 − β2
1− 2β
1+β 2
1 + β2 e2 e2
Frest = · = γ 2
(1 + β 2
) , (405)
1 − β 2 4πϵ0 r2 4πϵ0 r2
√
where γ ≡ 1/ 1 − β 2 .
The relative speed of the proton rest frame and the√ lab frame is βc, so the force in the
lab frame is smaller than Frest by a factor 1/γ = 1 − β 2 . (Remember, the force is
always largest in the rest frame of the particle on which it acts.) The repulsive force
on each of the protons in the lab frame is therefore
Frest e2
Flab = = γ(1 + β 2 ) . (406)
γ 4πϵ0 r2
This is the correct total force in the lab frame. As stated in the problem, it is not
equal to γe2 /4πϵ0 r2 . The task now is to explain the force as the sum of the electric
and magnetic forces in the lab frame.
As mentioned in the problem, Eq. (5.15) tells us that the repulsive electric force in
the lab frame is Flab,E = γe2 /4πϵ0 r2 . This must not be the whole force, because
it doesn’t equal the correct total force in Eq. (406). Apparently there must be an
additional repulsive force in the lab frame that equals the difference between these
two repulsive forces. This difference is
e2 e2 e2
Flab − Flab,E = γ(1 + β 2 ) − γ = γβ 2
. (407)
4πϵ0 r2 4πϵ0 r2 4πϵ0 r2
This is exactly the force produced by a magnetic field with strength B = (β/c)γe/4πϵ0 r2 .
(This is consistent with the Lorentz transformations in Chapter 6.) This field does
125
indeed produce the desired force, because the proton is moving with speed βc through
this B field, which yields a magnetic force with magnitude
γβ 2 e2
Flab,B = qvB = e(βc)B = . (408)
4πϵ0 r2
This is the quantity that appears on the right-hand side of Eq. (407), so that equation
is correctly the statement that
The direction of the magnetic force will be correct (repulsive) provided that the B
field at the location of each proton, due to the other proton, points out of the page.
Note that the electric and magnetic forces in the lab frame have exactly the same
magnitudes that they had in the second example in Section 5.9, because the speeds
are the same in the two setups. But the difference in direction of the top charge’s
motion in the two setups causes its magnetic field to point in the opposite direction.
The qv × B forces on both charges switch sign, so in the present setup the magnetic
force is added to the electric force instead of subtracted from it. The total force in the
lab frame is therefore different in the two setups.
5.30. Transformations of λ and I
The speed of the charges in the new frame is given by βk′ = (βk + β)/(1 + βk β). The
γ factor associated with this speed is γk′ = γk γ(1 + βk β) (see below). The density in
the rest frame of the charges is λk /γk , so the density in frame F ′ is
( ) ( )
λk ′ λk β βIk
λ′k = γk = γk γ(1 + βk β) = γ λk + (λk βk c) = γ λk + , (410)
γk γk c c
as desired. Since these two transformations are linear in Ik and λk , they hold for
any corresponding linear combinations of the Ik and λk (for example, 2I1 − 7I5 + 3I8
and 2λ1 − 7λ5 + 3λ8 ). In particular, they hold for the sums of the Ik and the λk .
But these sums are just the total current I and the total density λ. So the Lorentz
transformations hold for the total I and λ, as we wanted to show.
Here is the algebra that produces γk′ :
1 1 + βk β
γk′ = √ ( )2 = √
(1 +
2β 2 2 2
2β
k β + βk β ) − (βk +
2
kβ + β )
1− βk +β
1+βk β
1 + βk β
= √ = γk γ(1 + βk β). (412)
(1 − βk2 )(1 − β 2 )
with respect to the direction of motion of the electrons. So for the left electron in
Fig. 5.26 we have θ′ = ϕ − α. This holds for the right electron too, provided that we
consistently measure ϕ with respect to the positive x axis (unlike how it is defined for
the right electron in Fig. 5.26). The angle α is fixed by the parameters in the setup;
the variable we will integrate over is ϕ. The range 0 ≤ ϕ ≤ π corresponds to the range
−∞ ≤ x ≤ ∞ on the wire.
If the charge q is a distance ℓ from the wire in the lab frame, then it is ℓ′ = ℓ/γ from the
wire in its own frame, due to length contraction. (Imagine a transverse stick of length ℓ
attached to the wire as it moves toward the charge in the charge’s frame, and consider
the moment when the end of the stick coincides with the charge.) In the charge’s
frame, the distance r′ from the charge q to an electron is r′ = ℓ′ / sin ϕ = ℓ/γ sin ϕ.
The position of an electron along the wire is given by x = −ℓ′ / tan ϕ = −ℓ/γ tan ϕ.
Taking the differential of this gives dx = ℓ dϕ/γ sin2 ϕ.
We now have a handle on all the necessary quantities, so we can use Eq. (5.15) to
find the force on the charge q (as measured in its own frame) due to a small interval
of the wire with length dx. The (negative) electron charge contained in this length
is (−λ0 )dx = −λ0 ℓ dϕ/γ sin2 ϕ. (There is no length contraction along the wire in the
charge’s frame.) We are concerned with the x component of the force, which brings
in a factor of cos ϕ, so from Eq. (5.15) we obtain
Note that the γ here is associated with the speed v of the charge q in the lab frame,
whereas the βe is the total speed of the electrons in the charge q frame, which is given
by βe2 c2 = v02 /γ 2 + v 2 .
To find the total force on the charge q in its own frame, we need to integrate Eq. (413)
from ϕ = 0 to ϕ = π. The integral is given in Appendix K, so the total horizontal
force from all of the electrons in the wire is
π
γqλ0 (1 − βe2 ) (2 − βe2 ) sin ϕ + βe2 sin(2α − ϕ)
F =− √ . (414)
4πϵ0 ℓ
2(1 − βe2 ) 1 − βe2 sin2 (α − ϕ) 0
In the numerator, the sin ϕ term is zero at both limits, and sin(2α − π) = − sin 2α.
In the denominator, sin(α − π) = − sin α, but we’re squaring this, so the minus sign
doesn’t matter. The denominators are therefore equal at both limits. Hence we obtain
twice the value at the upper limit:
√ √
From tan α = γv/v0 we have sin α = γv/ γ 2 v 2 + v02 and cos α = v0 / γ 2 v 2 + v02 .
127
γ 2 v 2 + v02 γv · v0
γqλ0 γ 2 c2 γ 2 v 2 + v02
F = √
2πϵ0 ℓ γ 2 v 2 + v02 γ 2 v 2
1−
γ 2 c2 γ 2 v 2 + v02
qλ0 vv0 1 γqλ0 vv0
= √ = . (416)
2πϵ0 ℓc2
1 − v /c
2 2 2πϵ0 ℓc2
This is the total force on the charge q in its own frame. The force is positive, so
it points rightward, consistent with the field lines in Fig. 5.26. Now, back in the
lab frame, λ0 v0 equals the current I in the wire (which is directed to the left, since
the electrons are moving to the right). Transforming the above force to the lab frame
involves dividing by the γ factor associated with the speed v, which is the γ factor that
appears in Eq. (416). So we end up with a rightward directed force with magnitude
qvI/2πϵ0 ℓc2 , as desired. As mentioned in the text, if the magnetic field B points out
of the page, then v × B points to the right. So the force can be interpreted as the
qv × B force from a magnetic field with strength I/2πϵ0 ℓc2 .
128 CHAPTER 5. THE FIELDS OF MOVING CHARGES
Chapter 6
(a) p2
dvx qB dvy qB
= vy and =− vx . (420)
dt γm dt γm
129
130 CHAPTER 6. THE MAGNETIC FIELD
Taking the derivative of the first of these equations, and then substituting in the value
of dvy /dt from the second, gives
( )2
d2 vx qB
=− vx . (421)
dt2 γm
This is a simple-harmonic-oscillator type equation, for which the general solution takes
the form,
qB
vx (t) = A cos(ωt + ϕ), where ω = . (422)
γm
The first of the equations in Eq. (420) then quickly gives vy (t) = −A sin(ωt + ϕ). A
and ϕ are arbitrary constants, determined by the initial conditions. However, if the
momentum p = γmv is given, then vx and vy must each have an amplitude of p/γm.
Hence A = p/γm.
The period is 2π/ω = 2πγm/qB, in agreement with the result in part (a). Integrating
vx (t) and vy (t) to find x and y gives (up to arbitrary additive constants, which only
affect the position of the center of the circle)
( ) A( )
x(t), y(t) = sin(ωt + ϕ), cos(ωt + ϕ) . (423)
ω
This describes a circle with radius R = A/ω = (p/γm)/(qB/γm) = p/qB, in agree-
ment with the result in part (a).
(It was fine to set the factor of v in the numerator equal to c, but we of course can’t
set v = c inside the γ factor!) The time to complete one revolution is
as desired. The minus sign in the relation a = −aẑ comes from the given orientation
of the current, along with the right-hand convention. Since N = m × B is a vector
relation, its validity can’t depend on our specific choice of coordinate system. So if it
132 CHAPTER 6. THE MAGNETIC FIELD
it true for a particular choice of axes (as we just demonstrated), then it must be true
for any choice.
Let’s now look at the net force on the loop. If B is uniform over the loop, then the
net force is ∫ ∫ (∫ )
dF = I dl × B = I dl × B. (429)
loop
∫
But dl = 0 because we have a closed loop. The net force is therefore zero. Note
that this result holds even if the loop isn’t planar.
6.35. Determining c
If we follow the wires around in Fig. 6.42, we see that the voltage across both capacitors
is equal to E0 cos 2πf t. Let’s first determine the magnetic force between the rings. Ne-
glecting the inductance (the subject of Chapter 7) and resistance of the two rings and
leads, the charge on capacitor C2 at any time t takes the form, Q2 = E0 (cos 2πf t)C2 .
The current through C2 is then I = dQ2 /dt = −2πf E0 C2 sin 2πf t, with positive de-
fined as flowing into the left terminal of C2 , or equivalently out of the right terminal
and into the rings. The two rings are in series, so this current flows through each.
If h ≪ b we are justified in computing the magnetic force between the rings as if they
were parallel straight wires, with force per unit length µ0 I 2 /2πh. The length is 2πb,
so the magnetic force pulling the upper ring down (note that the currents are in the
same direction, so the force is attractive) is
µ0 I 2 µ0 (2πf E0 C2 )2 b
Fm = (2πb) = sin2 2πf t. (430)
2πh h
The time average of sin2 2πf t is 1/2, so the average attractive magnetic force between
the rings is
2µ0 π 2 f 2 E02 C22 b
Fm = . (431)
h
Now let’s determine the electric force between the capacitor plates. As mentioned
above, the potential difference between the plates is E0 cos 2πf t. The field between
the plates is therefore E = (E0 cos 2πf t)/s. The downward electric force on the upper
plate can be obtained in various ways. From Section 3.7 the force is the energy density
times the area, which gives Fe = (ϵ0 E 2 /2)(πa2 ). Alternatively, the force is the charge
times the average of the fields on either side of the plate, namely E/2 (see Section 1.14),
which gives (σπa2 )(E/2). (Equivalently, the force is the charge times the field from
the other plate.) These two expressions agree because σ = Eϵ0 . Using the above
expression for E, and noting that the time average of cos2 2πf t is 1/2, we find the
average attractive electric force between the rings to be
( )2
ϵ0 πa2 E0 cos 2πf t ϵ0 πa2 E02
Fe = =⇒ F e = . (432)
2 s 4s2
E02 C12
Fe = . (433)
4πϵ0 a2
133
When the forces are balanced (which might be brought about by varying C2 ), we have
as desired. Note that a constant voltage wouldn’t be useful here, because the capacitors
would quickly reach their maximum charge, which would mean the current would be
zero. The alternating voltage allows there to be (except at discrete moments during
each cycle) both a nonzero charge on the capacitor C1 and a nonzero current in the
rings.
Let’s see what some reasonable numbers give. If a = 0.1 m, b/h = 10, and C2 /C1 =
√
106 , we find that the righthand side equals 1/ µ0 ϵ0 = c = 3 · 108 m/s when f =
−1
60.2 s . So given all the other parameters, we can determine c by sweeping though
frequency values until we find the f (which is 60.2 s−1 in the present case) that makes
things balance. If the current rings consist of N turns each, this magnifies the magnetic
force by a factor N 2 (the magnetic field is N times as large, and there are N times as
many loops that it acts on), which decreases the necessary value of f (or C2 /C1 , etc.)
by a factor N .
or 30 gauss, and it points to the left. A more remarkable fact (see Exercise 6.38) is
that the field is 30 gauss pointing to the left not only at P but everywhere inside the
cylindrical hole.
µ0 Ir
B · 2πr = µ0 Ir =⇒ B = . (438)
2πr
If we want B to be independent of r, then we need Ir to be proportional to r. Ir is
found by integrating the current density J(r):
∫ ∫ r
Ir = J da = J(r′ ) · (2πr′ dr′ ). (439)
0
It is easiest to guess and check the form of J(r′ ). If J(r′ ) is proportional to 1/r′ , then
it takes the form of J(r′ ) = α/r′ , so
∫ r
Ir = (α/r′ )(2πr′ dr′ ) = 2παr, (440)
0
The above “1/r” result for the current density is the same result that holds for the
charge density in the case of the electric field due to a charged cylinder or sphere. In
both of these cases the electric field is independent of r if the density ρ is proportional
to 1/r.
Note that even though the current density diverges at r = 0, the actual current does
not. There is a finite amount of current in any cross section with radius r, and it is
given (by construction) by Ir = 2παr. Any ring (at any radius) with thickness dr
contains the same amount of current, dI = 2πα dr.
We can also solve this exercise by using the differential form of Ampere’s law, ∇ ×
B = µ0 J. Since B points tangentially and has a uniform value, it can be written as
B = B0 θ̂. Equation F.2 in Appendix F then gives
1 ∂(rB0 ) B0
∇×B= ẑ = ẑ. (442)
r ∂r r
Setting this equal to µ0 J ẑ gives J = B0 /(µ0 r), consistent with the 1/r dependence
we found above. The factor of B0 /µ0 here equals the α from above.
6.40. The pinch effect
If the conduction electrons are forced closer to the axis, there will be uncompensated
negative charge near the axis. This will generate an inward radial electric field E that
pushes outward on the electrons, preventing further constriction when the outward
electric force balances the inward magnetic force, that is, when eE = evB =⇒ E = vB.
The magnetic field at radius r is Br = µ0 Ir /2πr, where Ir is the current contained
within radius r. Assuming no redistribution of the charge, Ir is given by Ir = πr2 J,
where J = nev is the current density (n is the number of electrons per unit volume,
and v is the drift velocity). The B field is therefore Br = µ0 (πr2 nev)/2πr = µ0 rnev/2.
Suppose that the cloud of electrons at radius r is squeezed inward by a small distance
∆r. The cylinder of radius r will now contain, per unit length, an excess of negative
charge in the amount of ∆λ = (ne)(2πr ∆r); this is the volume charge density times
the cross-sectional area. This causes an inward electric field equal to Er = ∆λ/2πϵ0 r =
ne ∆r/ϵ0 . The condition for equilibrium in then (using µ0 ϵ0 = 1/c2 )
ne ∆r µ0 rnev ∆r µ0 ϵ0 v 2 v2
Er = vBr =⇒ =v =⇒ = = 2. (443)
ϵ0 2 r 2 2c
In solid conductors we always have v/c ≪ 1. In metal conduction, v/c is seldom much
greater than 10−10 , so (∆r)/r ≈ 10−20 is too small to detect. In highly ionized gases,
however, the “pinch effect,” as it is called, can be not only detectable but important.
If the effect were large enough to measure, a Hall probe in the spirit of Fig. 6.33 could
be used, with one lead connected to the axis (by drilling a thin tube in the rod), and
the other lead connected to the surface of the rod. If v ≈ 10−3 m/s and B ≈ 1 T,
the resulting E ≈ 10−3 V/m would be large enough to generate a measurable voltage
difference.
6.41. Integral of A, flux of B
Using Stokes’ theorem, along with B = ∇ × A, we have
∫ ∫ ∫
A · ds = ∇ × A · da = B · da = Φ, (444)
C S S
as desired. This relation is similar to Ampere’s law because the differential form of
that law, µ0 J = ∇ × B, takes the same form as the above B = ∇ × A relation.
136 CHAPTER 6. THE MAGNETIC FIELD
From inspection, a few choices for A that satisfy these equations are A = (0, B0 x, 0),
or (−B0 y, 0, 0), or (−B0 y/2, B0 x/2, 0). In general, any vector of the form (−ay, bx, 0)
works if a + b = B0 . And even more generally, adding on any vector with zero curl
also works.
and it points in the positive θ̂ direction. The θ̂ vector equals (−y/r, x/r, 0) because
this vector has length 1 and has zero dot product with the radial vector (x, y, 0). So
the Cartesian components of B are
y µ0 Iy x µ0 Ix
Bx = − B = − , and By = B= . (447)
r 2πr02 r 2πr02
∂Az ∂Az
B = ∇ × A = x̂ − ŷ = 2A0 yx̂ − 2A0 xŷ. (448)
∂y ∂x
To see why the integral along the axis should indeed be equal to µ0 I, consider the
closed path shown in Fig. 110, which involves a semicircle touching the points z = ±r.
Assume that r ≫ b. Along the z axis, Bz behaves like 1/z 3 for z ≫ b. And |B| also
behaves like 1/r3 along the (large) semicircle. Accepting that this is true (see below),
r
then since the length of the semicircle is proportional to r, the line integral along the
semicircle is at least as small (in order of magnitude) as r/r3 = 1/r2 , which goes to
zero as r → ∞. We can therefore ignore the return semicircular path. So the line
integral along the whole loop (which encloses a current I) equals the line integral along Figure 110
the z axis, in the r → ∞ limit.
Let’s now argue why |B| behaves like 1/r3 for large r. Consider the point at the “side”
of the semicircle in Fig. 110. In order of magnitude, the field at this point, due to
the ring, is the same as the field due to a square with side b. But the field due to
the square has contributions from two opposite sides (the sides perpendicular to the
r̂ vector) that nearly cancel, because the current moves in opposite directions along
these sides. The Biot-Savart law says that each side gives a contribution of order
1/r2 . Taking the difference of these contributions is essentially the same as taking a
derivative, and the derivative of 1/r2 is proportional to 1/r3 , as desired. Additionally,
the two sides parallel to the r̂ vector also happen to produce a contribution of order
1/r3 ; see Problem 6.14. At points in between the axis and the “side” point on the
semicircle, there will be various angles that come into play. But these simply bring in
factors of order 1 that morph the 1/z 3 result on the axis to the 1/r3 result at the side dθ P
point, so they don’t change the general 1/r3 result. θ
r b
6.45. Field from an infinite wire φ
Consider a small piece of the wire at angle θ, subtending an angle dθ, as shown in dl
I
Fig.111. If r is the distance from a given point P to the small piece, then Fig.112 shows
that the length of the piece is dl = r dθ/ cos θ. But r equals b/ cos θ, so dl = b dθ/ cos2 θ. Figure 111
(This can also be obtained by taking the differential of l = b tan θ.) In the Biot-Savart
law, the cross product between dl and r̂ brings in a factor of sin ϕ, which is the same
as cos θ. If the current points rightward, then we have (with ẑ pointing out of the to P
page) r dθ
θ
∫ ∫
µ0 I dl × r̂ µ0 I π/2 (b dθ/ cos2 ) cos θ dl =
r dθ
____
B = = ẑ cosθ
4π r2 4π −π/2 (b/ cos θ)2
∫ π/2
µ0 I π/2 µ0 I µ0 I
= ẑ cos θ dθ = ẑ sin θ = ẑ , (450) Figure 112
4πb −π/2 4πb −π/2 2πb
which agrees with the standard result obtained more much quickly via Ampere’s law. B A
z
C D
6.46. Field from a wire frame
P y
(a) In Fig. 113 the contributions to the field at P from the diagonally opposite edges x
AB and EF cancel, as do those from CD and GH. The pair BC and F G G
H
produces a field that points in the y direction at P , as does the pair HA and
F E
DE. So the total magnetic field at P points in the positive y direction.
Figure 113
(b) Imagine superposing on the given setup the current in the two horizontal squares
shown in Fig. 114. The currents along six of the edges cancel, and we end up with B A
the desired square loop of current. But the field at P due to the two squares in z
Fig. 114 is zero, due to a symmetry argument. (Imagine rotating the setup by C
D
180◦ around either the x or y axis. The setup is unchanged, so the magnetic field P y
must point along both the x and y axes. The zero vector is the only vector with x
G H
F E
Figure 114
138 CHAPTER 6. THE MAGNETIC FIELD
this property.) Or you can just note that diagonally opposite edges combine to
give zero field at P , as we saw in part (a). Therefore, since we added on zero
field at P , the field at P in the original setup must be the same as the field at P
in the case of the single square loop.
