Chap 21 SM PDF
Chap 21 SM PDF
Chap 21 SM PDF
A carbon atom can become a carbon ion if it has one or more of its
electrons removed during a process called ionization. What is the net charge on a
carbon atom that has had two of its electrons removed? (a) +e, (b) e, (c) +2e,
(d) 2e
Determine the Concept If two electrons are removed from a carbon atom, it will
have a net positive charge of +2e. (c ) is correct.
3
You have a positively charged insulating rod and two metal spheres on
insulating stands. Give step-by-step directions of how the rod, without actually
touching either sphere, can be used to give one of the spheres (a) a negative
charge, and (b) a positive charge.
Chapter 21
(a) Two point particles that have charges of +4q and 3q are separated
by distance d. Use field lines to draw a visualization of the electric field in the
neighborhood of this system. (b) Draw the field lines at distances much greater
than d from the charges.
Determine the Concept (a) We can use the rules for drawing electric field lines
to draw the electric field lines for this system. Two field lines have been assigned
to each charge q. (b) At distances much greater than the separation distance
between the two charges, the system of two charged bodies will look like a
single charge of +q and the field pattern will be that due to a point charge of +q.
Eight field lines have been assigned to the single charge.
(a)
(b)
+ 4q
3q
+q
Two positive point charges that are equal in magnitude are fixed in
place, one at x = 0.00 m and the other at x = 1.00 m, on the x axis. A third positive
point charge is placed at an equilibrium position. (a) Where is this equilibrium
position? (b) Is the equilibrium position stable if the third particle is constrained
to move parallel with the x axis? (c) What about if it is constrained to move
parallel with the y axis? Explain.
Determine the Concept
(a) A third positive charge can be placed midway between the fixed positive
charges. This is the only location.
(b) Yes. The position identified in (a) is one of stable equilibrium. It is stable in
the x direction because, regardless of whether you displace the third positive
charge to the right or to the left, the net force acting on it is back toward the
midpoint between the two fixed charges.
(c) If the third positive charge is displaced in the y direction, the net force acting
on it will be away from its equilibrium position. Hence, the position midway
between the fixed positive charges is one of unstable equilibrium in the y
direction.
9
Chapter 21
+
+
+
+
+
+
10
Three point charges, +q, +Q, and Q, are placed at the corners of an
equilateral triangle as shown in Figure 21-33. No other charged objects are
nearby. (a) What is the direction of the net force on charge +q due to the other
two charges? (b) What is the total electric force on the system of three charges?
Explain.
Determine the Concept The forces
acting on point charge +q are shown in
the diagram. The force acting on point
charge +q due to point charge Q is
along the line joining them and directed
toward Q. The force acting on point
charge +q due to point charge +Q is
along the line joining them and directed
away from point charge +Q.
r
F+Q
+q
r
FQ
1
+Q
(a) Because point charges +Q and Q are equal in magnitude, the forces due to
these charges are equal and their sum (the net force on charge +q) will
be to the right. Note that the vertical components of these forces add up to zero.
(b) Because no other charged objects are nearby, the forces acting on this system
of three point charges are internal forces and the net force acting on the system is
zero .
11
A positively
charged particle is free to move in a region with an
r
electric field E . Which statements must be true?
(a)
r
The particle is accelerating in the direction perpendicular to E .
r
The particle is accelerating in the directionrof E .
The particle is moving in the direction of E .
The particle could be momentarily at rest.
r
The force on the particle is opposite the directionrof E .
The particle is moving opposite the direction of E .
r
Er is zero at the midpoints of all four sides of the square.
Er is zero at the center of the square.
E is zero midway between the top two charges and midway between the
bottom two charges.
r
Determine the Concept E is zero wherever the net force acting on a test charge
is zero.
(a) False. A test charge placed at these locations will experience a net force.
Chapter 21
(b) True. At the center of the square the two positive charges alone produce a net
electric field of zero, and the two negative charges alone also produce a net
electric field of zero. Thus, the net force acting on a test charge at the midpoint
of the square is zero.
(c) False. A test charge placed at either of these locations will experience a net
force.
13
[SSM] Two point particles that have charges of +q and 3q are
separated by distance d. (a) Use field lines to sketch the electric field in the
neighborhood of this system. (b) Draw the field lines at distances much greater
than d from the charges.
Determine the Concept (a) We can use the rules for drawing electric field lines
to draw the electric field lines for this system. In the field-line sketch weve
assigned 2 field lines to each charge q. (b) At distances much greater than the
separation distance between the two charges, the system of two charged bodies
will look like a single charge of 2q and the field pattern will be that due to a
point charge of 2q. Four field lines have been assigned to each charge q.
