Physics 416G: Solutions For Problem Set 6: Due: October 9, 2015
Physics 416G: Solutions For Problem Set 6: Due: October 9, 2015
Physics 416G: Solutions For Problem Set 6: Due: October 9, 2015
Please read Griffiths chap. 3, sec. 3.4, 3.1 and 3.3, pages 113-166 except 124-130.
We covered the material from slightly different point of view in the class.
Please show all the details of your computations including intermediate steps.
1 Problem 1
If the electric field in some region is given by the expression (in spherical coordinate)
kh i
E(~r) = 3r + 2 sin cos sin + sin cos ,
r
for some constant k, what is the charge density?
~ E=
Sol: The charge density can be directly computed using the Gauss law
o .
~ E = 0 1 r r2 Er + 1 (sin E ) + 1 (E )
= o
r2 r sin r sin
1 2 3k 1 2k sin cos sin 1 k sin cos
= 0 r r + sin +
r2 r r sin r r sin r
k0 3k 0
= 2 3 + 2 sin (2 cos2 sin2 ) sin = 2 1 + sin (2 cos2 1)
r r
3k0
= 2 [1 + sin cos(2)] .
r
This exercise is for practice the derivatives in the spherical coordinate.
2 Problem 2
Three point charges are located at a distance d from the origin, q at (0, d, 0), q at (0, d, 0), and q
at (0, 0, d).
a) Draw the rough picture of the charge configuration and write down the volume charge distribution
function (~r0 ).
b) Consider the multipole expansion in the three lowest orders and compute the potential at the location
|~r| which is much bigger than d, |~r| d. Express your answer in the spherical polar coordinate.
Hint: Please first write it down the general formula of the multipole expansion in terms of the charge
distribution, which was written down in class. In Griffiths, they are given in Problem 3.52, p. 165, in
the 4th edition.
c) Find the approximate electric field at points far from the origin. Express your answer in spherical
coordinates, and include the three lowest orders in the multipole expansion.
Sol: a) The charge density is
Thus
2 2 2 2 2
1 q qd cos 2 sin cos 5 sin sin + 4 cos
(~r) = + + qd +
40 r r2 2r3
2 2
1 q qd cos 2 sin (3 cos(2) 2) + 4 cos
= + + qd + ,
40 r r2 2r3
where we use the spherical polar coordinate
x = r sin cos + cos cos sin , y = r sin sin + cos sin + cos , z = r cos sin .
~ we get
c) By taking a derivative E = ,
3 Problem 3
During the lecture, we learned that the potential, in the presence of Azimuthal symmetry (potential
is independent of the angle ), can be expressed as
X Bl
(r, ) = Al rl + l+1 Pl (cos ) ,
r
l=0
which should be valid for each l because the Legendre polynomial Pl (cos ) are orthogonal each other.
To prove this, we multiply Pl0 (cos ) and integrate to use the orthogonality properties, given by
Z
2
d(cos )Pl (cos )Pl0 (cos ) = ll0 .
0 2l + 1
Thus we have
Bl = Al R2L+1 .
To determine Al , we use the fact that the surface charge produces the discontinuity in the electric
field, only the normal component of the electric field with respect to the surface. This is radial direction.
We can express this as n E = r E = r . Thus we have
0
(r out r in ) = ,
r=R 0
which gives
X Bl X 0
(l + 1) l+2 Pl (cos ) lAl Rl1 Pl (cos ) = ,
R 0
l=0 l=0
X 0
(2l + 1)Al Rl1 Pl (cos ) = , (2)
0
l=0
where we use Bl = Al R2L+1 . By multiplying Pl0 (cos ) and using the orthogonality properties, we get
Z
1
Al = l1 0 ()Pl (cos )d(cos ) .
2R 0 0
Until now we use a general formula that will hold with the Azimuthal symmetry in spherical polar
coordinate.
For 0 () = k cos = kP1 (cos ), we know that only A1 will be non-zero due to the orthogonality
property of Legendre polynomial. Thus
Z
1 k
Al = l1 kP1 (cos )Pl (cos )d(cos ) = l1 .
2R 0 0 30
Thus we determine the potential as
k k
in (r, ) = rP1 (cos ) = r cos rR,
30 30
kR3 1 kR3 1
out (r, ) = 2
P1 (cos ) = cos rR.