If this field equals 0.5 gauss = 5 · 10−5 T, then I must be (taking the earth’s radius to
be R = 6 · 106 m)
√ √
5 5 RBz 5 5(6 · 106 m)(5 · 10−5 T)
I= = = 2.7 · 109 A, (454)
µ0 4π · 10 −7 kg m
C2
which is huge. For comparison, the peak current in a bolt of lightning is on the order y
of 105 A. (x,y)
6.52. Right-angled wire
I
Consider the contribution from the positive x-axis part of the wire. In the notation r
of Fig. 115, the Biot-Savart law gives
θ
∫ ∫ ∫ x
µ0 I dl × r̂ µ0 Iẑ ∞ sin θ dx′ µ0 Iẑ ∞ y dx′ dl
B= = = . (455) I
4π r2 4π 0 r2 4π 0 r3
Figure 115
If y is positive (or negative), then this B points in the positive (or negative) ẑ direction,
that is, out of (or into) the page. Writing r in terms of the Cartesian coordinates gives
(with x′′ ≡ x′ − x, and using the integral table in Appendix K)
∫ ∞ ∫
µ0 Iy dx′ µ0 Iy ∞ dx′′
Bz = =
4π 0 [y 2 + (x′ − x)2 ]3/2 4π −x [y 2 + x′′2 ]3/2
( )
∞
µ0 Iy x′′ µ I 1 x
= = 0
+ √ . (456)
4π y 2 [y 2 + x′′2 ]1/2 −x 4π y y x2 + y 2
If you write out the contribution from the positive y axis, you will find that it takes
the same form, except with x and y reversed. We therefore end up with the desired
result.
We can check a limit: If x ≫ y, we should end up with the field at a distance y from
an infinite wire lying along the x axis. And indeed, if x ≫ y, two of the terms in
the result in Eq. (6.98) are negligible, and two of the terms are essentially equal to
(µ0 I/4π)(1/y). So we end up with µ0 I/2πy, as expected.
x (x,y)
6.53. Superposing right angles y
An infinite straight wire carrying current in the positive direction along the x axis is
the superposition of the two right-angled wires shown in Fig. 116. So we need to find Figure 116
the fields at (x, y) due to these two wires. The field from the right wire is just the
field derived in Exercise 6.52. The field from the left wire is the same as the field at
the point shown in the setup in Fig. 117 (we have simply rotated the setup in Fig. 116
by 90◦ ). But this field can be obtained by taking the result from Exercise 6.52 and
setting the x value equal to y, and the y value equal to −x, as shown. So the field due
to the left right-angled wire is
( ) -x
µ0 I 1 1 y −x
Bz = + + √ + √ . (457)
4π y −x −x y 2 + x2 y y 2 + x2 y
When we add this to the field from the right wire given in Eq. (6.98), the last three Figure 117
terms in Eq. (457) cancel. We therefore end up with Bz = (µ0 I/4π)(2/y) = µ0 I/2πy,
as desired, because y is the distance from the wire (the x axis).
140 CHAPTER 6. THE MAGNETIC FIELD
β
z R
F current
B current out into page
Bz of page
l/2
Bx loop B Bz
F
Bx
Figure 118
from the side. This is consistent with conservation of angular momentum, because
the straight wire will gain angular momentum (relative to, say, an origin chosen to be
the center of the square) as it moves to the right. This angular momentum will have
a clockwise sense, consistent with the fact that the total angular momentum of the
system remains constant.
6.55. Helmholtz coils
Let the symmetry axis of the setup be the z axis, and let the centers of the rings be
located at z = ±b/2. If the currents in the rings are equal and point in the same
direction, then from Eq. (6.53) the field along the axis at position z is given by
1 1
Bz (z) ∝ + 2 . (459)
[a2 + (z + b/2)2 ]3/2 [a + (z − b/2)2 ]3/2
If we expand this function in a Taylor series around z = 0, the first derivative and all
other odd derivatives are zero at z = 0, because Bz (z) is an even function of z, due
to the symmetry of the setup. So the function will be most uniform near z = 0 if the
second derivative is zero there. The deviations will then be of order z 4 . That is, the
Taylor series will look like Bz (z) = Bz (0) + Cz 4 + · · · . Differentiating the first term
in Eq. (459) twice and evaluating the result at z = 0 yields
5(z + b/2)2 − [a2 + (z + b/2)2 ] 3(b2 − a2 )
3 2 2 7/2 = 2 . (460)
[a + (z + b/2) ] z=0 [a + b2 /4]7/2
The second derivative of the second term in Eq. (459) simply involves replacing b/2
with −b/2, so we end up with the same result, because there are no odd powers of b in
Eq. (460). We therefore see that the second derivative is zero at z = 0 if a = b. You
can show that if a = b, the field halfway from z = 0 to the plane of each ring (that is,
at z = ±b/4) is only 0.4% smaller than the field at z = 0. And at z = ±b/8 the field
is only 0.03% smaller. Two coils arranged with a = b are called Helmholtz coils.
A continuity argument shows why there must exist a point where the second derivative
of Bz (z) equals zero. If the rings are far apart (for example, if b = 4a), then the plot
of Bz consists of two bumps, as shown in Fig. 119. But if the rings are close together
(for example, if b = a/4), then they act effectively like one ring with twice the current,
so there is just one bump. The second derivative at z = 0 is positive in the former
case, and negative in the latter, so somewhere in between it must be zero.
Figure 119
142 CHAPTER 6. THE MAGNETIC FIELD
(a) The resistance of the winding in the small solenoid is 10 times that of the large
solenoid. This is true because the resistance is given by R = ρL/A, and the small
wire is 1/10 as long, with 1/100 the cross-sectional area. So if we apply the same
voltage of 120 V to the small solenoid, we get 1/10 the current. This is just what
is needed to produce a magnetic field equal to that in the large solenoid, because
the field is proportional to nI, and the small coil has 10 times as many turns per
unit length, each with 1/10 the current. Equivalently, the surface current density
J is the same in the small coil (it has 10 times as many wires per unit length,
with 1/10 the current in each), and J is all that matters for a solenoid, because
the field inside is B = µ0 J .
Symbolically, we have B ∝ nI = n(V /R) = nV /(ρL/A) = (V /ρ)(nA/L). But
the quantity nA/L is dimensionless (the units are m−1 m2 /m), so it can’t depend
on the length scale of the solenoid. Therefore, if V and ρ are the same in both
setups, then B is also the same.
(b) The power is P = IV , so it is smaller by a factor of 10 in the smaller solenoid,
because V is the same and I is smaller by a factor of 10. But the smaller solenoid
has only 1/100 the surface area (because area is proportional to length squared),
so it will be harder to keep it cool; any cooling mechanism operates by interacting
with the surface of the wire.
P . Now, r̂ has a z component, but the r̂ vector associated with the corresponding
patch defined by the thin cone on the other side of P has the opposite z component.
These components therefore yield canceling contributions to the total magnetic field.
So we need only worry about the component of r̂ that lies in the xy plane. Let’s call
this vector r̂xy .
We need to compute the cross product of dl and r̂xy , both of which lie in the xy
plane. (The resulting cross product will therefore point in the z direction, so we have
just proved that the B field from the solenoid must be longitudinal.) In general, dl
has a component parallel to r̂xy and a component perpendicular to r̂xy . The parallel
component yields zero in the cross product dl × r̂xy , so we need only worry about the
component perpendicular to r̂xy . In other words, if we project the area of the patch
onto the (vertical) plane orthogonal to r̂xy , then the cross product dl × r̂xy remains
the same. We can do the same with the other patch in the same cone.
We therefore need to compare the Biot-Savart contributions from the two “projected”
patches of area defined by a given cone. If the projected patches are distances r1 and
r2 from the point P , then their areas are proportional to r12 and r22 , because areas are
proportional to length squared, and because the patches cut the line from P at the
same angle (perpendicular, by construction).
Now, if we imagine a small rectangular patch (any patch can be built up from rectan-
gles), the I dl product in the Biot-Savart law is proportional to the area, because I is
proportional to the height dh of the rectangle (since I = J dh), and because dh dl is
the area of the rectangle. The numerators in the Biot-Savart law for the corresponding
patches are therefore proportional to r12 and r22 . These factors exactly cancel the r2
in the denominator of the Biot-Savart law. So the magnitudes of the contributions
from the two patches equal. And since the currents flow in opposite directions in
the projected patches, the contributions therefore cancel. The entire solenoid can be
considered to be built up from small patches subtended by cones, so the external field
is zero, as desired.
If the solenoid isn’t convex, then a given cone may define 4, 6, 8, etc., patches. But
there will still be equal numbers of patches having currents in each direction (which
plane can be traced to the fact that P has the property of being outside the solenoid), so
(top view)
the sum of the contributions will still be zero.
6.61. Rectangular torus
Consider two loops of current that are located symmetrically with respect to a plane
through the axis of the torus; a top view is shown in Fig. 121. At any point on this
plane (inside or outside the torus) the vector sum of the field due to Loop 1 and the
field due to Loop 2 is perpendicular to the plane (or is zero). You can check this by
Loop 1 Loop 2 looking at the Biot-Savart contributions from corresponding little pieces of the two
I B1 I loops; the components parallel to the plane cancel. In general, the B1 and B2 vectors
shown also have components perpendicular to the page, but you can show that these
B2
components are equal and opposite.
This result is actually true for two similar loops of any (planar) shape carrying equal
Figure 121 currents in the same orientation; the cross section of the torus doesn’t have to be
rectangular. The same Biot-Savart reasoning involving corresponding little pieces
holds.
The entire coil can be decomposed into pairs of loops located symmetrically with
respect to a given plane. Hence the total magnetic field at any point must be perpen-
dicular to the plane containing that point and the axis (or be zero). In other words,
the field points in the circumferential direction.
145
To find the magnitude of the field, we can use Ampere’s law. By symmetry, the
magnetic field must have the same magnitude B everywhere on a circle of radius r
around the axis. The line integral of B around this circle equals µ0 times the current
enclosed. Since B points in the tangential direction, the line integral equals 2πrB.
If the circle doesn’t lie inside the torus, the current enclosed is zero. This is true
because either the disk defined by the circle doesn’t intersect the torus, in which case
the current enclosed is clearly zero; or the disk does intersect the torus, in which case
a current of N I passes through the disk in one direction, but another N I also passes
through in the other direction. Therefore, B = 0 everywhere outside the torus.
On the other hand, if the circle lies inside the torus, the current enclosed is N I,
because the disk defined by the circle intersects only the inner boundary of the torus.
Therefore, 2πrB = µ0 N I =⇒ B = µ0 N I/2πr inside the torus. This expression for
B holds for a torus of any (uniform) cross section. Note that B depends only on r,
and not on the “height” inside the torus.
In the limit where b − a ≪ a, the curvature of the torus is negligible, so we essentially
have an infinite straight solenoid with rectangular cross section. The field should
therefore equal µ0 nI (see the solution to Problem 6.19), where n is the number of
turns per unit length. And indeed, in the above result, n = N/2πr is the number of
turns per unit length (where r is essentially equal to both a and b), so we do obtain
B = µ0 nI. F
θ2 E
θ0 θ1
6.62. Creating a uniform field
C D
Since 10 milligauss is about 2% of the earth’s field, we need a compensating field of
approximately 0.55 gauss that is uniform to about 2% over the region of interest. Let’s
try a solenoid 1 meter long and 50 cm in diameter; see Fig. 122. To see whether it Figure 122
meets the requirement, compare the field at the center C with the field on the axis 15
cm from the center, at D. From Eq. (6.56) we have
field at C 2 cos θ0
= . (464)
field at D cos θ1 + cos θ2
The ratio of the fields is therefore 1.024. So the deviation is about 2.4%. This is a little
too large for comfort, especially as we have no easy way to estimate the deviation at
off-axis points such as E. Let’s lengthen the solenoid to 1.2 meters. The denominators
in the above arctans are now 60, 45, and 75, respectively, and you can quickly show
that the ratio of the fields is now 1.013.
We expect the departure from uniformity in the radial direction to be of the same
magnitude, roughly, as the variation in the axial direction. But it has the opposite
sign; the field strength at E is larger than that at C. This makes sense, because in the
limit where the solenoid is very squat, so that it basically looks like a ring, we know
that the field increases (without bound, in fact) as we move away from the center
toward the circumference. An exact calculation of the field strength very close to the
solenoid at F , which involves an elliptic integral, shows it to be 1.4% greater than the
field strength at C, in the case of the 1.2 meter solenoid.
146 CHAPTER 6. THE MAGNETIC FIELD
The number of ampere turns, N I, required to make the field of the solenoid at C equal
to the earth’s field, 5.5 · 10−5 T, is found from Eq. (6.56). The number of turns per
unit length is n = N/(1.2 m), so we have
( )
µ0 N/(1.2 m) I
(2 cos θ0 ) = 5.5 · 10−5 T
2
(1.2 m)(5.5 · 10−5 T)
=⇒ N I = ( )
P2 P1 4π · 10−7 kgC2m (0.923)
= 57 ampere-turns. (466)
Figure 123
6.63. Solenoids and superposition
(a) Imagine adding a similar solenoid on the left, as shown in Fig. 123. This exactly
doubles the field strength at P2 . But now the field strengths at P2 and P1 are
approximately equal, because both points lie well inside a fairly long solenoid,
the field at P2 being slightly stronger. Therefore, the original field at P2 must
have been slightly more than half the field at P1 .
This “more than half” result is consistent with the extreme case where the
solenoid is very short, basically just a ring. In this case the fields at P2 and
α2 P1 are essentially equal, both taking on the value of the field at the center of a
α1 ring, namely µ0 I/2r. So the field at P2 is certainly more than half of the field at
P2 P1 P1 .
A less elegant way of solving this exercise is to use Eq. (6.56). The field at the
Figure 124 center is proportional to 2 cos α1 , and the field at the end is proportional to cos α2
(plus cos 90◦ , which is zero), where these angles are defined in Fig. 124. For small
angles, both of these cosines are essentially equal to 1, hence the ratio of 1/2 in
the fields. But cos α2 > cos α1 , hence the “more than half.”
(b) Let’s assume (in search of a contradiction) that there exists a field line that
crosses the line GH with a vertical component, as shown in Fig. 125(a). Imagine
flipping the solenoid upside down to obtain the situation in Fig. 125(b), and then
reversing the direction of the current (so that it now has the same direction as
in Fig. 125(a)) to obtain the situation in Fig. 125(c). Note that the field at the
given point on the line GH has a downward component in both figures (a) and
(c) (or upward in both, if we had initially drawn it upward).
I I
I
G H G H
Figure 125
Now join the two semi-infinite solenoids in figures (a) and (c) end to end, thereby
creating an infinite solenoid. By superposition, the fields simply add, so we
end up with a downward component at the given point along GH. But this is
a contradiction, because we know that the field of an infinite solenoid is zero
outside the solenoid. We conclude that the field due to the semi-infinite solenoid
147
at the given point must have had zero vertical component. In other words, it was
horizontal, as we wanted to show.
(c) The argument used in part (a), applied to the semi-infinite solenoid, shows that
the axial component of the field, at any point on the end face is exactly B0 /2,
where B0 is the (uniform) field throughout the inside the corresponding infinite
solenoid. This is true because adding another semi-infinite solenoid simply dou-
bles the axial field at any point on the end face (see the reasoning in part (b)), and
cancels the radial field, resulting in a purely axial field. As far as the flux goes,
when calculating the flux through the end face, only the axial field component is
involved. Therefore, the flux must be exactly half the interior flux.
(d) From the reasoning in part (c), a given flux tube that starts with area A far back
in the solenoid must flare out as it approaches the end face, so that its cross section
there (where the axial field is half as large as the field far back in the solenoid)
has area 2A and thus the same amount of flux. (There can be no net flux into
or out of the tube, since div B = 0.) In the special case of an axial tube√with
circular cross section everywhere, this√tells us that πr12 = 2πr02 =⇒ r1 = 2 r0 .
Of course, this holds only if r0 < R/ 2, where R is the radius of the solenoid.
Otherwise, the field line exits the solenoid before it reaches the end. B'
B0 P B P'
Remark: The arguments used in parts (b) and (c) lead to more general statements
about the field of a semi-infinite solenoid. Consider two points P and P ′ symmetrically
located with respect to the end plane, as shown in Fig. 126. The fields B and B′ are
related as follows: The radial components of B and B′ are equal. The sum of the axial Figure 126
components of B and B′ is equal to B0 if P lies inside the solenoid, or to zero if P lies
outside the solenoid (that is, above the top “edge” of the solenoid in the figure). The
conclusions of parts (b) and (c) follow in the special case where P and P ′ coincide.
(a) The total energy of each proton is 3 GeV, so√ the γ factor is√3 (because the total
energy of a particle is γmc2 ). Hence β = 1 − 1/γ 2 = 8/9 = 0.943. The
current is 10−3 C/s, so I = λv gives the linear charge density as
I 10−3 C/s
λ= = = 3.53 · 10−12 C/m. (467)
v 0.943 · 3 · 108 m/s
The electric field 1 cm from the axis of the beam is
λ 3.53 · 10−12 C/m
E= = ( ) = 6.35 V/m. (468)
2πϵ0 r 2π 8.85 · 10−12 kg
s 2 C2
m3 (0.01 m)
(c) In F ′ there is no current because the protons are at rest, so B = 0. The spacing
between the protons is “uncontracted,” so the density is smaller; it is λ′ = λ/γ.
The electric field is therefore E ′ = E/γ = 2.12 V/m.
(a) First solution: The electric field can be found via Eq. (5.15) in Chapter 5. If
θ = 90◦ , we get an extra factor of γ compared with the static case, so the electric
field at (3ℓ, 0, 0) is (with γ = 5/4 and ℓ = 1 m)
1 γe 1 γe 1 25 e
E = x̂ + x̂ = x̂ = 2 · 10−9 V/m. (472)
4πϵ0 ℓ2 4πϵ0 (3ℓ)2 4πϵ0 18 ℓ2
149
Second solution: Let F ′ be the lab frame, and consider the frame F traveling
upward with the left ion. Consider the field at (3, 0, 0) due to just the left ion. In
frame F there is no B field from the left ion, so the transformations in Eq. (6.76)
yield E′⊥ = γE⊥ . That is, the electric field due to the left ion is larger in F ′ (the
lab frame) by a factor γ. The same reasoning holds for the field due to the right
ion (because the direction of the relative velocity of the frames doesn’t matter in
γ), so the solution proceeds as above.
(b) Consider the frames F and F ′ defined above. Eq. (6.76) gives B′⊥ = −(γ/c2 )v ×
E⊥ , where v is the velocity of F ′ (the lab frame) with respect to F (the ion
frame). For the left ion traveling upward, we have v = −vŷ, where v = 3c/5. So
the magnetic field in F ′ (the lab frame) at (3ℓ, 0, 0) due to the left ion is (using
µ0 = 1/ϵ0 c2 )
( )
γ x̂e γµ0 ev
B′⊥,left = − 2 (−vŷ) × = −ẑ . (473)
c 4πϵ0 (3ℓ)2 4π(3ℓ)2
Similarly, the magnetic field due to the right ion (for which F ′ moves with velocity
v = vŷ with respect to F ) is ẑγµ0 ev/4πℓ2 . The sum is
( )
′ µ0 e e µ0 2c e
B⊥,total = ẑ (γv) 2 − = ẑ = 3.2 · 10−18 T. (474)
4π ℓ (3ℓ)2 4π 3 ℓ2
6.68. Force on electrons moving together
(a) Since there is no transverse length contraction, the distance between the electrons
in the frame in which they are at rest is still r . The force on each electron is
repulsive, and the magnitude is simply e2 /4πϵ0 r2 . To transform this force to the (lab frame F')
lab frame, we can use the fact that the transverse force on a particle is always
y
greatest in the rest frame of the particle. It is smaller in any other frame by the
-e
factor γ. The repulsive force in the lab frame is therefore (1/γ)(e2 /4πϵ0 r2 ).
x
(b) In the lab frame F ′ , consider the E ′ and B ′ fields at the location of the bottom r
electron arising from the top electron in Fig. 127. The electric field points upward E'
toward the top electron, with magnitude γe/4πϵ0 r2 . This follows from Eq. (5.15) -e
FB
in Chapter 5, with θ = 90◦ . Alternatively, it follows from the transformations in B'
Eq. (6.76); if F is the rest frame of the electrons, then B = 0, so the E′ field in FE
the lab frame is given by E′⊥ = γE⊥ = γ(e/4πϵ0 r2 )ŷ.
The transformations in Eq. (6.76) also give the B ′ field. Since B = 0 in the
electrons’ frame F , the B′ field in the lab frame F ′ is given by B′⊥ = −γ(v/c2 ) × Figure 127
E⊥ . The v here is the velocity of the lab frame F ′ with respect to the electrons’
frame F . So v points leftward, and the righthand rule gives the direction of B′⊥
as pointing out of the page. The magnitude is B ′ = (v/c2 )(γe/4πϵ0 r2 ).
All of the fields and forces are shown in Fig. 127 (when calculating the forces,
remember that electrons are negatively charged). The resulting force on the
electron is the sum of the eE ′ repulsive electric force and the evB ′ attractive (as
you can verify from the righthand rule) magnetic force. The net repulsive force
is therefore
( )
γe v γe v2 γe2 e2
eE ′ − evB ′ = e 2
− ev 2 2
= 1 − 2 2
= , (475)
4πϵ0 r c 4πϵ0 r c 4πϵ0 r γ4πϵ0 r2
in agreement with the result in part (a). If you forgot to consider the magnetic
field in the lab frame, then you would have a “paradox” where the γ appears
in the denominator of the correct force in part (a), but in the numerator in the
incomplete (that is, just electric) force in part (b).
150 CHAPTER 6. THE MAGNETIC FIELD
(c) In the limit v → c, we have γ → ∞, so the force in the lab frame goes to zero.