(a)
(b)
2q
14
Three equal positive point charges (each charge +q) are fixed at the
vertices of an equilateral triangle with sides of length a. The origin is at the
midpoint of one side the triangle, the center of the triangle on the x axis at x = x1
and the vertex opposite the origin is on the x axis at x = x2. (a) Express x1 and x2 in
terms of a. (b) Write an expression for the electric field on the x axis a distance x
from the origin on the interval 0 x < x2. (c) Show that the expression you
obtained in (b) gives the expected results for x = 0 and for x = x1.
1
2
+ q3
30
P
x
x1
12 a
+ q2
x2
+q
x22 + ( 12 a ) = a 2 x2 =
2
3
6
1
2
r
r
r
r
E P = E 2 + E3 + E 4
(1)
r
kq
kq
E 2 = 2 2 r2, P = 3 2 r2, P
r2, P
r2, P
where
r2, P = (x x2 ) i = x 23 a i
and
r2, P = x2 x
r
Substituting for q2, r2, P , and r2,P
yields:
x1
x1 = 12 a tan 30
a
tan 30 =
x1 =
3
2
r
E2 =
kq
3
2
ax
3
2
kq
ax
(x
)
3
2
a i
Chapter 21
r
kq r
kq r
E 3 = 2 3 r3, P = 3 3 r3, P
r3, P
r3, P
where
r
r3, P = (x 0) i + (0 12 a ) j = x i 12 aj
and
(x 0)2 + (0 12 a )2
r3, P =
r
Substituting for q3, r3, P , and r3,P
r
E3 =
yields:
r
Proceed similarly for E 4 to obtain:
r
E4 =
kq
(x
(x
+ a
1
4
kq
+ a
1
4
2 32
= x 2 + 14 a 2
(x i aj )
2 32
1
2
(x i + aj )
1
2
r r
r
Substituting for E2 , E3 , and E4 in equation (1) and simplifying yields:
r
EP =
kq
3
2
ax
= kq
i +
1
3
2
ax
(x
kq
2
+ a
1
4
) (x
2
2 32
2x
2
+ 14 a
(x i aj )+
1
2
2 32
(x
kq
2
+ a
1
4
2 32
(x i + aj )
1
2
r
(c) Evaluating E P (0) yields:
r
kq
1
2(0)
i =
E P (0) = kq
+
i
2
32
2
2
1
3
23 a
4a
2 a
Note that, because the electric fields due to q3 and q4 cancel each other at the
origin and the resultant field is that due to q2, this is the expected result.
( ) ( )
r
Evaluate E P
( )
( a ) to obtain:
3
6
( a)
i = kq 3 + 3 i = 0
2
32
a 2 a 2
2
3
3
3
+ 14 a 2
6 a
2 a 6 a
Note that, because the point at the center of the equilateral triangle is equidistant
from the three vertices, the electric fields due to the charges at the vertices cancel
and this is the expected result.
r
EP
( a ) = kq
3
6
) (( )
3
6
r
A molecule has a dipole moment given by p . The molecule is
r
r
momentarily at rest with p making an angle with a uniform electric field E .
Describe the subsequent motion of the dipole moment.
15
Determine the Concept The dipole moment rotates back and forth in oscillatory
motion. The dipole moment gains angular speed as it rotates toward the direction
of the electric field, and loses angular speed as it rotates away from the direction
of the electric field.
True or false:
16
(a)
(b)
The electric field of a point charge always points away from the charge.
The electric force on a charged particle in an electric field is always in the
same direction as the field.
Electric field lines never intersect.
All molecules have dipole moments in the presence of an external electric
field.
(c)
(d)
Comment:
10
Chapter 21
F = 0 to a proton to obtain:
Fbinding Felectrostatic = 0
kq 2
kq 2
Felectrostatic = 0 Fbinding = 2
2
r
r
Fbinding =
(8.988 10
)(
m)
N m 2 / C 2 1.602 10 19 C
(10
15
0.2 kN
11
Picture the Problem We can use the definition of electric field to express E in
terms of the work done on the ionizing electrons and the distance they travel
between collisions. We can use the ideal-gas law to relate the number density of
molecules in the gas and the scattering cross-section to the mean free path
and, hence, to the electric field.