30 r 30 r2
b) We canP use the results from the part a). Let us use the equation (2).
Then 0 = 0 l=0 (2l + 1)Al Rl1 Pl (cos ) . Now we can use one of the two equation for the potential,
inside or outside, of the sphere. Let us consider (1). This gives
X
in (r = R, ) = V0 () = k cos = Al Rl Pl (cos ) ,
l=0
Z Z
2l + 1 2l + 1
Al = V0 ()Pl (cos )d(cos ) = k P1 (cos )Pl (cos )d(cos )
2Rl 0 2Rl 0
2l + 1 2 k
= k l1 = l1 .
2Rl 2l + 1 R
Thus
3k0 3k0
0 = P1 (cos ) = cos .
R R
3k0
This is consistent with the part a), which can be check if you plug k = R into the result of a).
4 Problem 4 : Quadrupole moment (Bonus problem, optional)
a) Place two charges +q at two diagonal corners of a square (a, a, 0) and two minus charges q
at the two other diagonals of a square (a, a, 0), where the upper R and lower signs are correlated.
Evaluate the primitive quadrupole moment components Qij = 21 ri rj (~r)d3 x and use the result to
write down the asymptotic electrostatic potential in Cartesian coordinates.
b) Write the primitive electric quadrupole tensor explicitly in the form Q = iQij j, where 1 = x, 2 =
y, 3 = z and Qij are the corresponding matrix elements. Do not write down the vanishing components
for the situation of part (a). The remaining parts of this problem exploit the fact that your formula
is written in Cartesian coordinates, but gives a valid representation of the quadrupole tensor in any
coordinate system.
c) Express the Cartesian unit vectors x, y, z in terms of the spherical polar unit vectors r, , . Sub-
stituting these expressions into the expression for the quadrupole tensor you wrote down in part (b),
determine all nine matrix elements of Q in spherical polar coordinates, i.e., Qrr , Qr , etc.
d) Write down the expression for the corresponding electric potential in spherical polar coordinates
(where now 1 = r, 2 = , 3 = ) without substituting the matrix elements of Q. Write down only the
non-vanishing terms.
e) Combining the results of parts (c) and (d), compute the electric potential and compare with the
answer from part (a).
f) Explain why you cannot begin in spherical coordinates and use the formula for Qij quoted in part
(a) to reproduce the results of part (d).
Sol: a) The components of the primitive electric quadrupole moment tensor are Qij = 12 ri rj (~r)d3 x.
R
Since
(x, y, z) = q(z) ((x a)(y a) + (x + a)(y + a)) q(z) ((x a)(y + a) + (x + a)(y a)) .
From the delta functions, we see that all four terms contribute equally to both Qxy and Qyx . Therefore,
0 1 0
Q = 2qa2 1 0 0 .
0 0 0
The potential produced by this quadrupole is
3ri rj ij r2 xy
(~r) = Qij = 12qa2 2 .
r5 (x + y 2 )5/2
b) The primitive quadrupole tensor is Q = 2qa2 (xy + yx).
c) Writing the Cartesian unit vectors in terms of the spherical polar unit vectors gives
x = r sin cos + cos cos sin , y = r sin sin + cos sin + cos , z = r cos sin .
Substituting these into part (b) and simplifying yields the nine matrix elements of Q in spherical polar
coordinates:
Qrr = 4qa2 sin2 cos sin , Q = 4qa2 cos2 cos sin , Q = 4qa2 sin cos ,
Qr = Qr = 4qa2 sin cos cos sin , Qr = Qr = 4qa2 sin (cos sin ) ,
Q = Q = 4qa2 cos (cos sin ) .
d) Since r = rr in spherical polar coordinates, all terms involving r and r will vanish. Thus
3ri rj ij r2 2r2 r2 r2
radial (~r) = Qij = Qrr Q Q .
r5 r5 r5 r5
e) Substituting the appropriate results from part (c) into part (d) gives
r2
radial (~r) = 4qa2 sin cos (2 sin2 cos2 + 1) .
r5
Since x = r sin cos and y = r sin sin , the electric potential is
xy
radial (~r) = 12qa2 2 .
(x + y 2 )5/2
This is the same as the result in a).
f) The definition of the components of the quadrupole tensor in part (a) is valid in Cartesian coordinates
only.