Figure 128 Bottom-stick frame: In this frame, the situation is shown in Fig. 128. The
charge q and the top stick are now moving. The density of the top stick is γλ due
to length contraction. The top stick produces a current (γλ)v, so the magnetic field
is µ0 (γλv)/2πr, and it points into the page at the location of the charge q. The
magnetic force qvB therefore points upward with magnitude qv · µ0 (γλv)/2πr. Using
µ0 = 1/ϵ0 c2 , this force in the bottom-stick frame (bsf) can be written as FB,1
bsf
=
2 2
(v /c )γλq/2πϵ0 r.
bsf bsf
The electric forces are quickly found to be FE,1 = γλq/2πϵ0 r downward and FE,2 =
λq/2πϵ0 r upward. The total force in the bottom-stick frame is therefore
bsf bsf bsf bsf
Ftot = FB,1 + FE,1 + FE,2
( )
v 2 γλq γλq λq
= + − +
c2 2πϵ0 r 2πϵ0 r 2πϵ0 r
( ( 2 ) )
v λq
= γ 2 −1 +1
c 2πϵ0 r
( )
1 λq
= − +1 . (477)
γ 2πϵ0 r
This is positive, so the total force is upward. It is 1/γ times the total force in the lab
frame. This is correct, because the force on a particle is largest in the rest frame of
the particle (the lab frame here); it is smaller in any other frame by the factor 1/γ.
6.70. Drifting motion
If there is a frame in which the electric field is zero, then we know from Exercise 6.29
that the ion moves in a circle in that frame. Let F be the lab frame, and consider
the frame F ′ that moves in the positive y direction with speed v = (0.1 m)/(1 µs) =
105 m/s = c/3000. Since F ′ moves with the average velocity of the ion, it is the only
frame in which the ion could possibly be moving in a circle (because in any other
frame the ion would drift away). F sees F ′ moving with velocity vŷ, so if we demand
that the electric field be zero in F ′ , then Eq. (6.76) tells us how E⊥ and B⊥ in the
lab frame F must relate to each other:
E′⊥ = γ(E⊥ + v × B⊥ )
( )
=⇒ 0 = γ E⊥ + (vŷ) × (0.6 T)ẑ
=⇒ E⊥ = −(105 m/s)(0.6 T)(ŷ × ẑ)
= −(6 · 104 V/m)x̂. (478)
151
Note that E⊥ points in the (negative) x direction, whereas the drift of the motion is
in the y direction. See Problem 6.26 for more details.
axis charge on
charge on electrode stationary electrode
rotating disk stationary glass plate
---------------------
+++++++++++++++++++++ rotating disk (0.5 cm thick)
+++++++++++++++++++++
---------------------
0.6 cm space
3.9 cm
4.45 cm
10.55 cm
12 cm
19.45 cm
Figure 129
The left end of the charged region on the rotating disk is determined by the left end
of the electrode on the glass plate, which is at radius 4.45 cm. The right end of the
charged region is determined by the right end of the disk, which is at radius 10.55
cm. The mean radius of the charged part of the disk is therefore r = (4.45 cm +
10.55 cm)/2 = 7.5 cm. So the average velocity of the charges is
(Actually, this is the velocity at the average radius, and not the average velocity of
all points in the disk. But we’re just doing things roughly, so this distinction isn’t
important. For that matter, we could just pick a round number like r ≈ 0.1 m.) The
effective surface current density is then
J = σv = (1.5 · 10−5 C/m2 )(28.7 m/s) = 4.3 · 10−4 C/(s m). (482)
The combined effect of the two current sheets, each with surface current density J (in
the same direction), is to produce a horizontal field B = µ0 J both above and below
the disk. So
( )( )
B = 4π · 10−7 kg m/C2 4.3 · 10−4 C/(s m) = 5.4 · 10−10 T, (483)
152 CHAPTER 6. THE MAGNETIC FIELD
or 5.4 · 10−6 gauss. In terms of the various quantities, a few equivalent symbolic
expressions for the magnetic field are
2πrνµ0 ϵ0 V 2πrνV vV
B= = = 2 . (484)
d c2 d c d
We have used the speed at the mean radius to estimate the field strength B imme-
diately above the disk in that region. For a more accurate calculation of the field
strength to be expected at the location of the magnetometer needle, one could divide
the disk into circular rings and integrate over the whole distribution. That is what
Rowland did.
I V /R V /(ρL/A) V 1V
J= = = = = = 1.25 · 104 A/m2 .
A A A ρL (0.016 ohm-m)(0.005 m)
(486)
The drift velocity is then
The induced electric field is Et = vB = (39 m/s)(0.1 T) = 3.9 V/m. The Hall voltage
across the ribbon of width 0.002 m is therefore (3.9 V/m)(0.002 m) = 7.8 · 10−3 V, or
7.8 millivolts. Symbolically, the Hall voltage equals V Bw/ρLne, where w is the width.
Chapter 7
Electromagnetic induction
Solutions manual for Electricity and Magnetism, 3rd edition, E. Purcell, D. Morin.
morin@physics.harvard.edu (Version 1, January 2013)
153
154 CHAPTER 7. ELECTROMAGNETIC INDUCTION
the area enclosed by the loop is changing. The maximum rate of swept area equals
the maximum speed times ℓ. So the maximum induced voltage is
( )
dΦ
Emax = = Bℓvmax = 2πBℓνx0
dt max
= 2π(0.5 T)(0.018 m)(2000 s−1 )(3 · 10−4 m) = 0.034 V. (490)
This result depends linearly on all four of the given quantities, which makes intuitive
sense.
We can also solve this exercise by looking at the qvB magnetic force on the charges
in the wire. Multiplying this force by the distance ℓ along the wire over which it acts,
and dividing by q to obtain the work per charge, gives a voltage difference of vBℓ, as
above. This is maximum when v is maximum, since B and ℓ are constants.
B 2 b2 vπr2
F = . (491)
ρℓ
(You can check that this does indeed have units of force.) Since we are ignoring the
inertia of the frame, the applied
∑ force must be exactly∑equal to the magnetic force, in
magnitude. (If m = 0, then F = ma implies that F = 0.) For any particular F
that you pick, Eq. (491) can be solved for the velocity v that the frame will have. We
can now answer the various questions.
Eq. (491) implies that twice the force means twice the velocity. So a force of 2 N will
pull the frame out in half the time, or 0.5 sec.
Keeping everything else the same, doubling ρ means halving F (there is half as much
current). So a brass frame will be pulled out in 1 sec by a force of 0.5 N.
Doubling the radius increases F by a factor 22 = 4 (there is four times as much
z current). So a 1 cm aluminum frame will be pulled out in 1 sec by a force of 4 N. (We
current (side view) effectively have four of the original frames stacked on top of each other, each of which
out of page requires 1 N.)
h
B 7.25. Sliding loop
b/2 b/2
x In Fig. 130 the y axis points into the page. We’ve arbitrarily chosen the current in the
B v wire to flow in the negative y direction (out of the page), but the sign doesn’t matter
since all we care about is the magnitude of the emf.
√ At the leading edge of the square
Figure 130 loop, the magnitude of B is µ0 I/2πr, where r = h2 + (b/2)2 . Only the z component
matters in the flux, and this brings in a factor of (b/2)/r. So
µ0 I b/2 µ0 Ib
Bz = = . (492)
2πr r 4π(h2 + b2 /4)
155
At the trailing edge, Bz has the opposite sign. If the loop moves a small distance
v dt, there is additional positive flux through a thin rectangle with area b(v dt) at the
leading edge, and also less negative flux through a similar rectangle at the trailing
edge. Both of these effects cause the upward flux to increase. Therefore,
The flux is increasing upward. So for our choice of direction of the current in the
wire, the induced emf is clockwise when viewed from above, because that creates a
downward field inside the loop which opposes the change in flux. For h = 0 (or in
general for h ≪ b) E reduces to 2µ0 Iv/π. This is independent of b because the field at
the leading and trailing edges decreases with b, while the length of the thin rectangles
at these edges increases with b.
You can show that our result for E has the correct units, either by working them out
explicitly, or by noting that E has the units of B (which are the same as µ0 I/2πr)
times length squared divided by time, which correctly gives flux per time.
(a) Let v be the instantaneous velocity of the bar. The area of the circuit increases
at a rate b(v dt)/dt = bv, so the induced emf is E = dΦ/dt = Bbv. The current
is therefore I = E/R = Bbv/R. The general expression for the force on piece of
wire (the bar in our setup) is F = IBb, which yields B 2 b2 v/R here. So F = ma
gives (including the minus sign because the force opposes the motion, as you can
check with Lenz’s law and the right-hand rule)
∫ t 2 2 ∫ v ′
dv B 2 b2 v dv B b dv
−F = m =⇒ − =m =⇒ − dt′ = ′
dt R dt 0 mR v0 v
( )
B 2 b2 t v mR
=⇒ − = ln =⇒ v = v0 e−t/T , where T ≡ 2 2 . (494)
mR v0 B b
(You can check that T has units of time.) We see that the velocity decreases
exponentially, so technically the rod never stops moving (in an ideal world).
This exponential decay of v is a familiar result for forces that are proportional
to (the negative of) v.
(b) The total distance traveled in the limit t → ∞ is
∫ ∞ ∫ ∞ ∞
−t/T
−t/T v0 mR
x= v dt = v0 e dt = −v0 T e = v0 T = B 2 b 2 . (495)
0 0 0
(a) The magnetic field inside the solenoid is B(t) = µ0 nI(t) = µ0 nI0 cos ωt. Fara-
day’s law applied to the given ring yields
dΦ dB
E =− = −πr2 = πr2 µ0 nI0 ω sin ωt. (497)
dt dt
With the given positive direction of I, the right-hand rule gives the positive
direction of B as upward, and then also gives the positive direction of E as
counterclockwise when viewed from above (as for I). The current in the loop is
Iloop (t) = E/R = (πr2 µ0 nI0 ω/R) sin ωt.
(b) The force on a little piece of the ring is F (t) = Iloop (t) dl × B. With positive I
counterclockwise and positive B upward, this force is radial and equals
πr2 µ0 nI0 ω πr2 µ20 n2 I02 ω dl
F (t) = sin ωt · dl · µ0 nI0 cos ωt = sin ωt cos ωt. (498)
R R
The force is radially outward if this quantity is positive, inward if it is negative.
Since sin ωt cos ωt = (1/2) sin(2ωt) we see that the force is maximum outward
when ωt = π/4 (plus multiplies of π), and maximum inward when ωt = 3π/4
(plus multiplies of π).
(c) Since the force lies in the horizontal plane, it serves only to stretch/shrink the
ring (negligibly, if the ring is rigid).
Let’s now estimate roughly how large the resistance must be to make the effect of the
current in the loop negligible. The current in the loop at any instant is I ′ = E/R.
This causes a field B ′ and a flux Φ′ through the loop. Because E is changing with
time as the loop moves away from the wire, Φ′ is changing too, resulting in an extra
emf E ′ , which we have so far ignored. The question is, how large must R be so that
E ′ is negligible compared with E?
As a very rough estimate, we have B ′ ≈ µ0 I ′ /2πd, where d is a typical dimension of
the loop, say, d = 5 cm. The flux1 is then Φ′ ≈ B ′ A = (µ0 I ′ /2πd)wℓ, where ℓ is the
length of the loop (we could set ℓ ≈ w ≈ d here since we’re being rough, but we’ll keep
them separate). The (very rough) time characteristic of the change in Φ′ is τ = h/v,
where h is the mean distance from the loop to the wire (20 cm), and v is the speed of
the loop. So in order of magnitude, we have (using I ′ = E/R)
dΦ′ Φ′ µ0 I ′ wℓ v µ0 Ewℓv
E′ = ′
≈ = · = . (501)
dt τ 2πd h 2πRhd
Our goal is to have E ′ ≪ E, which is equivalent to
µ0 wℓv µ0 wℓv
≪ 1 =⇒ R ≫
2πRhd 2πhd
( )
4π · 10−7 kgC2m (0.08 m)(0.1 m)(5 m/s)
=⇒ R ≫ = 8 · 10−7 Ω. (502)
2π(0.2 m)(0.05 m)
This is roughly equal to 10−6 Ω, so the condition that the current in the loop is
negligible (more precisely, the condition that E ′ ≪ E) can be written as R ≫ 10−6 Ω.
This is a rather small resistance, so this bound is easily satisfied by a typical copper
wire.
We can alternatively write the condition in Eq. (502) as
µ0 wℓ 1 h
· ≪ . (503)
2πd R v
But from the definition of the self-inductance L, the above expression for Φ′ yields
L = µ0 wℓ/2πd. So the condition can be written as L/R ≪ τ . In other words, the
inductive time constant of the loop itself, L/R, should be short compared with the
time scale of the change of the externally induced emf.
Note that in the case where all of the above lengths (w, ℓ, h, d) are of the same order
of magnitude (which is the case here), the condition reduces to the simple expression
(ignoring the 2π): R ≫ µ0 v. We therefore see that the smallness of the above lower
bound on R (in SI units) comes from the smallness of µ0 (in SI units).
7.30. Work and dissipated energy
The induced emf is E = wv(B1 − B2 ), so the current is I = E/R = wv(B1 − B2 )/R.
From Lenz’s law it is counterclockwise when viewed from above. From the righthand
rule, the forward-directed force that must be applied to the loop to balance the retard-
ing magnetic force and keep the loop moving at constant speed is F = IB1 w −IB2 w =
Iw(B1 − B2 ). Using (B1 − B2 ) = IR/wv, the rate at which work is done is therefore
( )
IR
F v = Iw(B1 − B2 )v = Iw v = I 2 R, (504)
wv
1 Strictly speaking, we should expect the flux of B ′ to involve a factor like ln(r
loop /rwire ); see the com-
ments at the end of Section 7.8. But unless the wire is extremely thin, this logarithm won’t be a very large
number.
158 CHAPTER 7. ELECTROMAGNETIC INDUCTION
which agrees with the expression in Eq. (504), and hence equals I 2 R.
(a) The electric field caused by the changing flux is directed clockwise around the
∫b
solenoid, so the line integral a E · ds along path 1 equals +E0 . The voltage
Figure 133
159
∫b
difference Vb − Va ≡ − a E · ds therefore equals −E0 . Path 2 encloses zero
changing flux, so Va − Vb equals zero along path 2. We are assuming that a and
b are infinitesimally close to each other.
(b) We now have a simple electrostatic setup, so the voltage differences are path
independent. Point a at the higher potential, so we have Vb − Va = −E0 , and
Va − Vb = E0 (along any paths).
(c) The answers here are the same as in part (b), except with −E0 replaced with
−E0 /N , and E0 replaced with E0 /N . So in the N → ∞ limit, the potential
differences are zero.
The results are compiled in this table:
Path 1 Path 2
Vb − Va Va − Vb
Case (a) −E0 0
Case (b) −E0 E0
Case (c) 0 0
Cases (a) and (b) are equivalent for path 1, but not for path 2. Cases (a) and (c) are
equivalent for path 2, but not for path 1. Cases (b) and (c) are different for both paths.
The total voltage drop around a closed path is zero in cases (b) and (c) (consistent
with the fact that these are electrostatic setups), but nonzero in case (a).
7.33. Getting a ring to spin
The flux through the ring at any given time is Φ = πa2 B. From Faraday’s law, the
induced emf is E ∫= dΦ/dt = πa2 (dB/dt) (ignoring the sign). But by definition, the
emf is also E = E · ds = 2πaE. The tangential field is therefore E = E/2πa =
(a/2)(dB/dt). A little element of charge dq feels a tangential force E dq, and hence
a torque Ea dq. So the total torque is τ = Eaq = (qa2 /2)(dB/dt). The final angular
momentum acquired by the ring is therefore
∫ ∞ ∫ ∞ 2 ∫
qa dB qa2 0 qa2 B0
L= τ dt = dt = dB = − . (506)
0 0 2 dt 2 B0 2
We haven’t been keeping track of signs, so the minus sign here doesn’t mean much.
The correct statement to make is that Lenz’s law implies that the ring will spin
in the direction that has a positive right-hand-rule relation to the direction of the
initial magnetic flux. So you can quickly show that if q is positive, the direction of
L is the same as the direction of B0 . The angular momentum can be written as
L = Iω = (ma2 )ω, so we have
L qB0
ω= = , (507)
ma2 2m
as desired. We see that L depends only on the net change in B, and not on the
rate at which this change comes about. Intuitively, if B changes more slowly, then
E (and hence the torque) is smaller, so the angular momentum increases at a lesser
rate. But the process takes longer, so the increase goes on for a longer time. These
two competing effects exactly cancel.
Note that ω is independent of a. This is due to the fact that E is proportional to a,
which means that the linear acceleration in the tangential direction, and hence the
tangential speed v at a given time, is proportional to a. The angular velocity, ω = v/a,
160 CHAPTER 7. ELECTROMAGNETIC INDUCTION
Figure 135 on either side of the switch, and a spark might jump across. But the existence of coil #2 allows the current
in coil #1 to stop.
50 A
10-3 s
Figure 136
161
(a) In Fig. 7.40(a), if I2 is increasing we have an increasing upward flux through the
top circuit. This will induce an E1 in a direction to drive current that makes a
downward flux; this direction is opposite to the positive direction assigned to E1 .
Hence the sign in the first equation must be a “−.” A similar argument shows
that the sign in the second equation is also a “−.”
Had we assigned the opposite positive directions for I2 and E2 , the sign in front
of M in both equations would be a “+.”
(b) In Fig. 7.40(b), the current I is the same in both coils, so I = I1 = I2 . And the
emfs add, so the total emf is E = E1 + E2 . If we add the two given equations, we
obtain
dI dI dI dI dI
E = −L1 −M − L2 −M = −(L1 + L2 + 2M ) . (510)
dt dt dt dt dt
The circuit is therefore equivalent to a single coil with L′ = L1 + L2 + 2M .
In Fig. 7.40(c), the current is now I = I1 = −I2 , and the emf is E = E1 − E2 . If
we subtract the two given equations, we obtain
dI d(−I) d(−I) dI dI
E = −L1 −M + L2 +M = −(L1 + L2 − 2M ) . (511)
dt dt dt dt dt
The circuit is therefore equivalent to a single coil with L′′ = L1 + L2 − 2M . Since
M is positive, we see that L′ is larger than L′′ .
(c) A circuit with L < 0 would violate Lenz’s law; it would be unstable. That is, an
increasing current would cause an emf that would increase the current even more.
This would violate conservation of energy. Therefore, we must have L′′ ≥ 0, which
implies that M ≤ (L1 + L2 )/2 for any pair of circuits. Equality is achieved if
we have two identical coils wound right on top of each other. (An even stronger
inequality, M 2 ≤ L1 L2 , can be derived by considering coils connected in parallel.
This is indeed stronger, due to the arithmetic-geometric-mean inequality.)
Another (less enlightening) way of obtaining the M ≤ (L1 + L2 )/2 result is
the following. The energy stored in a system of two inductors is U = L1 I12 /2 +
L2 I22 /2+M I1 I2 (see Problem 7.10). As you can check, this energy can be written
as ( )
L1 + L2 L2 2
U= − M I12 + M (I12 + I1 I2 ) − (I − I22 ). (512)
2 2 1
This holds for any I1 and I2 . In particular, if I2 = −I1 , then only the first of
the three terms is nonzero. Since the energy must be positive, we must therefore
have M ≤ (L1 + L2 )/2.
flux through the outer ring comes not only from the field lines pointing upward in the
interior of the inner ring, but also from the field lines pointing downward in the region
between the rings. The latter flux partially cancels the former flux. The larger the
outer ring is, the larger this canceling effect is, and so the smaller the net flux is. The
field lines within the dotted curves yield a net flux of zero through the outer ring, so
it is only the lines in the central region that contribute to the net flux. The larger the
outer ring is, the smaller this central region is.
ring 1
ring 2
Figure 137
∂Φ21 µ0 πR22 I1
∆Φ21 = ∆R1 = − ∆R1 . (513)
∂R1 2R12
Now consider a current I2 in the inner ring. Let B2 be the desired field due to the
inner ring at the location of the outer ring at radius R1 . If we expand the outer ring
by ∆R1 , the flux Φ12 through this ring decreases by the amount of flux in the annular
region between radii R1 and R1 + ∆R1 (Exercise 7.37 explains why it is a decrease).
The area of this region is 2πR1 ∆R1 , so the decrease in flux is ∆Φ12 = −B2 2πR1 ∆R1 .
Now, if our theorem Φ21 /I1 = Φ12 /I2 always holds, in particular if it holds for any
value of R1 , then it must also hold if the Φ’s are replaced with ∆Φ’s. That is, we must
also have
∆Φ21 ∆Φ12 µ0 πR22 B2 2πR1 ∆R1 µ0 R22 I2
= =⇒ − 2 ∆R1 = − =⇒ B2 = . (514)
I1 I2 2R1 I2 4R13
We can pick any radius for the outer ring, so in more general notation we can write
B = µ0 R2 I/4r3 at any point at radius r in the plane of a ring with radius R and
current I, if r ≫ R. Note that this result can be written as B = µ0 (πR2 )I/4πr3 =
(µ0 /4π)(IA/r3 ), where A is the area of the ring. We will have more to say about this
form of B in Chapter 11.