W = K = Fs
F = qE
1
n
W=
n=
E=
qE
E =
nW
q
N
P
=
V kT
PW
qkT
(1)
(0.10 nm )101.325 10
N
(1.0 eV )1.602 10 19 J
2
eV
m
2.4 10 6 N/C
19
23 J
1.602 10 C 1.381 10
(300 K )
K
E=
E T 1
EP
12
Chapter 21
Charge
A plastic rod is rubbed against a wool shirt, thereby acquiring a charge
21
of 0.80 C. How many electrons are transferred from the wool shirt to the plastic
rod?
Picture the Problem The charge acquired by the plastic rod is an integral
number of electronic charges, that is, q = ne(e).
Relate the charge acquired by the
plastic rod to the number of
electrons transferred from the wool
shirt:
q = ne ( e ) ne =
ne =
q
e
0.80 C
1.602 10 19
C
electron
)(
[SSM]
Picture the Problem We can find the number of coulombs of positive charge
there are in 1.00 kg of carbon from Q = 6nCe , where nC is the number of atoms in
1.00 kg of carbon and the factor of 6 is present to account for the presence of 6
protons in each atom. We can find the number of atoms in 1.00 kg of carbon by
setting up a proportion relating Avogadros number, the mass of carbon, and the
molecular mass of carbon to nC. See Appendix C for the molar mass of carbon.
Q = 6nC e
N m
nC mC
nC = A C
=
M
NA M
Q=
13
6 N A mC e
M
atoms
19
6 6.022 10 23
(1.00 kg ) 1.602 10 C
mol
Q=
= 4.82 10 7 C
kg
0.01201
mol
24
Suppose a cube of aluminum which is 1.00 cm on a side accumulates a
net charge of +2.50 pC. (a)What percentage of the electrons originally in the
cube was removed? (b) By what percentage has the mass of the cube decreased
because of this removal?
Picture the Problem (a) The percentage of the electrons originally in the cube
that was removed can be found from the ratio of the number of electrons removed
to the number of electrons originally in the cube. (b) The percentage decrease in
the mass of the cube can be found from the ratio of the mass of the electrons
removed to the mass of the cube.
(a) Express the ratio of the electrons
removed to the number of electrons
originally in the cube:
Qaccumulated
N rem
e
=
N ini
N electrons N atoms
per atom
V
mcube
= Al cube
mAl atom
mAl atom
N atoms =
mAl atom =
M Al
NA
(1)
14
Chapter 21
N atoms =
N rem
N ini
AlVcube
M Al
NA
AlVcube N A
M Al
Qaccumulated
e
=
AlVcube N A
N electrons
M Al
per atom
Qaccumulated M Al
N electrons AlVcube eN A
per atom
N rem
:
N ini
=
g
N ini electrons
3
23 atoms
19
13
2.70 3 (1.00 cm ) 1.602 10 C 6.022 10
mol
atom
cm
1.99 10 15 %
(b) Express the ratio of the mass of
the electrons removed to the mass of
the cube:
N rem =
Qaccumulated
e
mrem
=
mcube
=
mrem
=
mcube
Qaccumulated
melectron
e
AlVcube
Qaccumulated melectron
e AlVcube
mrem
:
mcube
(2.50 pC)(9.109 10 31 kg )
(1.602 10
19
C 2.70 3 1.00 cm 3
cm
5.26 10 19 %
15
25
During a process described by the photoelectric effect, ultra-violet
light can be used to charge a piece of metal. (a) If such light is incident on a slab
of conducting material and electrons are ejected with enough energy that they
escape the surface of the metal, how long before the metal has a net charge of
+1.50 nC if 1.00 106 electrons are ejected per second? (b) If 1.3 eV is needed to
eject an electron from the surface, what is the power rating of the light beam?
(Assume this process is 100% efficient.)
Picture the Problem (a) The required time is the ratio of the charge that
accumulates to the rate at which it is delivered to the conductor. (b) We can use
the definition of power to find the power rating of the light beam.
t =
q
q
=
I
dq dt
1.50 nC
1h
= 9.363 10 3
electrons
C
3600 s
6
19
1.00 10
1.602 10
s
electron
= 2.60 h
(b) The power rating of the light
beam is the rate at which it delivers
energy:
P=
E
t
E = E per
electron
P=
E per
electron
I electron t
I electron t
= E per
electron
I electron
eV
1.602 10 -19
P = 1.3
eV
electron
J
electrons
13
1.00 10 6
= 2.1 10 W
s
16
Chapter 21
Coulombs Law
A point charge q1 = 4.0 C is at the origin and a point charge q2 = 6.0
26
C is on the x-axis at x = 3.0 m. (a) Find the electric force on charge q2. (b) Find
the electric force on q1. (c) How would your answers for Parts (a) and (b) differ if
q2 were 6.0 C?