163
We have neglected the fact that the field inside the solenoid is not constant. It de-
creases near each end, to the point where the flux through the last turn is only about
half the flux through a turn in the middle (see Exercise 6.63). This means that we
have overestimated the inductance; the true value is smaller than the above approx-
imate result. We might expect the error to be on the order of the diameter divided
by the length, because the diameter should determine the length scale of the region
near the end where the field differs appreciably from the idealized B = µ0 nI value.
This is 5% in the present example. The actual error is only about 2% in this case
(which is consistent with 5%, as far as order of magnitude goes), as you can discover
by referring to tables that give exact values for the inductance of cylindrical coils.
VA VB The circuit in Fig. 140 is a simple RL circuit, so as time goes on, the current
VB
10 V 1 ms 2 ms equals I(t) = I0 e−(R/L)t , where I0 = 0.2 A and where the time constant L/R equals
(0.1 H)/(200 Ω) = 5 · 10−4 s = 0.5 millisec. The potentials at A and B are propor-
tional to I, so they decrease like e−(R/L)t . After 0.5 millisec they have decreased by
VA a factor 1/e = 0.37, and after 1 millisec by 1/e2 = 0.14. After 5 millisec the factor is
1/e10 = 4.5 · 10−5 , which is negligible. The plots are shown in Fig. 141. We have only
-30 V plotted up to t = 2 millisec, because the curves are essentially zero after that. Note
the discontinuity in VA .
Figure 141
7.42. RL circuit
( )
From Eq. (7.69) the current is I(t) = I0 1 − e−(R/L)t , where I0 = E0 /R. In the
problem at hand,
E0 12 V R 0.01 Ω
I0 = = = 1200 A and = = 20 s−1 . (516)
R 0.01 Ω L 0.5 · 10−3 H
So the time scale is L/R = 0.05 s. The current reaches a value of (0.9)I0 when
At this time, the current is I = (0.9)(1200) = 1080 A, so the energy stored in the
magnetic field is
1 2 1
LI = (0.5 · 10−3 H)(1080 A)2 = 292 J. (518)
2 2
The instantaneous power delivered by the battery is E0 I, but since I is changing we
must perform an integral to find the energy delivered by the battery between t = 0
and t = 0.115 s:
∫ ∫
t=0.115 s ( t
′)
E0 I(t′ ) dt′ = E 0 I0 1 − e−(R/L)t dt′
0 0
( ) t
′ L −(R/L)t′
= E 0 I0 t + e
R
( 0
)
L −(R/L)t L
= E 0 I0 t + e −
R R
( )
= (12 V)(1200 A) 0.115 s + (0.05 s)(0.1) − (0.05 s)
= 1008 J. (519)
From conservation of energy, apparently 1008 J − 292 J = 716 J has been dissipated
in the resistor. The task of Problem 7.15 is to show that the energy delivered by
the battery does indeed equal the energy stored in the magnetic field plus the energy
dissipated in the resistor, at any general time t.
The volume of the galaxy is πr2 h = π(5 · 1020 m)2 (1019 m) ≈ 8 · 1060 m3 , so the total
energy contained in the magnetic field is (3.6 · 10−14 J/m3 )(8 · 1060 m3 ) ≈ 3 · 1047 J.
The radiation power of starlight in the galaxy is 1037 J/s, so the magnetic energy is
worth (3 · 1047 J)(1037 J/s) = 3 · 1010 s ≈ 1000 years.
7.45. Magnetic energy near a neutron star
The energy density is
B2 (1010 T)2
= ( ) = 4 · 1025 J/m3 . (523)
2µ0 2 4π · 10−7 kgC2m
7.47. A dynamo
The device on the bottom is the dynamo. To see why, suppose a current I is flowing in
the coil, in the clockwise direction in the upper circular loop. Such a current produces
a magnetic field B pointing downward through the disk. With v the velocity of any
part of the rotating disk, v × B is a vector pointing radially outward. Positive charges
in the disk will be pushed outward, negative charges pushed inward. Either effect
causes current to flow in the direction postulated. If we assume the opposite direction
for the coil current I, the field B and hence the cross product v × B will be reversed,
and the force will again be in the direction to sustain or increase the current. This
conclusion is independent of the sign of the mobile charges. You can quickly verify
that in the device on the top, the effect of the qv × B force on the charges in the disk
is to decrease whatever current happened to be flowing in the coil.
See if you can formulate an unambiguous rule to distinguish the potential dynamo on
the bottom from the non-dynamo on the top, a rule that refers only to the relation
of disk rotation to coil configuration. Would a mirror image of the figure on the top
represent a dynamo?
A dynamo of this kind runs equally well with current in either direction. The current
can also be zero. However, in any circuit not at absolute zero there are slight random
motions of charge, or randomly fluctuating currents. Some fluctuation, tremendously
amplified by the “positive feedback” of the dynamo action, becomes the steady dynamo
current. It retains the direction of its initial excitation. (In a conventional dc generator
there is some residual magnetic field in the iron poles, even at zero current, which
suffices to determine the eventual polarity.)
The magnitude of the current in this purely ohmic dynamo is determined by the input
mechanical power. The current will be such that the ohmic power loss in the coil and
disk is precisely equal to the applied torque times the shaft’s angular speed.
Chapter 8
Alternating-current circuits
Solutions manual for Electricity and Magnetism, 3rd edition, E. Purcell, D. Morin.
morin@physics.harvard.edu (Version 1, January 2013)
∆VC ∆VL UC UL
t=0 V0 −V0 CV02 /2 0
t = π/2ω 0 0 0 CV02 /2
Note: At t = 0 you can also work out the voltage across the inductor directly, to double
check that it equals −V0 . Using above form of√I(t), the voltage across the inductor
is −L(dI/dt) = −ω 2 LCV0 cos ωt. With ω = 1/ LC, this becomes −V0 cos ωt, which
equals −V0 at t = 0. However, it’s risky to trust this minus sign. The magnitude is
certainly correct, but it’s best to check the sign by thinking about things physically.
At t = 0 the current is zero but is increasing in the clockwise direction. The voltage
above the inductor must therefore be higher than the voltage below; this difference is
what causes the current to increase.
8.17. Amplitude after Q cycles
After Q cycles, the angle ωt equals 2πQ. But from Eq. (8.13) we know that Q = ω/2α.
So after Q cycles, the angle ωt equals 2π(ω/2α) = π(ω/α). The time t is therefore
given by t = π/α. The exponential factor e−αt that appears in I(t) and V (t) therefore
equals e−π as desired.
167
168 CHAPTER 8. ALTERNATING-CURRENT CIRCUITS
(a) The impedance of the 105 Ω resistor is much larger than the impedances of the
other circuit elements we will find below, so we can neglect the current through
the 105 Ω resistor, at least during the initial period right after the switch is closed.
During this period we therefore essentially have a series RLC circuit in the right
loop.
There are two things we can estimate by looking at the given plot: the frequency
of oscillation and the rate of decay of the signal. These two things will allow us
to calculate C and R, respectively.
In cases where the voltage doesn’t immediately become negligible after a few
oscillations, the 1/LC term in Eq. (8.9) √ dominates (see Problem 8.5), so the
frequency is essentially given by ω = 1/ LC. In the given plot, four cycles are
completed in 10−3 sec,1 so ω = 2π · 4/(10−3 s) = 2.5 · 104 s−1 . The capacitance is
then
1 1
C= 2 = −1
= 1.6 · 10−7 F. (532)
ω L (2.5 · 10 s )2 (0.01 H)
4
(b) The decay constant is α = R/2L, so the time for the voltage to decrease by
a factor of 1/e is t = 2L/R. From the figure, this time is about 0.5 · 10−3 s.
Therefore,
2L 2L 2(0.01 H)
t= =⇒ R = = = 40 Ω. (533)
R t 0.5 · 10−3 s
As promised, this is negligible compared with the 105 Ω resistor. Addition-
ally (although at this point we technically haven’t gotten to impedances in this
chapter), the impedances of the inductor and capacitor have magnitudes ωL =
(2.5 · 104 s−1 )(0.01 H) = 250 Ω and 1/ωC = 1/(2.5 · 104 s−1 )(1.6 · 10−7 F) = 250 Ω.
(These are equal because we are using ω 2 = 1/LC.) These are also negligible
compared with 105 Ω. But it was fine to just assume that here.
1 In the first printing of the book, the millisec span was drawn a little too long.
169
(c) After a long time, we essentially have just two resistors of 105 Ω and 40 Ω in series
with a 20 V dc battery. (No current passes through the capacitor in the eventual
direct-current steady state. And there is no voltage drop across the inductor for
constant current.) So the voltage across the oscilloscope is
( )
40
V40 Ω = (20 V) = 0.008 V, (534)
105 + 40
or 8 millivolts. The 40 in the denominator is of course inconsequential.
where we have used c2 = 1/µ0 ϵ0 . The fields are shown in Fig. 142. The electric
field spans the s gap. From Exercise 6.61, we know that the magnetic field points
tangentially around the toroid. The current flows “around” the toroid in the same Figure 142
manner as it did in Exercise 6.61, except that it doesn’t complete a full loop; the
charge simply piles up on the end of the inner cylinder and on the corresponding part
of the top face of the torus (which completes the capacitor). The charge oscillates
back and forth. At the instants when the electric field is maximum, the current is zero
so there is no magnetic field. At the instants when the magnetic field is maximum,
the charge on the capacitor is zero so there is no electric field.
8.21. Solving an RLC circuit
Let I1 and I2 be the loop currents in the left and right loops, respectively, with
clockwise taken to be positive. Let V be the voltage across the capacitor, taken to be
positive when the upper plate of the capacitor is positive. The statements involving
the three equal voltage drops are
Q dI2
V = , V = R′ (I1 − I2 ), V =L . (536)
C dt
But I1 = −dQ/dt, so if we take the second derivative of the first statement, and also
the first derivative of the second statement, we obtain
( )
d2 V 1 dI1 dV ′ dI1 dI2 dI2
2
=− , =R − , V =L . (537)
dt C dt dt dt dt dt
We can now eliminate I1 and I2 in favor of V by plugging the results for dI1 /dt and
dI2 /dt from the first and third equations into the second. The result is
( ) ( ) ( )
dV ′ d2 V V d2 V 1 dV 1
= R −C 2 − =⇒ + + V = 0. (538)
dt dt L dt2 R′ C dt LC
Using the same trial solution for V as in Eq. (8.4) (and Eq. (8.5)), substituting into
Eq. (538), and demanding that the coefficients of sin ωt and cos ωt independently
vanish, we obtain, in place of Eq. (8.7),
ω α 1
2αω − =0 and α2 − ω 2 − + = 0, (539)
R′ C ′
RC LC
170 CHAPTER 8. ALTERNATING-CURRENT CIRCUITS
Alternatively, note that Eq. (538) is identical to Eq. (8.2) if 1/R′ C = R/L, that is,
if R = L/R′ C. So we can quickly obtain the results in Eq. (540) by simply replacing
the R in Eqs. (8.8) and (8.9) with L/R′ C.
Now let’s assume that L, C, and Q are the same in the series and parallel circuits.
Since Q = ω/2α, we just need to equate the values of ω/2α for the two circuits. Using
Eqs. (8.8) and (8.9), along with Eq. (540), you can show that the values of ω/2α are
equal if 1/R′ C = R/L =⇒ R′ = L/RC. In view of the preceding paragraph, this is
no surprise, because the time dependence of V is exactly the same in the two circuits
if R′ = L/RC. And the energy (which appears in the definition of Q) is proportional
to V 2 .
R 600 Ω 4L 4 · 10−4 H 1
= = 3 · 106 s−1 and = = . (543)
2L 2 · 10−4 H 2
R C (600 Ω)2 (10−8 F) 9
Therefore,
( √ )
β1 = (3 · 106 s−1 ) 1 + 8/9 = 5.83 · 106 s−1 ,
( √ )
β2 = (3 · 106 s−1 ) 1 − 8/9 = 0.172 · 106 s−1 . (544)
The constants A and B are determined by the initial conditions. In the setup in
Fig. 8.4, the current is zero at t = 0 (because it can’t increase abruptly, due to
the inductor), so dQ/dt = 0. And since Q = CV , our initial condition is therefore
dV /dt = 0 at t = 0. From Eq. (542), the value of dV /dt at t = 0 is −β1 A − β2 B. This
equals zero if B/A = −β1 /β2 ≈ −34.
Since β2 is much smaller than β1 , the Be−β2 t term goes to zero much more slowly
than the Be−β1 t term. After a microsecond or so, the Be−β2 t term is essentially all
that is left.
1 1 i ( )
= − + (3.14 · 10−5 )i = 10−3 1 − 7.93i Ω−1 . (555)
Z 1000 125.6
Note that the effect of the capacitor is very small. The impedance is
1 103 (1 + 7.93i)
Z= = = (15.7 + 124i) Ω. (556)
10−3 (1 − 7.93i) 1 + 7.932
The inductor dominates the admittance, so Z is roughly equal to iωL. The frequency
is low enough so that the inductor lets current through easily compared with the
resistor and capacitor.
For 10 MHz we have ω = 2π(107 s−1 ) = 6.28 · 107 s−1 . Therefore,
1 1 i ( )
= − + (3.14 · 10−2 )i = 10−3 1 + 31.4i Ω−1 . (557)
Z 1000 1.256 · 10 5
1 103 (1 − 31.4i)
Z= = = (1.01 − 31.8i) Ω. (558)
10−3 (1 + 31.4i) 1 + 31.42
The capacitor now dominates the admittance, so Z is roughly equal to 1/iωC. The
frequency is high enough so that the capacitor lets current through easily compared
with the resistor and inductor.
174 CHAPTER 8. ALTERNATING-CURRENT CIRCUITS
To find the frequency for which |Z| is largest, Eq. (554) gives
1 1
Z= =⇒ |Z| = √ . (559)
1/R + i(ωC − 1/ωL) (1/R)2 + (ωC − 1/ωL)2
This is maximum when the denominator is smallest, that is, when
1 1 1
ωC = =⇒ ω = √ =√ = 106 s−1 , (560)
ωL LC (2 · 10−3 H)(5 · 10−10 F)
R(iωL) 1 iR2 ωL + Rω 2 L2 i
=R+ =⇒ 2 2 2
=R− . (565)
R + iωL iωC R +ω L ωC
Equating the real and imaginary parts on the left and right sides of this equation gives
Rω 2 L2 R2 ωL 1
=R and =− . (566)
R2+ ω 2 L2 R2
+ω L2 2 ωC
The left equation is true if R = 0 or if L = ∞ (or if ω = ∞, but it is understood
that we want to find a condition that works for all ω). In the right equation, the
negative sign implies that the only way the relation can be true is if both sides are
zero. So we must have C = ∞ and either R = 0 or L = ∞. (Again, ω = ∞ works
too.) The condition for the two circuits to have equal impedances is therefore: C = ∞
and either R = 0 or L = ∞. In the case where C = ∞ and R = 0, the impedance
of both circuits is zero; the capacitor and resistor are short circuits (the impedance
of the inductor is nonzero, but that doesn’t matter). In the case where C = ∞ and
L = ∞, the impedance of both circuits is R; the capacitor is a short circuit and the
impedance of the inductor is infinite.
Physically, the reason why there are only a couple special solutions, both of which
involve some infinities, is that the magnitude of the impedance of the parallel combi-
nation is less than or equal to R, while for the series combination it is greater than or
equal to R.
Note: if you solve this exercise by multiplying Eq. (565) through by (R + iωL)(iωC)
and simplifying, you must be careful. A spurious solution is introduced, and a valid
solution is missed.
8.31. Zero voltage difference
Both branches have the same voltage difference V0 , so the complex currents are given
by
V0 V0
IA = and IB = . (567)
R1 + 1/iωC R2 + iωL
The voltage at A is VA = V0 − IA (1/iωC), and at B is it VB = V0 − IB R2 . Therefore,
V0 V0 1
VB − VA = −IB R2 + IA (1/iωC) = − · R2 + ·
R2 + iωL R1 + 1/iωC iωC
V0 R2 V0
= − + . (568)
R2 + iωL iωCR1 + 1
This vanishes if
L
R2 (iωCR1 + 1) = R2 + iωL =⇒ iωCR1 R2 = iωL =⇒ R1 R2 = , (569)
C
as desired. Given a known capacitance C and known resistors R1 and R2 , we could
determine an unknown L by adjusting C, or R1 , or R2 , until we obtain VB − VA = 0.
But a real inductor generally has some resistance too, so we effectively have another
resistor r in series with R2 and L. This makes things more complicated.
-6
L R = 35 Ω C = 10 F
176 CHAPTER 8. ALTERNATING-CURRENT CIRCUITS
15.5 V
8.32. Finding L
The setup is shown in Fig. 143. We can quickly determine the amplitude of the current
10.1 V
(or the rms value, depending on what the voltmeter is calibrated to read; the final
Figure 143 value of L won’t depend on the choice). The frequency is ω = 2π(1000 s−1 ) = 6283 s−1 ,
so the V0 = I0 |Z| statement for the capacitor alone yields
I0
15.5 V = =⇒ I0 = (15.5 V)(6283 s−1 )(10−6 F) = 0.0974 A. (570)
ωC
Since the elements are in series, this is the current through all of the components in
the circuit. The V0 = I0 |Z| statement for the whole circuit then tells us that (ignoring
the units)
V0 10.1
I0 = =⇒ 0.0974 = √ =⇒ ωL − 1/ωC = ±97.6 Ω.
|Z| (35) + (ωL − 1/ωC)2
2
(571)
Note that there are two roots. We therefore have
1 97.6
L= ± =⇒ 0.0253 ± 0.0155 =⇒ L = 0.041 H or 0.0098 H. (572)
ω2 C ω
So L could be 41 mH or 9.8 mH. The amplitude of the voltage across the inductor alone
is I0 ωL, which gives 25.1 V and 6.0 V for the two possibilities. If we then measure the
voltage across the inductor and obtain 25.4 V, the second possibility is ruled out, and
we have reasonably good agreement with the computed value of 25.1 V.
1 1 1 1 0.64 · 10−6 iω
= + = +
Z 6
5000 1250 + 10 /(0.64)iω 5000 8 · 10−4 iω + 1
(1 + 8 · 10−4 iω) + 32 · 10−4 iω
=
5000(1 + 8 · 10−4 iω)
R2 5000 + 4iω
=⇒ Z = , (574)
R1 1 + 4 · 10−3 iω
C1
in agreement with an intermediate step for the top circuit.
Now let’s find the general rules for constructing the bottom circuit, given the top
circuit. Consider the general circuits shown in Fig. 144. If these two circuits are to
R3 have the same impedance for all values of ω, then in particular they must have the
same impedance in the ω → 0 and ω → ∞ limits. In the ω → 0 limit, no current
C2 goes through the capacitors, so we must have R1 + R2 = R3 (which is indeed true
R4
for the given circuits). In the ω → ∞ limit, the capacitor has no impedance, so we
Figure 144
177
Figure 145
178 CHAPTER 8. ALTERNATING-CURRENT CIRCUITS
8.35. RC circuit
i i
Z =R− = 2000 Ω − = (2000 − 2650i) Ω. (578)
ωC (377 s )(10−6 F)
−1
√
The magnitude is |Z| = 20002 + 26502 = 3320 Ω.
(b) The rms voltage is V = 120 V, so the rms current is
V 120 V
I= = = 0.036 A. (579)
|Z| 3320 Ω
There is no need for a factor of 1/2 since we are using rms values. We can
alternatively use the P = V I cos ϕ expression (where these are the rms values).
Here V = 120 V, I = 0.036 A, and tan ϕ = 2650/2000 which yields cos ϕ = 0.60.
These quantities yield P = 2.6 W, as desired.
(d) A voltmeter connected across the resistor will read
I
VC = = (0.036 A)(2650 Ω) = 95 V (rms). (582)
ωC
(e) The amplitudes of the voltages associated with the above rms values are 102 V
and 134 V. The voltages across the resistor and capacitor are 90◦ out of phase,
with the resistor ahead of the capacitor. (Remember, in general we have VL
ahead of VR ahead of VC .) So the pattern will be an ellipse, as shown in Fig. 146.
If the plates are connected in the natural way as shown, then the ellipse is traced
out counterclockwise. To see why, consider an instant when the current through
134 V the resistor is maximum downward, in which case the right plate of the tube is
B at a higher potential (so the electrons are deflected that way). The charge on
the capacitor is 90◦ out of phase with the current, so there is no charge on the
A capacitor at this moment. The voltage across the capacitor is therefore zero, so
102 V the electrons are at the point A in the figure.
A quarter cycle later, the top plate of the capacitor will have maximum charge, in
which case the top plate of the tube is at a higher potential (so the electrons are
deflected that way). And the current is zero at this moment, so the voltage across
Figure 146 the resistor is zero. The electrons are therefore at point B in the figure. We see
that the curve on the screen passes point B a quarter cycle after point A. So
the curve is traced out counterclockwise. On the other hand, if the connections
are made in the reverse manner for either of the elements, then the curve would
be traced out clockwise. If both connections are reversed, then the trace reverts
back to counterclockwise. Without being told which way the connections are
made, there is no way to know the direction of the trace.
179
R
= 3ω = 3 · 2π · 100 s−1 ≈ 2000 s. (584)
L
This is satisfied by, for example, R = 100 Ω and L = 0.05 H. The power is proportional
to V 2 . If ω ≪ R/L, then we can ignore the “1” in the denominator. So |Ṽ1 /V0 |2 ≈
ω 2 L2 /R2 ∝ ω 2 . Halving ω decreases this by a factor of 4.