Picture the Problem We can find the electric forces the two charges exert on
each by applying Coulombs law and Newtons 3rd law. Note that
r1, 2 = i because the vector pointing from q1 to q2 is in the positive x direction.
The diagram shows the situation for Parts (a) and (b).
r
F2,1
r
F1, 2
x, m
q2 = 6.0 C
q1 = 4.0 C
r
kq q
F1, 2 = 12 2 r1, 2
r1, 2
r
Substitute numerical values and evaluate F1, 2 :
r
(8.988 109 N m 2 /C2 )(4.0 C)(6.0 C) i =
F1, 2 =
(3.0 m )2
(24 mN ) i
r
r
(b) Because these are action-andF2,1 = F1, 2 = (24 mN ) i
reaction forces, we can apply
Newtons 3rd law to obtain:
(c) If q2 is 6.0 C, the force between q1 and q2 is attractive and both force
vectors are reversed:
r
8.988 109 N m 2 /C 2 (4.0 C)( 6.0 C)
F1, 2 =
i = (24 mN ) i
(3.0 m )2
and
r
r
F2,1 = F1, 2 =
(24 mN ) i
27
[SSM] Three point charges are on the x-axis: q1 = 6.0 C is at
x = 3.0 m, q2 = 4.0 C is at the origin, and q3 = 6.0 C is at x = 3.0 m. Find the
electric force on q1.
17
r
Picture the Problem q2 exerts an attractive electric force F2,1 on point charge q1
r
and q3 exerts a repulsive electric force F3,1 on point charge q1. We can find the net
electric force on q1 by adding these forces (that is, by using the superposition
principle).
r
F3,1
r
2 F2,1 -1
q1 = 6.0 C
q2 = 4.0 C
x, m
q3 = 6.0C
r
r
r
F1 = F2,1 + F3,1
r
kq q
F2,1 = 12 2 i
r2,1
r
kq q
F3,1 = 12 3 i
r3,1
r kq q
kq q
F1 = 12 2 i 12 3 i
r2,1
r3,1
( )
q
q
= k q1 22 23 i
r2,1 r3,1
r
Substitute numerical values and evaluate F1 :
r
4.0 C
6.0 C
i = 1.5 10 2 N i
28
A 2.0 C point charge and a 4.0 C point charge are a distance L
apart. Where should a third point charge be placed so that the electric force on
that third charge is zero?
Picture the Problem The third point charge should be placed at the location at
which the forces on the third point charge due to each of the other two point
charges cancel. There can be no such place except on the line between the two
point charges. Denote the 2.0 C and 4.0 C point charges by the numerals 2 and
4, respectively, and the third point charge by the numeral 3. Assume that the
2.0 C point charge is to the left of the 4.0 C point charge, let the +x direction be
to the right. Then the 4.0 C point charge is located at x = L.
18
Chapter 21
r
r
F4,3 + F2,3 = 0
or
F4,3 = F2,3
F4,3 =
F2,3 =
kq3 q 4
(L x )2
kq3 q 2
x2
kq3 q 4
kq3 q 2
x2
(L x )
or, simplifying,
q4
q
= 22
2
(L x ) x
2
(1)
x 2 + 2 Lx L2 = 0
x=
2 L 4 L2 + 4 L2
= L 2L
2
The root corresponding to the negative sign between the terms is extraneous
because it corresponds to a position to the left of the 2.0 C point charge and is,
therefore, not a physically meaningful root. Hence the third point charge should
be placed between the point charges and a distance equal to 0.41L away from the
2.0-C charge.
A 2.0 C point charge and a 4.0 C point charge are a distance L
29
apart. Where should a third point charge be placed so that the electric force on
that third charge is zero?
Picture the Problem The third point charge should be placed at the location at
which the forces on the third point charge due to each of the other two point
charges cancel. There can be no such place between the two point charges.
Beyond the 4.0 C point charge, and on the line containing the two point charges,
the force due to the 4.0 C point charge overwhelms the force due to the 2.0 C
point charge. Beyond the 2.0 C point charge, and on the line containing the two
19
point charges, however, we can find a place where these forces cancel because
they are equal in magnitude and oppositely directed. Denote the 2.0 C and 4.0
C point charges by the numerals 2 and 4, respectively, and the third point charge
by the numeral 3. Let the +x direction be to the right with the origin at the
position of the 2.0 C point charge and the 4.0 C point charge be located at
x = L.
Apply the condition for translational
equilibrium to the third point charge:
r
r
F4,3 + F2,3 = 0
or
F4,3 = F2,3
F4,3 =
F2,3 =
(1)
kq3 q 4
(L + x )2
kq3 q 2
x2
kq3 q 4
(L + x )
kq3 q 2
q4
q
= 22
2
2
x
(L + x ) x
x 2 2 Lx L2 = 0
x=
2 L 4 L2 + 4 L2
= L 2L
2
The root corresponding to the positive sign between the terms is extraneous
because it corresponds to a position to the right of the 2.0 C point charge and is,
therefore, not a physically meaningful root. Hence the third point charge should
be placed a distance equal to 0.41L from the 2.0-C charge on the side away
from the 4.0-C charge.
Three point charges, each of magnitude 3.00 nC, are at separate
30
corners of a square of edge length 5.00 cm. The two point charges at opposite
corners are positive, and the third point charge is negative. Find the electric force
exerted by these point charges on a fourth point charge q4 = +3.00 nC at the
remaining corner.
20
Chapter 21
Picture the Problem The configuration of the point charges and the forces on the
fourth point charge are shown in the figure as is a coordinate system. From the
figure it is evident that the net force on the point charge q4 is along the diagonal of
the square and directed away from point charge q3. We can apply Coulombs law
r
r
r
to express F1, 4 , F2, 4 and F3, 4 and then add them (that is, use the principle of
superposition of forces) to find the net electric force on point charge q4.
y, cm
5.00
r
F1, 4
q2 = 3.00 nC
r
F3, 4
0 q = 3.00 nC
3
q4 = 3.00 nC
r
F2, 4
q1 = 3.00 nC
5.00
x , cm
r
r
r
r
F4 = F1, 4 + F2, 4 + F3, 4
(1)
r
kq q
F1, 4 = 12 4 j
r1, 4
r
Substitute numerical values and evaluate F1, 4 :
r
3.00 nC
N m2
j = 3.236 10 5 N j
(3.00 nC )
F1, 4 = 8.988 109
2
2
C
(0.0500 m )
r
kq q
F2, 4 = 22 4 i
r2, 4
r
Substitute numerical values and evaluate F2, 4 :
r
3.00 nC
N m2
i = 3.236 105 N i
(3.00 nC )
F2, 4 = 8.988 109
2
2
C
(0.0500 m )
21
r
kq q
F3, 4 = 32 4 r3, 4 , where r3, 4 is a unit
r3, 4
r
r
r
Express r3, 4 in terms of r3,1 and r1, 4 :
= (0.0500 m ) i + (0.0500 m ) j
r
Convert r3, 4 to r3, 4 :
r
r
(0.0500 m ) i + (0.0500 m ) j = 0.707 i + 0.707 j
r3, 4 = r3, 4 =
r3, 4
(0.0500 m )2 + (0.0500 m )2
r
Substitute numerical values and evaluate F3, 4 :
N m2
3.00 nC
( 3.00 nC )
F3, 4 = 8.988 109
2
0.0500 2 m
) (
0.707 i + 0.707 j
= 1.14 10 5 N i 1.14 10 5 N j
r
Substitute numerical values in equation (1) and simplify to find F4 :
r
F4 = 3.24 10 5 N j + 3.24 10 5 N i 1.14 10 5 N i 1.14 10 5 N j
= 2.10 10 5
) (
) (
N )i + (2.10 10 N ) j
) (
This result tells us that the net force is 2.97 105 N along the diagonal in the
direction away from the 3.0 nC charge.
31
A point charge of 5.00 C is on the y axis at y = 3.00 cm, and a second
point charge of 5.00 C is on the y axis at y = 3.00 cm. Find the electric force
on a point charge of 2.00 C on the x axis at x = 8.00 cm.