The physical reason why V1 decreases with decreasing frequency is the following. For
very high frequency, the impedance of the inductor is very large. The impedance of
the resistor is negligible in comparison, so essentially all of the V0 voltage drop occurs
across the inductor, which is what V1 registers. On the other hand, for very low
frequency, the impedance of the inductor is very small; it is essentially a short circuit.
Therefore, essentially all of the V0 voltage drop occurs across the resistor. Very little
occurs across the inductor which, again, is what V1 registers.
As in Problem 8.13, adding on another RL loop would square the attenuation effect.
The voltage would then be proportional to ω 4 .
1/R
cos ϕ = √ . (585)
(1/R)2 + (ωC − 1/ωL)2
Equation (8.84) therefore gives the average power delivered to the circuit as
1
P = E0 I0 cos ϕ
2
1 √ 1/R
= E0 · E0 (1/R)2 + (ωC − 1/ωL)2 · √
2 (1/R) + (ωC − 1/ωL)2
2
1 E02
= . (586)
2R
The average power dissipated in the resistor is given by Eq. (8.80), where V0 is the
voltage across the resistor. But this voltage is simply E0 because we have a parallel
circuit. So
1 E02
PR = , (587)
2R
in agreement with Eq. (586).
180 CHAPTER 8. ALTERNATING-CURRENT CIRCUITS
(a) The impedance of the capacitor is ZC = 1/iωC. But since ω = 1/RC here, we
have ZC = −iR. Using the standard rules for adding impedances in series and
parallel, the total impedance of the circuit is
Alternatively, we can find the power dissipated by finding the amplitude of the
voltage across each of the resistors and then using PR = (1/2)VR2 /R for each.
(The resistors are the only places where power is dissipated.) The complex voltage
across the right resistor is simply E0 . The complex voltage across the left resistor
is E0 ZR /(ZR + ZC ) = E0 /(1 − i), because the complex voltages across the R and
C in series are proportional to their impedances. The magnitudes of these two
complex√voltages (which are the amplitudes of the two actual voltages) are E0
and E0 / 2. Adding the VR2 /2R powers for each resistor gives 3E02 /4R, as above.
Chapter 9
181
182 CHAPTER 9. MAXWELL’S EQUATIONS AND E&M WAVES
Since s ≪ b we can neglect the edge fields, in which case the displacement current Jd is
uniformly distributed in the gap. The integral of Jd over the area of the plates
∫ equals
the conduction current I in the wire (see Exercise 9.13). The fraction of Jd · da = I
that is enclosed in a circle through P , centered on the axis, is πr2 /πb2 . The integral
law applied to this circle therefore gives (with the conduction current J = 0 inside the
capacitor) ( 2)
r µ0 Ir
2πrB = µ0 I 2 + 0 =⇒ B = , (595)
b 2πb2
as desired. The similarity of this calculation to the calculation of the E field in Fig. 7.16
is the following. If we solve the problem straight from Maxwell’s equation, without
invoking the definition of the displacement current, we can write (with J = 0 inside
the capacitor)
∫ ∫
∂E dΦE
B · ds = µ0 ϵ0 · da =⇒ 2πrB = µ0 ϵ0 , (596)
C S ∂t dt
where ΦE is the flux of the electric field through the given surface. This equation
is exactly analogous to Faraday’s law of induction, which we used in the example of
Fig. 7.16 (among many other places),
∫ ∫
∂B dΦB
E · ds = − · da =⇒ 2πrE = − . (597)
C S ∂t dt
The similarity arises because of the symmetry of the two “curl” Maxwell’s equations;
and also because there is no current J of real electric charges inside the capacitor in the
present problem, and likewise there is no current of real magnetic charges in Fig. 7.16
(or anywhere else) because magnetic monopoles don’t exist (as far as we know).
v dt
9.16. Changing flux from a moving charge
A time dt later, the whole field pattern has moved a distance v dt to the right. So
v r
all of the electric flux that initially passed through the circle still passes through, but
θ
some additional flux now passes through. This additional flux is the flux that passes
q θ
through the left circle in Fig. 147 but doesn’t pass through the right circle. (If you
imagine the left circle riding along with the field pattern, it becomes the right (fixed)
circle after a time dt.) Equivalently, the additional flux is the flux that passes through
Figure 147 the cylindrical ring between the two circles, with radius r and width v dt. The area of
this cylinder is 2πrv dt, so the change in electric flux through the fixed right circle is
dΦE = (2πrv dt)E sin θ, where the sin θ factor comes from the fact that in finding the
flux through the cylinder, we are concerned only with the component perpendicular to
the x axis. Maxwell’s equation is therefore satisfied if the magnitudes E and B satisfy
∫
1 dΦE 1 v
B · ds = 2 =⇒ 2πrB = 2 2πrvE sin θ =⇒ B = 2 E sin θ. (598)
c dt c c
In view of the given relation B = (v/c2 ) × E, this is indeed true. The magnitude
is correct because the cross product between v and E generates the sin θ. And the
sign is correct because the flux increases to the right, which by the right-hand rule
corresponds to B pointing in the counterclockwise direction when viewed from the
right. This is consistent with the direction of B obtained from B = (v/c2 ) × E.
9.17. Gaussian conditions
In free space, the two “curl” Maxwell’s equations in Gaussian units are
1 ∂B 1 ∂E
∇×E=− and ∇×B= . (599)
c ∂t c ∂t
183
(b) Using the “Faraday” Maxwell equation, the same argument with a loop perpen-
dicular to the page (lying in the horizontal plane) gives
∫
dΦB
E · ds = − =⇒ Eℓ = −(−Bℓv) =⇒ E = Bv. (609)
dt
The minus sign in the flux comes from the fact that if the loop is traced out in the
counterclockwise direction when viewed from above, the right-hand rule defines
185
Figure 150
186 CHAPTER 9. MAXWELL’S EQUATIONS AND E&M WAVES
What do the B field lines look like for general x and y values? Again, since the tangent to
the field line is in the direction of B, we have the general relation,
dy By sin kx cos ky
= =− . (612)
dx Bx cos kx sin ky
Separating variables and integrating gives ln(cos ky) = − ln(cos kx) + D, where D is a con-
stant. Exponentiating gives cos kx cos ky = C, where C = eD is another constant. Small
values of C yield near-squares close to the boundary of the box, and values close to 1 yield
the small near-circles close to the origin we found above. Note that the cos kx cos ky = C
curves of the B field lines are also the curves of constant Ez , which we found above and
plotted in Fig. 149. This can be traced to the fact that if E has only a z component, then
∇ × E is perpendicular to ∇Ez , as you can verify.
E2
S= =⇒ E 2 = (377 Ω)(10−8 W/m2 ) =⇒ Erms ≈ 0.002 V/m, (613)
377 Ω
or 2 millivolts/meter.
9.25. Microwave background radiation
As shown in Section 9.6, the average energy density U of a sinusoidal electromagnetic
wave is U = ϵ0 E02 /2 = ϵ0 Erms
2
. So we have
U 4 · 10−14 J/m3
2
Erms = = = 4.5 · 10−3 V2 /m2 =⇒ Erms = 0.067 V/m. (614)
ϵ0 8.85 · 10−12 kg
s2 C2
m3
If the 1 kilowatt radiated by the transmitter is spread out over a sphere of radius R,
then the power density at radius R equals S = (103 W)/4πR2 . The energy density is
then U = S/c. We therefore want
1 103 W
· = 4 · 10−14 J/m3 =⇒ R = 2600 m, (615)
c 4πR2
or 2.6 km. However, the power is undoubtedly emitted in at least a somewhat directed
manner, so the distance from an actual radio transmitter would be larger than this.
9.26. An electromagnetic wave
1 1( s2 C2 ) 3
ϵ0 E02 = 8.85 · 10−12 (10 V/m)2 = 4.4 · 10−6 J/m3 . (619)
2 2 kg m3
The power density equals the energy density times the speed, so
1
S= ϵ0 E02 c = (4.4 · 10−6 J/m3 )(3 · 108 m/s) = 1300 J/(m2 s). (620)
2
9.27. Reflected wave
Let Ei be the amplitude of the oscillating electric field of the incident wave, and√Er
that of the reflected wave. If half of the incident energy is reflected, then Er = Ei / 2.
As in Section 9.5, the incident electric wave is described by Eq. (9.30) with E0 → Ei ,
and the reflected wave is described by Eq. (9.29) with E0 → Er . Using the trig sum
formulas, you can check that the sum of these two waves is
2πy 2πct 2πy 2πct
E = ẑ(Ei + Er ) sin cos + ẑ(Ei − Er ) cos sin . (621)
λ λ λ λ
At points where y equals λ/4, 3λ/4, 5λ/4, etc., the second term is zero, and the
sin(2πy/λ) factor in the first term equals ±1. So E oscillates with amplitude Ei + Er .
At these points the two oscillating electric fields are, and remain at all times, in phase.
(In the setup in Fig. 9.10 the mirror was a perfect conductor, so Er = Ei .)
Similarly, at points where y equals 0, λ/2, λ, 3λ/2, etc., the first term is zero, and
the cos(2πy/λ) factor in the second term equals ±1. So E oscillates with amplitude
Ei − Er . At these points the two oscillating electric fields are, and remain at all times,
exactly 180◦ out of phase. (In Fig. 9.10 these points have E = 0 at all times, because
Er = Ei .) Note that at these points E reaches its maximum value a quarter cycle
before or after the maximum at the points in the previous paragraph, due to the
sin(2πct/λ) versus cos(2πct/λ) dependence.
√
In our case with Er = Ei / 2, the ratio of maximum amplitude observed to minimum
amplitude observed is √
Ei + Er 1 + 1/ 2
= √ = 5.83. (622)
Ei − Er 1 − 1/ 2
9.28. Poynting vector and resistance heating
The electric field inside the wire is given by E = J/σ. Since the curl of E is zero,
we can draw a thin rectangular loop along the surface to show that the electric field
right outside the wire is also E = J/σ (and it points in the direction of the current,
of course). The magnetic field right outside the wire points tangentially with the
usual magnitude of B = µ0 I/2πR, where R is the radius of the wire. E and B are
perpendicular, and you can show with the right-hand rule that the Poynting vector
S = E × B/µ0 points radially into the wire. So the direction is correct; the energy
in the wire increases, consistent with the fact that it heats up. The magnitude of S
equals
1 1 J µ0 I JI
S= EB = = . (623)
µ0 µ0 σ 2πR 2πRσ
188 CHAPTER 9. MAXWELL’S EQUATIONS AND E&M WAVES
To obtain the power flux into the wire through the surface, we must multiply by 2πRℓ,
where ℓ is the length of a given section of the wire. So the total energy flow per time
into a length ℓ of the wire is
JI JI (I/A)I ℓ ρℓ
Pℓ = S · 2πRℓ = 2πRℓ = ℓ= ℓ = I2 = I2 = I 2 R, (624)
2πRσ σ σ σA A
where R is the resistance of the length ℓ of the wire. We have used the fact that the
resistivity ρ is given by ρ = 1/σ. As desired, Pℓ equals the rate of resistance heating
in the length ℓ of the wire. Pℓ can also be written as I(IR) = IV , of course, where V
is the voltage drop along the length ℓ of the wire.
Alternatively, we never actually had to use the J/σ form of E. A quicker method is:
1 µ0 I
Pℓ = S · 2πRℓ = E · 2πRℓ = IEℓ = IV, (625)
µ0 2πR
b because V = Eℓ.
I
E 9.29. Energy flow in a capacitor
r S
B
We can find the magnetic field inside the capacitor by integrating the ∇ × B =
µ0 J + ϵ0 µ0 ∂E/∂t Maxwell equation over a disk of radius r. Since J points upward in
Fig. 151, the upper plate is positive, so ∂E/∂t points downward. We therefore have
Figure 151
(using Stokes’ theorem)
∫
∂E ∂E
B · ds = µ0 I − ϵ0 µ0 (area) =⇒ B(2πr) = µ0 I − ϵ0 µ0 (πr2 )
∂t ∂t
µ0 I ϵ0 µ0 r ∂E
=⇒ B = − . (626)
2πr 2 ∂t
It will be helpful to write I in terms of E (or rather ∂E/∂t). We have
The total energy per time (that is, the power) flowing out of a cylinder of radius r is
S times the area 2πrh of the cylinder (where h is the separation between the plates).
1 Very close to the wire, the electric field actually points upward due to surface charges on the wire, but
a little farther away the downward field from the capacitor dominates. See the remark at the end of the
solution to Problem 9.10.
189
where we have used ω = 2π/T and 1/µ0 ϵ0 = c2 . As desired, this result is small if
the period T much larger than r/c, which is (half) the time it takes light to travel
across the capacitor disks. As in Problem 9.6, we have ignored the high-order feedback
effects between E and B. These effects are negligible if the current doesn’t change too
quickly. E B
as desired. We evaluated the trig integral here by writing sin3 θ as sin θ(1 − cos2 θ).
Alternatively, you can use the integral table in Appendix K.
E ′2 − c2 B ′2 = (E∥′2 + E⊥
′2
) − c2 (B∥′2 + B⊥
′2
)
= (E∥2 − c2 B∥2 ) + (E′⊥ · E′⊥ − c2 B′⊥ · B′⊥ ), (633)
When we add the previous two equations, the two middle terms will cancel if E⊥ ·
(v × B⊥ ) = −B⊥ · (v × E⊥ ). This is indeed true, because each of these “dot-cross”
products can be transformed into the other by cyclicly permuting the vectors (which
doesn’t change anything) and then reversing the order of the cross product (which
brings in a minus sign). The sum of Eqs. (635) and (636) is therefore
= E⊥ 2
− c2 B⊥ 2
. (637)
as desired. Alternatively, it doesn’t take too long to solve this exercise by explicitly
writing out the squares of all the components of E and B in Eq. (6.74).
Chapter 10
191
192 CHAPTER 10. ELECTRIC FIELDS IN MATTER
be better called a tray) or too tall (in which case it would be better called a tube), the
dependence of the capacitance on the exact dimensions is fairly weak. (If the height
is h = nd, then you can show that the capacitance is proportional to (n + 1/4)/n2/3 .)
The capacitance of a sphere is 4πϵ0 r, so a sphere will have a capacitance of 9.2·10−10 F
if r = 8.3 m. The diameter is then 16.6 m, or about 54 feet.
10.17. Maximum energy storage
The maximum field is 14 kilovolts/mil, which in volts/meter equals
1.4 · 104 V
Emax = = 5.5 · 108 V/m. (642)
2.54 · 10−5 m
The capacitance of the Mylar-filled capacitor is κϵ0 A/s. The energy stored is still
Cϕ2 /2, so the maximum possible energy density is
energy 1 1 1 κϵ0 A 1 1
= Cϕ2 = (Es)2 = κϵ0 E 2 (643)
volume 2 V 2 s As 2
1 ( 2 2 )
s C
= (3.25) 8.85 · 10−12 (5.5 · 108 V/m)2 = 4.4 · 106 J/m3 .
2 kg m3
The maximum energy per kilogram of Mylar is therefore
To determine how high the capacitor could lift itself, let the entire mass of the capacitor
be m. Then 3m/4 of this is Mylar, so conservation of energy gives
(3/4)((3100 J/kg)
E = mgh =⇒ (3100 J/kg)(3m/4) = mgh =⇒ h = = 240 m.
9.8 m/s2
(645)
The D cell in Exercise 4.41 had an energy storage of 1.8 · 105 J/kg, which is about
60 times as much as the Mylar capacitor. However, the capacitor can deliver all the
stored energy in less than a microsecond!
10.18. Partially filled capacitors
The second capacitor in the figure consists of two capacitors in series; you can imagine
the boundary between them to be two plates with charge Q and −Q superposed. Both
of these capacitors have plate separation s/2 and area A, so the capacitances are (with
the two halves labeled by “v” for vacuum and “d” for dielectric) Cv = ϵ0 A/(s/2) and
Cd = κϵ0 A/(s/2). Since C0 = ϵ0 A/s, we have Cv = 2C0 and Cd = 2κC0 . Problem 3.18
gives the rule for adding capacitors in series, so the desired capacitance is (with “S”
for series)
1 1 1 1 1 2κ
= + = + =⇒ CS = C0 . (646)
CS Cv Cd 2C0 2κC0 κ+1
The third capacitor in the figure consists of two capacitors in parallel. They both
have plate separation s and area A/2, so the capacitances are Cv = ϵ0 (A/2)/s and
Cd = κϵ0 (A/2)/s. These can be written as Cv = C0 /2 and Cd = κC0 /2. Problem 3.18
gives the rule for adding capacitors in parallel, so the desired capacitance is (with “P”
for parallel)
C0 κC0 1+κ
CP = Cv + Cd = + = C0 . (647)
2 2 2
193
If κ = 1, then both CS and CP are equal to C0 , as they should be. For any other
value of κ (greater than 1, of course), both CS and CP are larger than C0 . This makes
sense because the effect of a dielectric is to partially cancel the existing charge, which
means that more charge must be added if the same potential is to be maintained.
If κ = ∞, then CS = 2C0 and CP = ∞. The former makes sense because the dielectric
is actually a conductor in this case, so we effectively have a vacuum capacitor with
separation s/2. The latter makes sense because the capacitance of a conductor is
infinite, since any charge you dump on it will be neutralized by the shifting of charges
within the conductor. So the left half of the third capacitor has infinite capacitance.
10.19. Capacitor roll
Let w and s stand for the width and thickness of the materials. Then the various
constants in the problem are (after converting to meters) κ = 2.3, wp = 5.72 · 10−2 m,
sp = 2.54·10−5 m, wa = 5.08·10−2 m, and sa = 1.27·10−5 m. We want the capacitance
to be C = 5 · 10−8 F. The capacitance of a parallel-plate capacitor (with width w,
length ℓ, and separation s) in the presence of a dielectric is
κϵ0 A κϵ0 wℓ Cs
C= = =⇒ ℓ = . (648)
s s κϵ0 w
If we have the tape stretched out straight, with the polyethylene sandwiched between
two strips of aluminum, then we need the length of this linear aluminum capacitor to
be
Csp (5 · 10−8 F)(2.54 · 10−5 m)
ℓ= = ( ) = 1.23 m. (649) polyethylene
κϵ0 wa (2.3) 8.85 · 10−12 kg m3 (5.08 · 10
s2 C2 −2 m)
If we simply rolled up the capacitor into a spiral, then one aluminum strip would
touch the other. This would ruin the capacitor, because we need the two strips to
aluminum
have opposite charge. So we need to add on a second layer of polyethylene, as shown
in Fig. 153. (We don’t need a third layer of polyethylene on bottom.) So it appears Figure 153
that the total length of each kind of tape should be 2(1.23 m) = 2.46 m. However,
from Problem 3.21 and Exercise 3.57, we effectively have twice as much area in the
capacitor, because the two sides of each strip of aluminum tape act like two different
sheets. So ℓ only needs to be half of the above 1.23 m. Hence ℓ = 0.61 m, and the
total length of each kind of tape is 2(0.61 m) = 1.23 m.
To calculate the diameter of the rolled-up capacitor, note that the area (in the plane
of the page) of the capacitor in Fig. 153 is (2sp + 2sa )(ℓ) = (7.62 · 10−5 m)(0.61 m) =
4.7 · 10−5 m2 . This area doesn’t change when we roll up the capacitor, so the radius
of the roll is given by
πr2 = 4.7 · 10−5 m2 =⇒ r = 3.9 · 10−3 m. (650)
The diameter is therefore a little less than 0.8 cm. If √
you missed the above factor of
1/2 in ℓ, the diameter would be larger by a factor of 2, but it would still be in the
right ballpark.
10.20. Work in a dipole field
From Eq. (10.15) the potential energy, per unit charge, in the field of a dipole is
ϕ = p cos θ/4πϵ0 r2 . Hence
√ √
p p(1/ 2) 2p
ϕA = and ϕB = √ = . (651)
4πϵ0 a2 4πϵ0 (a/ 2)2 4πϵ0 a2
√
The work done per unit charge is therefore ϕB − ϕA = ( 2 − 1)(p/4πϵ0 a2 ).
194 CHAPTER 10. ELECTRIC FIELDS IN MATTER
(a) Let the origin be located at the −2q charge (although the choice doesn’t affect the
result, because
√ the net
√ charge is zero). Then p points downward with magnitude
p = 2 · q( 3 d/2) = 3 qd.
(b) p = 0, by symmetry.
(c) Let the origin be located at the upper left corner (although, again, the choice
doesn’t matter). Then px √= qd + q(2d) = 3qd, and py = (−q)(−d) + (2q)(−d) =
−qd. So p has magnitude 10 qd and points diagonally rightward and downward
at an angle of tan θ = −1/3 =⇒ θ = −18.4◦ below the horizontal.
(a) From Eq. (10.18) the field at a point 3 m away in the plane of the plates is
p 7.5 · 10−9 C m
E= = ( ) = 2.5 V/m. (652)
4πϵ0 r3 4π 8.85 · 10−12 kg
s2 C2
m3 (3 m)
3
If the upper plate is the positively charged one, this field points downward.
(b) From Eq. (10.18) the field at a point in the direction perpendicular to the plates
is E = p/2πϵ0 r3 . This is just twice the field in part (a), so at 3 m away it equals
5 V/m. If the upper plate is the positively charged one, this field points upward
(both above and below the capacitor).
We should check that our far-field dipole approximation is in fact valid. The
capacitance of a parallel plate capacitor is C = ϵ0 A/s, so the area is
If the plates are square, then they are about 0.65 m on a side. The largest
√ distance
from the center to a point on the plates is therefore (0.65 m)/ 2 = 0.46 m.
The given distance of 3 m is reasonably large compared with this, so our dipole
approximation is a fairly good one. However, if s or C were increased enough,
z then the length scale of the capacitor would be on the order of 3 m.
P1
θ 10.23. Dipole field plus uniform field
r
y
p r We want the field of the dipole to point in the −ŷ direction with magnitude 1.5 ·
105 V/m. So we are interested in two points in the yz plane like the ones shown in
√ (10.17) tells us that 3 cos θ − 1 = 0. Hence
2
P2 Fig. 154. Since
√ we need Ez = 0, Eq.
cos θ = ±1/ 3, and then sin θ = ± 2/3. We want the points with opposite signs in
Figure 154 these two trig relations (the other two points yield fields in the +ŷ direction). From
Eq. (10.17) we have Ey = 3p sin θ cos θ/4πϵ0 r3 . Therefore,
√ √ √
3p sin θ cos θ 3(6 · 10−10 C m)(± 2/ 3 )(∓1/ 3 )
r3 = = ( ) . (654)
4πϵ0 Ey 4π 8.85 · 10−12 kg m3 (−1.5 · 10 V/m)
s2 C2 5
195
This yields r = 0.037 m. The coordinates of the lower right point are then
x = r sin θ = r0 sin3 θ,
z = r cos θ = r0 sin2 θ cos θ. (656)
Therefore,
dx
= 3r0 sin2 θ cos θ,
dθ
dz
= r0 (2 sin θ cos2 θ − sin3 θ) = r0 sin θ(2 cos2 θ − sin2 θ)
dθ
= r0 sin θ(3 cos2 θ − 1). (657)
The ratio of these derivatives gives the slope of the tangent to the r = r0 sin2 θ curve
as
dz 3 cos2 θ − 1
= . (658)
dx 3 sin θ cos θ
This is the same as the ratio Ez /Ex as given by Eq. (10.17). The tangent to the
r = r0 sin2 θ curve therefore points in the same direction as the E field, as we wanted
to show.
Alternatively, we can work with polar coordinates, as we did in Section 2.7.2. With
respect to the local r̂-θ̂ basis, the slope of the field-line curve is
1 dr 1 2 cos θ
= 2 · 2r0 sin θ cos θ = . (659)
r dθ r0 sin θ sin θ
But from Eq. (10.18) this equals Er /Eθ . So again, the tangent to the r = r0 sin2 θ
curve points in the same direction as the E field.
as desired.
196 CHAPTER 10. ELECTRIC FIELDS IN MATTER
The monopole and dipole moments are zero, so Eq. (12.469) gives the potential at
position r as (1/4πϵ0 )(r̂ · Qr̂/2r3 ). If r̂ = (0, 0, 1),
√ then we quickly find r̂ · Qr̂ = 6ea ,
2
makes sense, because the given point is equidistant from the upper +e and right −e
charges, yielding zero potential. Likewise for the lower +e and left −e charges.
10.27. Pascal’s triangle and the multipole expansion
(a) If we make a copy of one of the configurations, negate all of its charges, shift it one
unit to the left relative to the original, and then add its charges to the original,
then we end up with the next configuration in the series. This is demonstrated
in Fig. 155.
(b) For the octupole, the potential at a point P on the axis is (ignoring the factor of
q/4πϵ0 )
( )
1 3 3 1 1 3 3 1
ϕP = − + − = 1− + − .
r r + a r + 2a r + 3a r 1 + a/r 1 + 2a/r 1 + 3a/r
(663)
We can expand this with the Taylor series 1/(1 + ϵ) ≈ 1 − ϵ + ϵ2 − ϵ3 . With
ϵ ≡ a/r we have
1[
ϕP ≈ 1 − 3(1 − ϵ + ϵ2 − ϵ3 )
r
+ 3(1 − 2ϵ + 22 ϵ2 − 23 ϵ3 )
]
− 1(1 − 3ϵ + 32 ϵ2 − 33 ϵ3 ) . (664)
Collecting terms with the same power of ϵ, the sum in brackets can be written as
1+ϵ0 (−3 · 10 + 3 · 20 − 1 · 30 )
−ϵ1 (−3 · 11 + 3 · 21 − 1 · 31 )
+ϵ2 (−3 · 12 + 3 · 22 − 1 · 32 )
−ϵ3 (−3 · 13 + 3 · 23 − 1 · 33 ). (665)
197
monopole
1
dipole
-1 1
-1 1
quadrupole
1 -1
1 -2 1
octupole
-1 2 -1
-1 3 -3 1
24-pole
1 -3 3 -1
Figure 155
You can quickly check that only the ϵ3 term has a nonzero coefficient; the sum
equals 6ϵ3 . Remembering the 1/r out front in Eq. (664), bringing back in the
factor of q/4πϵ0 , and using ϵ ≡ a/r, the potential at P is
q 1 3 q 6a3
ϕP ≈ (6ϵ ) = . (666)
4πϵ0 r 4πϵ0 r4
As promised, this is proportional to 1/r4 . And the units are (charge)/[ϵ0 ·(length)],
which are correct.
∑3 ( )
You can see that each line in Eq. (665) takes the form of (−ϵ)m k=0 (−1)k k3 k m ,
for m = 0, 1, 2, 3. (The first line takes this form if we let 00 equal 1.) The sum is
nonzero only for m = 3.
To prove the theorem mentioned in the problem, start with the “(1 − x)N =
[binomial expansion]” equation, and then repeat the process m times of alter-
nately taking derivatives and multiplying by x (which will generate the mth
power of the k’s in the above sum), and then set x = 1. The left-hand side
will vanish as long as we haven’t taken more than N − 1 derivatives. If we
take N derivatives, the left-hand side (and hence the right-hand side) will equal
(−1)N N !. It’s easy to see how all of this works out if you do the procedure for,
say, N = 4.
Let’s be quantitative. From Eq. (10.26) the x component of the force on a dipole is
Fx = p · ∇Ex , and likewise for the other components. Let’s first look at the y force
due to the field from the left dipole. This field is Ey = −p/4πϵ0 (b + x)3 at points on
the x axis near the origin. The gradient of this has only an x component, and it is
∂Ey 3p
= . (667)
∂x x=0 4πϵ0 b4
Therefore,
3p 3p2
Fy = p · ∇Ey = px (∇Ey )x = p 4
= . (668)
4πϵ0 b 4πϵ0 b4
Now consider the x force due to the field from the right dipole. This field is Ex =
2p/4πϵ0 (b − x)3 at points on the x axis near the origin. The gradient of this has only
an x component, and it is
∂Ex 6p
= . (669)
∂x x=0 4πϵ0 b4
Therefore,
6p 6p2
Fx = p · ∇Ex = px (∇Ex )x = p 4
= . (670)
4πϵ0 b 4πϵ0 b4
Since Fy /Fx = 1/2, the field points up to the right at an angle tan
√ θ = 1/2 =⇒
θ = 26.6◦ with respect to the x axis. The magnitude is F = 3 5p2 /4πϵ0 b4 =
(6.71)p2 /4πϵ0 b4 .
Alternatively, you could work out the force from scratch, by letting the dipole consist
of two charges ±q at positions x = ±ℓ/2, with qℓ = p. If you explicitly calculate the
forces on the two charges due to the left and right dipoles, to leading order in ℓ, you
will end up with the above values of Fx and Fy . The differences in the forces on the
two charges will effectively give the above gradients of the E’s.
(a) If we bring in the right dipole from infinity, with it pointing leftward, this requires
no work, because the leftward displacement is always perpendicular to the vertical
field from the left dipole. But when we rotate the right dipole 90◦ to the desired
orientation, this requires pE worth of work; see Eq. (10.22). Since E = p/4πϵ0 d3 ,
the work required is W = p2 /4πϵ0 d3 .
Note that if we bring in the right dipole from infinity, with it pointing upward,
then we cannot say that the leftward displacement is always perpendicular to the
field from the left dipole, because this field is not vertical at points off the x axis.
If we imagine the right dipole to consist of two point charges slightly above and
slightly below the x axis, then at the location of these charges, the field from the
left dipole has a slight horizontal component (rightward above the x axis, and
leftward below). Positive work must be done to move each charge against this
field. You can be quantitative about this if you want.
(b) From reasoning similar to that in part (a), the work required is W = −p2 /4πϵ0 d3 .
(c) If we bring in the right dipole from infinity, with it pointing upward, this requires
no work, because the works for the charges on the two ends of the dipole cancel.
But when we rotate it 90◦ to the desired orientation, this requires −pE worth of
work, from Eq. (10.22). Since E = 2p/4πϵ0 d3 along the axis of the left dipole,
the work required is W = −2p2 /4πϵ0 d3 .
199
(d) From reasoning similar to that in part (c), the work required is W = 2p2 /4πϵ0 d3 .
The task of Problem 10.3 is to directly calculate the above potential energies by
looking at the point charges that make up the dipoles.
4πϵ0 Ea3
∆z = , (672)
e polarizable atoms
which agrees with Eq. (10.27). The hydrogen atom won’t actually remain spherically A B
symmetric, but that won’t affect the rough size of ∆z. α α
PA PB
10.31. Mutually induced dipoles
Consider the setup shown in Fig. 156. The field at B due to pA is 2pA /4πϵ0 r3 . r
Hence the induced dipole at B is pB = α(2pA /4πϵ0 r3 ). In a similar manner we find
pA = α(2pB /4πϵ0 r3 ). Substituting pB from the first of these relations into the second Figure 156
gives pA = 4α2 pA /(4πϵ0 )2 r6 . This is satisfied by pA = 0, or by any value of pA
provided that
( )1/3
4α2 α
r6 = =⇒ r = ≡ rc , (673)
(4πϵ0 )2 2πϵ0
where the “c” stands for critical (or cutoff). If r > rc , and if both dipole moments
are nonzero at a given instant, they will decay to zero. But if r < rc , and if both A B
dipole moments are nonzero at a given instant, they will increase until limited by the
nonlinearity of polarizability.
Atomic polarizabilities α/4πϵ0 are typically, in order of magnitude, an atomic volume;
see Section 10.5. So rc = (2·α/4πϵ0 )1/3 is on the order of an atomic radius. The object Figure 157
we are concerned with therefore might look something like what is shown in Fig. 157.
Whether the lowest state of this system is a spontaneously polarized structure cannot
be decided by considering only the interactions of dipoles. Ordinarily the lowest state
of two similar atoms would be symmetrical with pA + pB = 0. But we cannot exclude
the possibility that the symmetry is “spontaneously broken.”
10.32. Hydration
In a water molecule, the side with the two hydrogen atoms is positively charged, and
the side with the oxygen atom is negatively charged. So if the given ion is negative,
the hydrogen side will be closer to the ion.
The field of a dipole along its axis equals 2p/4πϵ0 r3 , so the force on the negative ion
will be attractive with magnitude 2ep/4πϵ0 r3 . The work required to move the ion to
infinity by applying an equal and oppositive force is therefore
∫ ∞
2ep dr ep
W = = . (674)
r0 4πϵ0 r
3 4πϵ0 r02
200 CHAPTER 10. ELECTRIC FIELDS IN MATTER
Alternatively, from Eq. (10.15) we know that the initial energy of the negative ion
is (−e)ϕ = (−e)(p/4πϵ0 r02 ). The work that must be done in bringing the ion to
infinity where ϕ = 0 is the negative of this, in agreement with the above result. With
p = 6.13 · 10−30 C-m and r0 = 1.5 · 10−10 m, we find
p 3.43 · 10−30 C m
E= = ( ) = 6.2 · 107 V/m. (676)
2πϵ0 z 3
2π 8.85 · 10−12 kg
s 2 C2
m3 (10 −9 m)3
And the magnitude at y = 10 angstroms is just half of this, or 3.1 · 107 V/m.
10.34. Hydrogen chloride dipole moment
If the electron distribution is spherically symmetric around the chlorine nucleus, then
we effectively have one proton and one electron separated by 1.28 angstroms. The
dipole moment is therefore
e To determine where the charge really is, we can treat the entire group of electrons like
17e -18e
a point charge of −18e∫located at their “center of gravity” (defined to be the origin
with respect to which rρ dv = 0). So we have the distribution shown in Fig. 158.
1.28.10-10 m
Let x be the distance from the chlorine nucleus to the electron center. Then with the
Figure 158 chlorine nucleus as our origin, we want x to satisfy
or 0.059 angstroms, which is about 1/22 of the way from the chlorine nucleus to the
proton. If it were 1/18 of the way (at 0.071 angstroms), which is the location of
the center of the positive charge, then p would be zero. This would be the case if the
electron from the hydrogen atom remained spherically symmetric around the hydrogen
nucleus (the proton).
10.35. Some electric susceptibilities
The susceptibility is given by χ = κ − 1, so χ = CN p2 /ϵ0 kT yields C = (κ −
1)ϵ0 kT /N p2 . The κ values for water and methanol in Table 10.1 are given for room
temperature (20◦ C), whereas the κ for ammonia is given for −34◦ C. The values of
kT at these temperatures are 4.0 · 10−21 J and 3.3 · 10−21 J, respectively. Knowing the
values of κ, kT , N , and p (and ϵ0 , of course), we can compute C = (κ − 1)ϵ0 kT /N p2 .
So we just need to find N for the various substances.
201
κ kT N p C
H2 O 80 4.0 · 10−21 J 3.3 · 1028 m−3 6.13 · 10−30 C-m 2.3
NH3 23 3.3 · 10−21 J 2.9 · 1028 m−3 4.76 · 10−30 C-m 1.0
CH3 OH 34 4.0 · 10−21 J 2.5 · 1028 m−3 5.66 · 10−30 C-m 1.5
10.36. Discontinuity in E⊥
The internal field is E = −P/3ϵ0 , so E⊥ in
= P cos θ/3ϵ0 . (As a double check, the factor
here is indeed cos θ because the field at the north pole is P/3ϵ0 , pointing downward.)
This component points inward in the upper hemisphere (and outward in the lower
hemisphere), because E points downward. The perpendicular external field is the
out
radial field from a dipole, E⊥ = Er = p0 cos θ/2πϵ0 r03 . This component points
outward in the upper hemisphere (and inward in the lower hemisphere). The effective
dipole moment p0 equals (4πr03 /3)P . Hence E⊥ out
= 2P cos θ/3ϵ0 . Due to the different
inward/outward directions of the vectors, the discontinuity in E⊥ is 2P cos θ/3ϵ0 −
(−P cos θ/3ϵ0 ) = P cos θ/ϵ0 = P⊥ /ϵ0 , as desired.
10.37. E at the center of a polarized sphere
Consider a horizontal ring at an angle θ down from the top of the sphere, with angular
span dθ. The area of this ring is 2π(R sin θ)(R dθ). Since the density is σ = P cos θ,
the charge in the ring is q = 2πP R2 sin θ cos θ dθ. A little bit of charge dq in a ring in
the upper hemisphere creates a diagonally downward field of dq/4πϵ0 R2 at the center
of the sphere. But by symmetry we are concerned only with the vertical component,
which brings in a factor of cos θ. Integrating over all the dq’s in a ring simply gives
the total charge q in the ring. The net field from the ring therefore points downward
with magnitude
2πP R2 sin θ cos θ dθ P sin θ cos2 θ dθ
cos θ = . (679)
4πϵ0 R2 2ϵ0
You can verify that this expression is valid for rings in the lower hemisphere too; all
contributions to the field point downward. Integrating over θ from 0 to π gives a total
magnitude of ∫ π π
P sin θ cos2 θ dθ P cos3 θ P
E= =− = , (680)
0 2ϵ0 6ϵ0 0 3ϵ0
as desired. The direction is downward.
10.38. Uniform field via superposition
(a) Applying Gauss’s law to a sphere of radius r inside the given sphere yields
q (4πr3 /3)ρ ρr
4πr2 E = =⇒ 4πr2 E = =⇒ E = . (681)
ϵ0 ϵ0 3ϵ0
positive
202 CHAPTER 10. ELECTRIC FIELDS IN MATTER
q (πr2 ℓ)ρ ρr
2πrℓE = =⇒ 2πrℓE = =⇒ E = . (684)
ϵ0 ϵ0 2ϵ0
The field points radially, so we have E = (ρ/2ϵ0 )r. From the same reasoning as
in part (b) (the picture looks exactly the same, when viewing along the axis), the
field at an arbitrary point P inside both distributions is
ρr1 (−ρ)r2 ρ ρs
E = E1 + E2 = + = (r1 − r2 ) = − . (685)
2ϵ0 2ϵ0 2ϵ0 2ϵ0
We again have P = ρs, so the field inside a cylinder with uniform transverse
polarization is E = −P/2ϵ0 .
Remark: A similar result holds if we kick things down another dimension and consider
a slab. For a slab, applying Gauss’s law to an internal box centered on the center-plane
of the slab, with thickness 2x and end-face area A, gives
q (2x · A)ρ ρx
E(2A) = =⇒ 2EA = =⇒ E = . (686)
ϵ0 ϵ0 ϵ0
The field points perpendicular to the faces of the slab, so we have E = (ρ/ϵ0 )x. The
same reasoning holds again, with two slightly displaced slabs. It’s even easier in this
case, because everything takes place in one dimension:
ρx1 (−ρ)x2 ρ ρs
E = E1 + E2 = + = (x1 − x2 ) = − . (687)
2ϵ0 ϵ0 ϵ0 ϵ0
We again have P = ρs, so the desired field is E = −P/ϵ0 . The only difference in the
E’s for the various objects we have considered is the numerical factor (3, 2, or 1) in the
denominator.
concerned, from the reasoning near the beginning of Section 10.9. From Eq. (10.18)
the tangential field of a dipole is p sin θ/4πϵ0 a3 , which gives E0 sin θ here. But this
is exactly what is needed to cancel the tangential component of the original uniform
field E0 (you can check that the direction is correct). The tangential component of
the total external field is therefore zero, as desired.
The field strength inside the dielectric sphere, which from Eq. (10.53) is 3E0 /(2 + κ),
goes to zero as κ → ∞. The sphere therefore becomes an equipotential, which is
correct for a conducting sphere. A sketch of the field lines outside the sphere is shown
in Fig. 160. E=0 E0
Using the above value of p, we see that the polarizability α, defined by p = αE0 , of a
perfectly conducting sphere equals 4πϵ0 a3 . The quantity that we normally work with,
α/4πϵ0 , therefore equals a3 for our conducting sphere of radius a. Since α/4πϵ0 =
0.66·10−30 m3 for hydrogen, a conducting sphere of equal polarizability has a radius of
(0.66 · 10−30 m3 )1/3 = 8.7 · 10−11 m. This is very close to the Bohr radius, 5.3 · 10−11 m.
Figure 160
10.40. Continuity of D
The E field due to the slab is the same as the E field due to two capacitor plates
with surface charge densities ±P . Both E and P are zero outside the slab, so the
external D is likewise zero. Our task is therefore to show that D is zero inside the
slab. And indeed, E = −P/ϵ0 (this is the field between two plates with densities ±P ),
so D = ϵ0 E + P = ϵ0 (−P/ϵ0 ) + P = 0.
10.41. Discontinuity in D∥
Inside the sphere, we have E = −P/3ϵ0 , so the displacement vector is D ≡ ϵ0 E + P =
ϵ0 (−P/3ϵ0 ) + P = 2P/3. The tangential component of this is
2P sin θ
D∥in ≡ Dθin = − . (688)
3
The minus sign here comes from the fact that P points toward the north pole, whereas
the positive θ̂ direction is defined to point away from the north pole.
Outside the sphere, E is the field due to a dipole with p0 = (4πR3 /3)P. From
Eq. (10.18) the tangential component of the dipole field is Eθ = p0 sin θ/4πϵ0 R3 . In
terms of P this becomes Eθ = P sin θ/3ϵ0 . Since P = 0 outside the sphere, the external
D is obtained by simply multiplying the external E by ϵ0 . Therefore
P sin θ
D∥out ≡ Dθout = . (689)
3
Comparing this with the D∥in in Eq. (688), we see that D∥ has a discontinuity of P sin θ.
D∥ increases by P sin θ in going from inside to outside. This makes sense, because we
know that Eθ is continuous across the boundary, so the discontinuity in the tangential
component of D ≡ ϵ0 E + P is simply the discontinuity in the tangential component
of P.
10.42. Energy density in a dielectric
With a dielectric present, the capacitance of a parallel-plate capacitor is C = κϵ0 A/s ≡
ϵA/s. The energy stored is still Cϕ2 /2, because it equals Qϕ/2 for all the same reasons
as in the vacuum case (imagine a battery doing work in transferring charge from one
plate to the other). So the energy density is
energy 1 1 1 ϵA 1 ϵE 2
= Cϕ2 = (Es)2 = , (690)
volume 2 V 2 s As 2
204 CHAPTER 10. ELECTRIC FIELDS IN MATTER
as desired. Since ϵ ≡ κϵ0 , this energy density is κ times the ϵ0 E 2 /2 energy density
without the dielectric. Basically, since C is κ times larger, so is the energy, and hence
the energy density. To see physically why the energy is larger, consider the case of
induced dipole moments, discussed in Section 10.5. The stretched atoms and molecules
are effectively little springs that are stretched, so they store potential energy. This
makes the total energy larger than it would be for the same equivalent charge on/at
the capacitor plates (free charge plus bound-charge layer).
In an electromagnetic wave in a dielectric, the energy density of the magnetic field is
still B 2 /2µ0 . (It would be B 2 /2µ in a magnetized material, but we’re assuming that
the material here is only electrically polarizable.) But from Eq. (10.83) the amplitudes
√
of the E and B fields are related by B0 = µ0 ϵ E0 . So B 2 /2µ0 = ϵE 2 /2. The E and
B energy densities are therefore equal, just as they are in vacuum.
The total wave in the empty space y < 0 is the sum of the incident and reflected
waves.
Let’s apply continuity of E and B at y = 0. After setting y = 0, we can cancel all the
sin ωt terms (which means that our results will hold for all t), but we must be careful
about the extra minus sign in sin(−ωt). We obtain
We also have
Ei = cBi , Er = cBr , Et = (c/n)Bt . (693)
Given the “i” quantities, we have four equations in four unknowns (the “r” and “t”
quantities). Eliminating the B’s quickly turns Eq. (692) into
(a) Equation (11.15) gives the field at position R along the axis of a dipole as Br =
µ0 m/2πR3 , so
(b) Equation (6.53) gives the field at position z on the axis of a current ring as
Bz = µ0 Ib2 /2(z 2 + b2 )3/2 . If R is the radius of the earth, then we have z = R
and b ≈ R/2, so in terms of R the field is Bz = µ0 I/(53/2 R). Therefore,
If we instead treat the current ring as a dipole with moment m = 8.1 · 1022 J/T,
then we have
m 4(8.1 · 1022 J/T)
m = I(πb2 ) =⇒ I = = = 2.5 · 109 A, (699)
π(R/2) 2 π(6.4 · 106 m)2
205
206 CHAPTER 11. MAGNETIC FIELDS IN MATTER
Note that this result can be written as ω(πR2 σ)R2 /4 = ωQR2 /4, where Q is the
total charge on the disk. If all of this charge were instead located on the rim, then
(since one revolution
) 2 takes a time of 2π/ω, the magnetic moment would be IπR2 =
Q/(2π/ω) πR = ωQR2 /2, which is twice the moment of the disk.
θ R dθ 11.14. Sphere dipole
Let the axis of rotation be vertical. Consider a strip located at angle θ, with width
Figure 161 dθ, as shown in Fig. 161. The speed of any point in this strip is v = ωr = ω(R sin θ).
The effective linear charge density of the strip, dλ, equals σ times the width, so
dλ = σ(R dθ). The current dI in the strip is then
ωQ sin θ dθ
dI = v dλ = (ωR sin θ)(σR dθ) = ωσR2 sin θ dθ = , (700)
4π
where we have used σ = Q/4πR2 . Alternatively, you can find dI by multiplying the
total charge in the ring, which is σ(2πR sin θ)(R dθ), by the number of revolutions per
second, which is ω/2π.
The magnetic moment of the ring is
ωQR2 sin3 θ dθ
dm = (dI)π(R sin θ)2 = . (701)
4
Integrating this over the whole sphere to obtain the total magnetic moment gives
(using the integral table in Appendix K or writing sin3 θ as sin θ(1 − cos2 θ))
∫ ( ) π
ωQR2 π 3 ωQR2 cos3 θ ωQR2 4 ωQR2
m= sin θ dθ = − cos θ + = · = .
4 0 4 3 4 3 3
0
(702)
Note that if all of the charge Q were located on the equator, then the current would
be I = (ω/2π)Q, and the magnetic moment would be m = IπR2 = (ωQ/2π)(πR2 ) =
ωQR2 /2. The m for a spinning shell is therefore 2/3 of the m for a spinning ring
with the same radius and total charge. Also, from Exercise 11.13 the m for a spinning
shell is 4/3 times the m for a spinning disk with the same radius and total charge. It
makes sense that this factor is larger than 1, because if the shell is projected onto the
equatorial disk, the density near the rim is larger than the density at the center.
11.15. A solenoid as a dipole
To estimate roughly the magnetic dipole moment of the solenoid, let us suppose that
it is equivalent to a point dipole that would produce, 20 cm away on its axis, a field
strength Bz equal to that at the end of the solenoid, namely 1.8 T. This is reasonable
because the magnetic field configuration near the end of the solenoid and beyond looks
not very different from a dipole field. On this assumption, Eq. (11.15) gives
However, there is actually no need to know this value of m, because we are interested
only in a rough estimate of the field at the location of the complaining physicists.
All we care about is that the field decreases like 1/r3 ; the exact nature of the radial
and tangential components isn’t critical. The physicists are 100 feet away, which is
about 30 meters. This is 150 times the 0.2-meter distance at which the field is 18, 000
gauss, so the desired field is smaller by a factor of 1503 . It is therefore roughly equal
207
to 18, 000/1503 ≈ 5 · 10−3 gauss. (If you want to find the two components, you can
simply plug the above value of m into Eq. (11.15), with θ = tan−1 (4/3) ≈ 53◦ .)
This field is about 100 times smaller than the earth’s field of about 0.5 gauss. If it
were perfectly steady it would not be noticed. But if it were frequently switched on
and off, it might cause trouble.
where we have used 2d ≈ a, and where we have ignored the d’s in the denominator
because they are small compared with r. Since the area of the trapezoid is essentially
equal to ba, we have B ≈ µ0 m/4πr3 , in agreement with the Bθ in Eq. (11.15) when
θ = 90◦ .
208 CHAPTER 11. MAGNETIC FIELDS IN MATTER
Note that the above result is first order in b. You can show that the approximations
we made (setting the Biot-Savart angle equal to 90◦ , using 2d ≈ a) involve errors that
are second order in b, so we were justified in ignoring them.
11.18. Field somewhat close to a solenoid
Consider a single loop of the solenoid at its middle. From Eq. (11.15), this little dipole
creates a field at P equal to µ0 m/4πℓ3 = µ0 (πR2 I)/4πℓ3 . Up to numerical factors, all
the other loops create this same field too; the only differences are some trig factors of
order 1. So if there are N loops in all, the total field at P behaves like
µ0 (πR2 I) R2
B∼N 3
= µ0 (N/ℓ)I 2 . (705)
4πℓ 4ℓ
Ignoring the factor of 1/4, and recalling that µ0 (N/ℓ)I ≡ µ0 nI is the field B0 inside
the solenoid, we obtain B ∼ B0 R2 /ℓ2 , as desired.
11.19. Using reciprocity
Let the dipole m(t) consist of a ring of area a carrying current I2 (t) = I2 cos ωt. So
m(t) = (I2 cos ωt)a =⇒ m0 = I2 a. If a current I1 (t) in C1 causes a field B1 (t) at
the location of the dipole ring, then the flux through the ring is Φring (t) = B1 (t) · a.
The coefficient of mutual inductance is therefore M = Φring /I1 = (B1 · a)/I1 . (We’ve
suppressed the t arguments here.) Due to the reciprocity theorem, this is also the M
going the other way. The emf in circuit C1 is therefore given by E1 (t) = −M dI2 (t)/dt,
where I2 (t) is the current in the dipole ring. Hence the emf in C1 is
B1 · a d(I2 cos ωt) B1 · a
E1 (t) = − =− (−ωI2 sin ωt)
I1 dt I1
ω ω
= B1 · (I2 a) sin ωt = B1 · m0 sin ωt ≡ E1 sin ωt, (706)
I1 I1
with E1 = (ω/I1 )B1 · m0 , as desired. I1 and B1 are technically functions of time. But
since they have the same time dependence, we can take them to be the amplitudes (a
vector amplitude in the B1 case).
11.20. Force between a wire and a loop
At an arbitrary point on the wire, the magnetic field from the square-loop dipole has
both an upward vertical (z) component and a horizontal component along the wire.
But the latter produces no force on the current in the wire, so we care only about
the z component. Since the current in the wire in Fig. 6.47 points into the page, the
right-hand rule gives the magnetic force on the wire as pointing to the right.
Equation (11.14) gives the z component of the dipole field, with m equal to m = I2 ℓ2 .
The rightward force on a little piece dx of the wire equals I1 Bz dx. With θ measured
away from the vertical axis, dx is given by the usual expression, dx = z dθ/ cos2 θ.
(See the reasoning in the paragraph following Eq. (1.37).) Also, the distance r to the
little piece is r = z/ cos θ. Integrating over the entire infinite wire, we find the total
rightward force on it to be
∫ ∞ ∫ ∞ ( )
µ0 I2 ℓ2 3 cos2 θ − 1
F = I1 Bz dx = I1 dx
−∞ −∞ 4π r3
∫
µ0 I1 I2 ℓ2 π/2 3 cos2 θ − 1 z dθ
= 3 2
4π −π/2 (z/ cos θ) cos θ
∫
µ0 I1 I2 ℓ2 π/2 µ0 I1 I2 ℓ2
= (3 cos 3
θ − cos θ) dθ = . (707)
4πz 2 −π/2 2πz 2
209
The integral here equals 2, as you can check with Mathematica or the integral table
in Appendix K. This force is consistent with the magnitude of the leftward force on
the square loop we found in Eq. (458) in Exercise 6.54, because z ≈ R for large z.
(a) Equation (11.23) gives the force on a dipole as F = ∇(m · B). The applied
force is the negative of this, so the associated work equals the line integral of
−∇(m · B). The line integral of the gradient of a quantity is simply the change
in that quantity. So the work required to move a dipole to infinity equals the
change in −m · B, which is 0 − (−m · B0 ) = m · B0 , because the field is zero at
infinity. B0 here is the field (due to all the other dipoles) at the initial location
of a dipole on a particular square. This result is consistent with the fact that the
energy of a dipole is given by −m · B.
Due to the opposite directions of the dipoles on the white and black squares, the
B fields from the four nearest neighbors (or two or three, if the dipole is near the
edge) point parallel to a given dipole m. So we expect that the sum of all 63 of
the m · B0 contributions to the work will be positive. That is, we expect all of
the dipoles to be bound.
Eq. (11.15) gives the magnetic field due to a dipole, in the plane of the dipole,
as µ0 m/4πr3 . The distance r between the various squares is found from the
Pythagorean theorem. For a given point labeled by the coordinates (a, b), where
a and b each run from 1 to 8, the sum of the m · B0 contributions to the work
equals
µ0 m2 ∑ ∑
8 8
(−1)a−i (−1)b−j
−
4πs3 i=1 j=1 (a − i)2 + (b − j)2 )3/2
( , (708)
with the caveat that the term with (i, j) = (a, b) is excluded from the sum. You
can check that the negative sign out front makes the overall sign correct. There
must be a way to cleanly exclude the (i, j) = (a, b) term in Mathematica, but
I can’t figure out what it is. At any rate, this program gets the job done (the
values of a and b can be changed):
a = 3; b = 2;
NSum[-(-1)^(a-i)(-1)^(b-j)/((a-i)^2+(b-j)^2)^(3/2), {i,1,a-1}, {j,1,8}]+
NSum[-(-1)^(a-i)(-1)^(b-j)/((a-i)^2+(b-j)^2)^(3/2), {i,a+1,8}, {j,1,8}]+
56
64
NSum[-(-1)^(a-i)(-1)^(b-j)/((a-i)^2+(b-j)^2)^(3/2), {i,a,a}, {j,1,b-1}]+
2.
NSum[-(-1)^(a-i)(-1)^(b-j)/((a-i)^2+(b-j)^2)^(3/2), {i,a,a}, {j,b+1,8}]
21
68
64
64
Up to a factor of µ0 m2 /4πs3 , the amounts of work for the various squares on the
2.
2.
91
82
chessboard are shown in Fig. 163. These entries are repeated in other parts of the 01
65
63
65
board; there are only ten independent entries. The most tightly bound dipoles
2.
2.
2.
are the ones on the “bishop’s pawn” and equivalent squares; there are eight such
28
71
56
4 0
21
22
27
56
squares. However, these are only negligibly more bound than the other interior
2.
2.
2.
1.
dipoles. The binding energies of the 36 interior dipoles differ from one another
by less than 1%. Figure 163
(b) The field from a magnetic dipole takes the same form as the field from an electric
dipole. And the force on a dipole does also, because there aren’t any outside
currents in the setup; see the discussion following Eq. (11.24). So the magnetic
force on each of our magnetic dipoles is conservative, just as the electric force on
an electric dipole is. So we can use the same reasoning we used back in Chapters
1 and 2 to say that the total work required to remove all of the dipoles far from
210 CHAPTER 11. MAGNETIC FIELDS IN MATTER
each other equals one half of the sum of all the individual works. The factor of
1/2 gets rid of the double counting in the energy. From Fig. 163 you can show
that the total work required is 77.67, in units of µ0 m2 /4πs3 .
Similarly, with the By field, if α is defined in the same way, the torque involves the
factor sin α. The integral of this gives a factor of − cos α. But since α starts at
zero and ends at π/2 − θ2 , this yields a factor of (1 − sin θ2 ) So the work done is
m2 (µ0 m1 sin θ1 /4πr3 )(−1 + sin θ2 ). We have put in a minus sign because the required
work is negative; we are making m2 be more parallel to the negative By ŷ field.
The total required work done (that is, the potential energy of the system) is the sum
of the above three results:
µ0 m1 m2 [ ]
W = sin θ1 − 2 cos θ1 cos θ2 + (− sin θ1 + sin θ1 sin θ2 )
4πr3
µ0 m1 m2 ( )
= 3
− 2 cos θ1 cos θ2 + sin θ1 sin θ2 , (711)
4πr
as desired. If θ1 = 0 and θ2 = 90◦ , then we have the configuration in Fig. 11.39(b),
and W is correctly
√ = θ2 ≡ θ, then the potential energy is positive
zero. Note that if θ1 √
if tan θ > 2 and negative if tan θ < 2. This cutoff angle is the angle for which
the field of one dipole is perpendicular to the other dipole at its location. This makes
sense, because if the dipoles are bought together in this configuration, then m · B is
always zero. So the energy is always zero, and no work is required.
• 8 lines are tilted at 45◦ with respect to the dipoles; these are the lines connected
to either of the two vertices on the z axis. The associated angles θ1 and θ2 in
Exercise 11.23 are both equal to 45◦ . And the distance r equals b.
• 4 lines form a square in the xy plane. The θ’s are 90◦ , and r = b.
√
• 2 lines are diagonals of the square in the xy plane. The θ’s are 90◦ , and r = 2b.
√
• 1 line is vertical along the z axis. The θ’s are 0◦ , and r = 2b.
Using the U = (µ0 m1 m2 /4πr3 )(sin θ1 sin θ2 −2 cos θ1 cos θ2 ) result from Exercise 11.23,
the potential energy is
( ( ))
µ 0 m2 1 1) ( ) 1 ( ) 1 (
U = 8 −2· + 4 1 − 2 · 0 + 2 3/2 1 − 2 · 0 + 1 3/2 0 − 2 · 1
4πb3 2 2 2 2
= 0. (712)
Remark: It isn’t obvious why the potential energy should be zero. However, as you can
check, it is also zero in the analogous cases of a tetrahedron, a cube with the dipoles aligned
parallel to an edge, and a cube with the dipoles aligned parallel to a long diagonal (this case
involves some tricky angles). So it is reasonable to conjecture that the potential energy is zero
in the case of any platonic solid, for any (common) orientation of the dipoles. Unfortunately,
I can’t think of a general proof. Note that our setup with magnetic dipoles is equivalent to
one with electric dipoles, because the forces and torques take the same form in electric and
magnetic dipoles (since there are no external currents involved; see the discussion following
Eq. (11.24)). An extreme case is the limit of an infinite number of infinitesimal electric
dipoles lying on the surface of a sphere (so we effectively have two opposite shells of charge
near each other). You can show that the potential energy of this system is zero. There must
be a general proof for the platonic solids, and perhaps other configurations with sufficient
symmetry. . .
212 CHAPTER 11. MAGNETIC FIELDS IN MATTER
From the discussion in Section 11.6, the work required to rotate the dipole by 90◦ is
mB (assuming that it starts aligned with the field). If we have 10 crystals, and if we
take the earth’s magnetic field to be 0.5 gauss, then the required work is
Fm vϵ0 µ0 m (m/c) v
= = · . (715)
Fe p p c
With the given values of p, m, and v, this becomes
Using 10−29 C m for p would make the result even smaller. And even if v ≈ c, the
ratio will still be small.
11.27. Diamagnetic susceptibility of water
The magnetic susceptibility is given by M = χm B/µ0 . (The M = χm H definition
would give essentially the same result, because χm will turn out to be very small; see
Exercise 11.38.) The magnetic dipole moment of a given volume V is m = MV =
χm BV /µ0 . The force on a dipole is therefore
∂Bz χm Bz V ∂Bz
F =m = . (717)
∂z µ0 ∂z
From Table 11.1 we have (being careful with the signs; we’ll take upward to be positive
for all quantities) Bz = 1.8 tesla, ∂Bz /∂z = −17 tesla/m, and F = 0.22 newtons. The
volume taken up by 1 kilogram of water (which is what the data in Table 11.1 are
given for, even though an actual sample would of course be much smaller) is 10−3 m3 ,
so Eq. (717) gives
( )
µ0 F 4π · 10−7 kgC2m (0.22 N)
χm = = = −9.0 · 10−6 . (718)
Bz V (∂Bz /∂z) (1.8 T)(10−3 m3 )(−17 T/m)
213
(a) We have χ = µ0 N m2 /kT . There are about 6 · 1023 nuclei in a gram of anything,
because the mass of a nucleon is about 1.67·10−24 g, while the mass of an electron
is negligible in comparison. So a cubic meter of water, which has a mass of 106
grams, contains 6 · 1029 protons. In water, 2 out of 18 nuclei are hydrogen
protons, so the number of protons per cubic meter is N = (2/18)(6 · 1029 m−3 ) =
6.7 · 1028 m−3 .
The proton magnetic moment is 1/700 of the electron magnetic moment, or
(9.3 · 10−24 A m2 )/700 = 1.3 · 10−26 A m2 . At room temperature, kT = 4 · 10−21 J.
So
µ0 N m2 (4π · 10−7 kg m
C2 )(6.7 · 1028 m−3 )(1.3 · 10−26 A m2 )2
χ= = = 3.6 · 10−9 .
kT 4 · 10−21 J
(719)
(b) The magnetization is given by M = χB/µ0 . (The M = χm H definition would
give essentially the same result, because the above χ is very small; see Exer-
cise 11.38.) The magnetic moment in a volume V is m = M V . V = 10−3 m3
here, so
m 4.3 · 10−6 A m2
I= = ≈ 9 · 10−4 A, (721)
a 5 · 10−3 m2
or 900 microamps.
If the material is moved along the axis from a point where the field is B1 to a point
where the field is B2 , the work done is (χm̃/2µ0 )(B12 −B22 ). If B2 is negligible compared
with B1 , then W = χB12 m̃/2µ0 . So the work per kilogram is χB12 /2µ0 , as desired.
From above, we have χ = µ0 F/(m̃B ∂B/∂z), so we can eliminate χ from the result
for W and write
( )
m̃ m̃ µ0 F FB
W = χB 2 = B2 = . (723)
2µ0 2µ0 m̃B ∂B/∂z 2 ∂B/∂z
214 CHAPTER 11. MAGNETIC FIELDS IN MATTER
From Table 11.1, the data for 1 kg of liquid oxygen are F = 75 N, B = 1.8 T, and
∂B/∂z = 17 T/m. So the work for 1 kg is
(75 N)(1.8 T)
W = = 4 J. (724)
2(17 T/m)
Equation (6.56) gives the field in the interior of a finite solenoid as B = (µ0 nI/2)(cos θ1 −
cos θ2 ), where the
√ angles are shown in Fig. 6.16. In the present case we have θ2 = π
and cos θ1 = z/ z 2 + r02 , where z is the distance from the end (with positive z being
inside the solenoid). So Eq. (6.56) gives
( )
µ0 nI z ∂B µ0 nI r02
B= 1+ √ =⇒ =
2 z 2 + r02 ∂z 2 (z 2 + r02 )3/2
∂2B µ0 nI −3r02 z
=⇒ = . (726)
∂z 2 2 (z 2 + r02 )5/2
Squaring yields
√
9z
4
9z
+ 9z 2 r02 = r04 − 6r02 z 2 + 4
=⇒ 15z 2 = r02 =⇒ z = r0 / 15. (728)
We have chosen the positive root because the negative root was introduced in the
squaring operation and isn’t a solution to Eq. (727). The desired point therefore lies
slightly inside the end of the solenoid.
11.31. Boundary conditions for B
From Problem 11.8(a) the external field is the field of a dipole with strength m =
(4πR3 /3)M . So from Eq. (11.15) the components are
µ0 m cos θ 2µ0 M cos θ
Brout = = ,
2πR3 3
µ0 m sin θ µ0 M sin θ
Bθout = = . (729)
4πR3 3
215
From Problem 11.8(b) the internal field is B = (2/3)µ0 M upward (assuming M points
upward). The radial component of this is Brin = (2/3)µ0 M cos θ, and the tangential
component is Bθin = −(2/3)µ0 M sin θ. The negative sign comes from the fact that the
field points toward the top of the sphere, which is the direction of decreasing θ.
As predicted, Br is continuous across the surface, and Bθ has a discontinuity of
µ0 M sin θ. This gives us the desired result of µ0 Jθ provided that the surface cur-
rent density Jθ equals M sin θ. And it does indeed, from the reasoning in the example
at the end of Section 11.8; the component of M parallel to the surface is M∥ = M sin θ.
In terms of the total charge Q = (4πR3 /3)ρ, this result can be written as B =
µ0 ωQ/4πR. This field is 5/2 as large as the field at the north pole; see Problem 11.9.
(a) From Problem 11.8(a), we know that the external magnetic field of a uniformly
magnetized sphere with radius R is the same as the field of a magnetic dipole
m located at the center, with magnitude m = (4πR3 /3)M . From Eq. (11.15)
the field at points on the axis of the dipole is radial and has magnitude Br =
µ0 m/2πR3 . Writing m in terms of M gives
( )( )
2 2 −7 kg m 5 J
Br = µ0 M = 4π · 10 7.5 · 10 = 0.628 T. (731)
3 3 C2 T m3
You should check that the units work out correctly. Note that this result is
independent of R.
(b) From Eq. (11.15) the field at points on the equator of the sphere is tangential
and has magnitude Bθ = µ0 m/4πR3 . This is half of the above Br , which yields
0.314 T.
(c) The force acting on each of the two uniformly polarized spheres must be the
same as if the other sphere were replaced by a point dipole m at its center,
because that leaves the external field unchanged. So we need to find the force
between two point dipoles separated by 2R. From Eq. (11.20) this force is F =
m ∂Bz /∂z, where ±ẑ are the directions of the dipoles. Since Bz = µ0 m/2πz 3 ,
we have ∂Bz /∂z = 3µ0 m/2πz 4 (in magnitude). So the attractive force between
216 CHAPTER 11. MAGNETIC FIELDS IN MATTER
the dipoles is
The units are easier to see if you write the units of M as C/(m s).
U 1010 eV
γ= = = 50. (733)
mc2 2 · 108 eV
So β ≡ v/c is essentially equal to 1. The magnitude of the momentum is
The magnetization is
M = N m = (1.5 · 1029 m−3 )(9.3 · 10−24 J/T) = 1.4 · 106 J/(T m3 ). (735)
From Eq. (11.55) the bound surface current on each face of the plate is J = M . But
we know from Section 6.6 that the magnetic field between two such sheets (with the
surface currents pointing in opposite directions) is µ0 J . So the magnetic field inside
the plate equals
r∫3 · I/r = Ir2 = m/π. And we are assuming that m is held constant. Therefore
B dv over the near region remains constant as r shrinks down.
∫
The average of, say, E over a macroscopic volume V equals ( E dv)/V . There will
be finite contributions to the numerator from the near fields of all the dipoles in the B
volume, no matter how small they are. So the near fields will necessarily contribute
to the average of E. Likewise for B.
11.36. Equilibrium orientations
First consider the more general case of the isosceles triangle in Fig. 164. If the dipoles
m1 and m2 make the same angle θ shown, then the sum of their fields at the top
vertex is horizontal. This follows from the form of the dipole field, but it also follows m1 θ θ m2
from a symmetry argument: Start with just m1 and B1 , and then rotate the setup
180◦ around the vertical axis to produce the m′1 and B′1 vectors shown in Fig. 165. Figure 164
Then negate m′1 to obtain m2 and B2 . The B1 and B2 fields make the same angles
above and below the horizontal, so the total B field points horizontally.
B1 B'1 B1
B2
B
B2
m1 m'1 m2 m1 m2
(a)
Figure 165
In the case of an equilateral triangle, we see that if the dipoles are arranged as shown
in Fig. 166(a), then each dipole will point in the direction of the field produced by the
other two. Since a dipole has the least energy when it points in the direction of the
field it lies in, this is the stable equilibrium configuration that the dipoles will assume.
Similarly in the case of other N -gons, we obtain the situations shown in Fig. 166(b,c).
The same symmetry argument that we used above can be used to show that each (b)
dipole points in the direction of the field due to all the other dipoles.
∇ · (H + M) = 0 and ∇ × H = 0. (738)
Now, the solution for E in the polarized sphere is completely determined by the Figure 166
polarization P, along with the two equations in Eq. (739). Likewise, the solution
218 CHAPTER 11. MAGNETIC FIELDS IN MATTER
(a) Ignore the rock for a moment and imagine that there is a current I in the coils.
Using the expression in Eq. (6.53) for the field due to a single ring, we find that
the field at the location of the rock due to the current in the coils is
µ0 Ir2
B = 2(1500) (744)
2(r2 + z 2 )3/2
B 1500(4π · 10−7 kg m/C2 )(0.06 m)2
=⇒ = = 0.0225 T/A.
I ((0.06 m)2 + (0.03 m)2 )3/2
Now let’s reintroduce the rock. Its angular frequency is ω = 2π(1740/(60 s)) =
182 s−1 . The vertical component of its dipole moment produces zero net flux
through the coils. But the horizontal component does, and this component oscil-
lates sinusoidally. So we effectively have the setup presented in Exercise 11.19,
with m0 pointing along the axis of the coils. So m0 is parallel to B1 in the
E1 = (ω/I1 )B1 · m0 result from that exercise. The magnetic moment of the rock
is therefore given by
E 1 10−3 V 1
m0 = · = · = 2.4 · 10−4 J/T. (745)
ω B/I 182 s−1 0.0225 T/A
(b) The electron magnetic moment is about 10−23 J/T, so the above magnetic mo-
ment corresponds to (2.4 · 10−4 )/10−23 = 2.4 · 1019 electrons. Each iron atom can
contribute 2 electron spins, so we need 1.2 · 1019 atoms. There are 56 nucleons
in an iron atom, so each atom has a mass of about 9 · 10−26 kg. The minimum
mass is therefore (1.2 · 1019 )(9 · 10−26 kg) ≈ 10−6 kg, or 0.001 g.
219
(a) From the B-H curve in Fig. 11.41(d), we find that B = 1.6 T corresponds to
H ≈ 5000 A/m. In the 20 cm gap, H = (1.6 T)/µ0 = 1.3 · 106 A/m. As a rough
estimate, the curve bcdea has a length of about 300 cm. So
∫ ∫ ∫
H · dl = H · dl + H · dl
gap iron
ρN L ρN (8 m)
R= = = N 2 ρ(53 m−1 ). (747)
A (0.15 m2 )/N
P = I 2 R = I 2 N 2 ρ(53 m−1 ) = (275, 000 A)2 (2·10−8 ohm-m)(53 m−1 ) = 8·104 J/s.
(748)
This is independent of N and I individually, because the product N I takes on a
given value.
(c) Another expression for the power is P = IV . If V = 400 V, then
P 8 · 104 J/s
I= = = 200 A. (749)
V 400 V
Therefore,
NI 275, 000 A
N= = ≈ 1400, (750)
I 200 A
or 700 turns per coil. The cross-sectional area of the wire is (1500 cm2 )/1400 ≈
1.1 cm2 . If the voltage source were instead, say, 800 V, then the current would be
100 A, so the number of turns would be 2800, and the cross-sectional area would
be roughly 0.5 cm2 .
220 CHAPTER 11. MAGNETIC FIELDS IN MATTER
Appendix F
A = xx̂ + 2yŷ
= r cos θ(r̂ cos θ − θ̂ sin θ) + 2r sin θ(r̂ sin θ + θ̂ cos θ)
= r̂r(cos2 θ + 2 sin2 θ) + θ̂r sin θ cos θ. (751)
You can check that this agrees with what you would obtain by projecting A onto
the r̂ and θ̂ unit vectors. Taking the divergence of A gives
1 ∂(rAr ) 1 ∂Aθ
∇·A = +
r (∂r r ∂θ )
1 ∂ r · r(cos2 θ + 2 sin2 θ) 1 ∂(r sin θ cos θ)
= +
r ∂r r ∂θ
= (2 cos2 θ + 4 sin2 θ) + (cos2 θ − sin2 θ)
= 3 cos2 θ + 3 sin2 θ = 3, (752) ∆θ is negative
y
C
as desired. ∆r r∆θ
F.2. Cylindrical divergence θ
A ∆x B
In Fig. 167, the first-order change in a function f in going from point A to point B
is ∆f = (∂f /∂x)∆x. But we can also imagine going from A to B in two steps along θ x
the radial and tangential segments shown, via point C. So we can also write ∆f as
(∂f /∂r)∆r + (∂f /∂θ)∆θ. Therefore,
Figure 167
∂f 1 ∂f
∆f = ∆r + (r∆θ)
∂r r ∂θ
∂f 1 ∂f
= (∆x cos θ) + (−∆x sin θ)
∂r r ∂θ
∆f ∂f 1 ∂f
=⇒ = cos θ − sin θ. (753)
∆x ∂r r ∂θ
But in the limit where A and B are close together, the left-hand side is just ∂f /∂x.
Since this equation holds for an arbitrary function f , we can erase the f , which yields
the first equation in Eq. (F.28). Similarly, breaking a vertical ∆y segment into the
∆r = ∆y sin θ and r∆θ = ∆y cos θ pieces yields the second equation in Eq. (F.28).
221
Aθ is negative
y
Ar C 222 CHAPTER 11. MAGNETIC FIELDS IN MATTER
Aθ
θ
We can use the same type of reasoning to find the relation between the Cartesian and
A Ax B cylindrical components of a vector. In Fig. 168, Ax is the horizontal component of
θ a vector A. But we can also imagine breaking up Ax x̂ into its radial and tangential
x components. The horizontal component of the radial vector Ar r̂ is Ar cos θ, and
the horizontal component of the tangential vector Aθ θ̂ is −Aθ sin θ (note that Aθ is
Figure 168 negative here). So we must have Ax = Ar cos θ − Aθ sin θ, in agreement with the
first equation in Eq. (F.29). Likewise, looking at the vertical component Ay gives
Ay = Ar sin θ + Aθ cos θ.
Alternatively, we can use the expressions for the r̂ and θ̂ unit vectors given in the
statement of Exercise F.1.
A = Ar r̂ + Aθ θ̂
= Ar (x̂ cos θ + ŷ sin θ) + Aθ (−x̂ sin θ + ŷ cos θ)
= x̂(Ar cos θ − Aθ sin θ) + ŷ(Ar sin θ + Aθ cos θ). (754)
The x and y components that we read off from this equation agree with those in
Eq. (F.29).
Now we must calculate ∇ · A = ∂Ax /∂x + ∂Ay /∂y + ∂Az /∂z. Using the above
expressions for the partial derivatives and the components, and ignoring the z term,
we have (with c ≡ cos θ and s ≡ sin θ)
( )
∂ 1 ∂
∇·A = cos θ − sin θ (Ar cos θ − Aθ sin θ)
∂r r ∂θ
( )
∂ 1 ∂
+ sin θ + cos θ (Ar sin θ + Aθ cos θ)
∂r r ∂θ
( ) ( )
∂Ar
∂A H 1 ∂Ar 1 1 ∂Aθ Z 1
+ −scHH + s2 Ar + s2
θ
= c2 − cs + scZ
Aθ
∂r ∂r r ∂θH r r ∂θ r Z
( ) ( )
∂Ar ∂A H 1 ∂Ar 1 1 ∂Aθ Z 1
+ csHH + c2 Ar + c2
θ
+ s2 + sc − csZ
θ
A
∂r ∂r r ∂θH r r ∂θ r Z
∂Ar 1 1 ∂Aθ
= + Ar + , (755)
∂r r r ∂θ
in agreement with the expression in Eq. (F.2) (with the first term expanded out). The
z term is the same.
This flux is different due to the facts that both the value of A1 and the area of the
face at (x1 + dx1 , x2 , x3 ) are different (in general) from what they are at (x1 , x2 , x3 ).
The volume of the box is f1 dx1 · f2 dx2 · f3 dx3 , so the net contribution from these two
223
1 ∂(A1 f2 f3 )
, (758)
f1 f2 f3 ∂x1
which agrees with the first term in Eq. (F.31). The other two pairs of faces work out
in exactly the same manner.
In spherical coordinates, f1 , f2 , f3 are equal to 1, r, r sin θ. So the given expression
becomes (with the indices 1, 2, 3 changed to r, θ, ϕ)
( )
1 ∂(r2 sin θAr ) ∂(r sin θAθ ) ∂(rAϕ )
∇·A = + +
r2 sin θ ∂r ∂θ ∂ϕ
2
1 ∂(r Ar ) 1 ∂(sin θAθ ) 1 ∂Aϕ
= + + , (759)
r2 ∂r r sin θ ∂θ r sin θ ∂ϕ
The first term is quickly found to be 16r2 (sin4 θ +cos4 θ). The second term equals
(letting sin θ → s and cos θ → c)
∂(4s3 c − 4c3 s)
r2 = r2 (12s2 c2 − 4s4 + 12c2 s2 − 4c4 ) = 24r2 s2 c2 − 4r2 (s4 + c4 ).
∂θ
(761)
Putting it all together gives
as desired.
In
∫ the special case where the cylinder is very small, we can write the volume integral
∇2f dV as (πr2 z)(∇2f )center . Therefore,
dfavg,r 1 r
= (πr2 z)(∇2f )center = (∇2f )center . (764)
dr 2πrz 2
Since (∇2f )center is a constant, we can integrate with respect to r to obtain
r2 2
favg,r = fcenter + (∇ f )center . (765)
4
Similarly, in 1D the area in Eq. (F.19) is the area 2yz of the two faces of the slab. So
we have ∫
dfavg,±x 1
= ∇2f dV. (766)
dx 2yz
In
∫ the special case where the slab is very small, we can write the volume integral
∇2f dV as (2xyz)(∇2f )center , assuming that the slab extends from −x to x. There-
fore,
dfavg,±x 1
= (2xyz)(∇2f )center = x(∇2f )center . (767)
dx 2yz
Integrating with respect to x gives
x2 2
favg,±x = fcenter + (∇ f )center . (768)
2
Note that the denominator in the second term in the results in Eqs. (768), (765), and
(F.25) equals 3d, where d is the dimension of the space.
The result in Eq. (768) makes sense, because in 1D, ∇2f is simply d2 f /dx2 , so the
result is consistent with what we obtain by adding together the Taylor series,
x2 ′′
f (x) = f (0) + xf ′ (0) + f (0) + · · · , (769)
2
evaluated at x and −x, and dividing by 2. The first-order terms cancel, while the
zeroth- and second-order terms add, yielding Eq. (768).
the function and its first two derivatives. So it must in fact be the correct function
(at least to second order).
When we take the average of the above expression over the surface of the given cube,
the x, y, and z terms vanish, because for every point with a given value of x, there
is a point with the value −x. For the same reason, the xy, xz, and yz terms vanish.
Our task therefore reduces to finding the average value of, say, x2 over the surface of
the cube. Two of the faces have x = ±ℓ, so the average value of x2 over these faces
is simply ℓ2 . For the other four faces, x ranges from −ℓ to ℓ, so the average of x2 is
(∫ℓ )
given by the integral −ℓ x2 dx /2ℓ, which equals ℓ2 /3. The average of x2 over all six
faces is therefore ( )
1 ℓ2 5ℓ2
x2 = 2 · ℓ2 + 4 · = . (771)
6 3 9
The same result holds for the averages of y 2 and z 2 , so from Eq. (770) the average of
f over the entire surface of the cube is
1 5ℓ2 5ℓ2 2
favg = fcenter + (fxx + fyy + fzz ) = fcenter + (∇ f )center , (772)
2 9 18
as desired. A sphere with radius ℓ lies completely inside the given cube of side 2ℓ.
From Eq. (F.25), this sphere would have a 1/6 in place of the 5/18. Consistent with
this, we have 5/18 > 1/6; the average value of f over the surface of the cube is larger
2
than the average value over the
√ surface of the sphere (assuming (∇ f )center is positive).
Also, a sphere with radius 3ℓ lies completely √ outside the given cube of side 2ℓ (it
touches its corers). This sphere would have a ( 3)2 /6 in place of the 5/18. Consistent
with this, we have 3/6 > 5/18.
226 CHAPTER 11. MAGNETIC FIELDS IN MATTER
Appendix H
e2 (v/t)2
·t
6πϵ0 c3 e2 4 e2 1 4 r0
2 = = = . (773)
mv 3πϵ0 mc3 t 3 4πϵ0 mc2 ct 3 ct
2
Note that ct is larger than vt, which is twice as large as the stopping distance vt/2.
So unless the electron stops within a distance that is of order r0 , the radiated energy
will be negligible compared with the initial kinetic energy.
H.2. Simple harmonic motion
(a) If the position is given by x = A cos ωt, then the acceleration is a(t) = d2 x/dt2 =
−Aω 2 cos ωt. The average of cos2 ωt over a period is 1/2, so the average of a2 is
A2 ω 4 /2. From Eq. (H.7) the average power radiated is then
e2 (A2 ω 4 /2) e2 A2 ω 4
P = = . (774)
6πϵ0 c3 12πϵ0 c3
(b) The velocity of the electron is v(t) = −Aω sin ωt, so the initial speed is v = Aω.
The initial energy of the oscillator is therefore U = mv 2 /2 = mA2 ω 2 /2. As
time goes on, the amplitude will decrease. But in terms of the amplitude at any
instant, U and P are given by the above expressions. They are therefore always
related by P = (e2 ω 2 /6πϵ0 mc3 )U . So with 6πϵ0 mc3 /e2 ω 2 ≡ T , we have
∫ U ∫ t ′
dU dU U dU ′ dt
= −P =⇒ =− =⇒ ′
= −
dt dt T U0 U 0 T
( )
U t
=⇒ ln =− =⇒ U (t) = U0 e−t/T . (775)
U0 T
So U falls to 1/e of its initial value after a time T = 6πϵ0 mc3 /e2 ω 2 . There
was actually no need to separate variables and integrate as we did, because we
know that the solution to the differential equation, dU/dt ∝ −U , is simply an
exponential.
Note that ωT , which is the number of radians of oscillation during the time T ,
can be written as
6πϵ0 mc3 c 6πϵ0 mc2 λ 3 3 λ
ωT = = = = , (776)
e2 ω ω e2 2π 2r0 4π r0
227
228 CHAPTER 11. MAGNETIC FIELDS IN MATTER
where λ is the wavelength of the emitted light (which satisfied λν = c), and r0
is the classical electron radius, r0 = e2 /4πϵ0 mc2 . So if λ is much larger than r0 ,
then many oscillations will occur before the amplitude decays significantly.
Dividing this result by the power density (power per unit area), ϵ0 E02 c/2, we find that
the area that receives an amount of power P is
e4
σ= . (778)
6πϵ20 m2 c4
This is the scattering cross section. In terms of the classical electron radius, r0 =
e2 /4πϵ0 mc2 , we can write σ as σ = (8π/3)r02 . Since r0 = 2.8 · 10−15 m, we have
σ = 6.6 · 10−29 m2 . As far as energy absorption goes, the electron looks like it takes
up this much area, from the wave’s point of view.
H.4. Synchrotron radiation
Let the electron’s velocity in the lab frame be v. Consider the inertial frame F ′
moving along with the electron at a given instant. Using the Lorentz transformations,
the electric field in F ′ is E′⊥ = γv × B⊥ . (B⊥ here is simply the B field in the lab
frame.) This electric field causes the electron to accelerate in frame F ′ (where its
initial velocity was zero) with an acceleration of a = eE⊥ /m. From Eq. (H.7) this
acceleration causes the electron to radiate energy in F ′ at a rate
e2 (eE⊥ /m)2 γ 2 e4 v 2 B 2 γ 2 e4 B 2
P′ = = ≈ , (779)
6πϵ0 c3 6πϵ0 m2 c3 6πϵ0 m2 c
where we have used the fact that β ≈ c, since we are told that the electron is highly
relativistic.
We now claim that the power is the same in both the frame F ′ and the lab frame F , in
which case transforming back to the lab frame doesn’t change the answer. This claim
is true because power is energy per time, and both energy U and time t transform the
same way under a Lorentz transformation; they are both the fourth (or first, depending
on the convention) component of a 4-vector. More precisely, in this particular case
the x-t Lorentz transformation gives ∆t = γ∆t′ , because ∆x′ = 0 in F ′ . And the p-E
Lorentz transformation (E here is energy, not electric field) gives E = γE ′ , because
p′ = 0 in F ′ . This then implies ∆E = γ∆E ′ . Therefore, ∆E/∆t = ∆E ′ /∆t′ =⇒
P = P ′.
As time goes on, we will need to continually pick new inertial frames F ′ that move
along with the electron. In any one of these frames the power equals the P ′ we found
above, so the power in the lab frame takes on the constant value P = P ′ .
Appendix J
229
230 CHAPTER 11. MAGNETIC FIELDS IN MATTER
mnet = f N m = (3.5 · 10−7 )(1.33 · 1025 )(1.41 · 10−26 J/T) = 6.6 · 10−8 J/T. (784)
If you want to be a little more precise, you can use the Boltzmann distribution.
Let N/2 + n protons point in the direction of B0 , and N/2 − n point in the
opposite direction. The difference in energy of these two states is 2mB0 . So in
thermal equilibrium we have
Actually, the coil in Fig. J.3 is more like a solenoid than a ring, and the water
almost fills it. But the result from Exercise 11.19, combined with the fact that
the field inside the solenoid is somewhat uniform, implies that the actual E0 won’t
be too much different from the one we calculated.