Picture the Problem The configuration of the point charges and the forces on
point charge q3 are shown in the figure as is a coordinate system. From the
geometry of the charge distribution it is evident that the net force on the 2.00 C
point charge is in the negative y direction. We can apply Coulombs law to
r
r
express F1,3 and F2,3 and then add them (that is, use the principle of superposition
22
Chapter 21
y, cm
3.00
q1 = 5.00 C
3.00
r
F2,3
q3 = 2.00C
8.00
x , cm
r
F1,3
q2 = 5.00C
r
r
r
F3 = F1,3 + F2,3
r
F1,3 = F cos i F sin j
F is given by:
N m2
(5.00 C )(2.00 C )
8.988 10 9
C 2
kq1 q3
F= 2 =
= 12.32 N
r
(0.0300 m )2 + (0.0800 m )2
and
3.00 cm
= 20.56
8.00 cm
= tan 1
The force that point charge q2 exerts
on point charge q3 is:
r
r
Substitute for F1,3 and F2,3 and
simplify to obtain:
r
F2,3 = F cos i F sin j
r
F3 = F cos i F sin j F cos i
F sin j
= 2 F sin j
r
F3 = 2(12.32 N )sin 20.56 j
= (8.65 N ) j
23
32
A point particle that has a charge of 2.5 C is located at the origin. A
second point particle that has a charge of 6.0 C is at x = 1.0 m, y = 0.50 m. A
third point particle, and electron, is at a point with coordinates (x, y). Find the
values of x and y such that the electron is in equilibrium.
Picture the Problem The positions of the point particles are shown in the
diagram. It is apparent that the electron must be located along the line joining the
two point particles. Moreover, because it is negatively charged, it must be closer
to the particle with a charge of 2.5 C than to the particle with a charge of
6.0 C, as is indicated in the figure. We can find the x and y coordinates of the
electrons position by equating the two electrostatic forces acting on it and solving
for its distance from the origin. We can use similar triangles to express this radial
distance in terms of the x and y coordinates of the electron.
y, m
q1 = 6.0C
0.50
xe
ye
r e
F1,e
r
F2,e
F1, e = F2, e
F1,e =
(r +
F2,e =
k q2 e
x, m
q2 = 2.5 C 1.0
(r +
kq1e
1.25 m
r2
q1
1.25 m
q2
r2
24
Chapter 21
6
(r +
r = 2.036 m
1.25 m
2.5
r2
and
r = 0.4386 m
Because r < 0 is unphysical, well
consider only the positive root.
Use the similar triangles in the
diagram to establish the proportion
involving the y coordinate of the
electron:
ye
2.036 m
y e = 0.909 m
=
0.50 m 1.12 m
xe
2.036 m
xe = 1.82 m
=
1.0 m 1.12 m
33
A point particle that has a charge of 1.0 C is located at the origin; a
second point particle that has a charge of 2.0 C is located at x = 0, y = 0.10 m;
and a third point particle that has a charge of 4.0 C is located at x = 0.20 m,
y = 0. Find the total electric force on each of the three point charges.
Picture the Problem Let q1 represent the charge of the point particle at the
origin, q2 the charge of the point particle at (0, 0.10 m), and q3 the charge of the
point particle at (0.20 m, 0). The following diagram shows the forces acting on
each of the point particles. Note the action-and-reaction pairs. We can apply
Coulombs law and the principle of superposition of forces to find the net force
acting on each of the point particles.
y, m
0.10
q2 = 2.0 C
F1,2
F2,1
q1 = 1.0 C
F3,1
F1,3
q3 = 4.0 C
x, m
0.20
F2,3
r
r
r
F1 = F2,1 + F3,1
r
r
kq2 q1
kq2 q1 r2,1 kq2 q1 r
F2,1 = 2 r2,1 = 2
= 3 r2,1
r2,1
r2,1 r2,1
r2,1
r
Substitute numerical values and evaluate F2,1 :
r
r
kq q r
F3,1 = 33 1 r3,1
r3,1
r
Substitute numerical values and evaluate F3,1 :
r
Substitute to find F1 :
r
F1 =
(0.90 N ) i + (1.8 N ) j
25
26
Chapter 21
r
r
r
F2 = F3, 2 + F1, 2
r
r
= F3, 2 F2,1
r
= F3, 2 (1.80 N ) j
r
r
because F1, 2 and F2,1 are action-and-
reaction forces.
Express the force that the point
particle whose charge is q3 exerts on
the point particle whose charge is q2:
r
kq q r
F3, 2 = 33 2 r3, 2
r3, 2
kq3 q 2
( 0.20 m ) i + (0.10 m ) j
r33, 2
r
Substitute numerical values and evaluate F3, 2 :
r
= ( 1.28 N ) i + (0.640 N ) j
Find the net force acting on the point particle whose charge is q2:
r
r
F2 = F3, 2 (1.80 N ) j = ( 1.28 N ) i + (0.640 N ) j (1.80 N ) j
=
( 1.3 N ) i (1.2 N ) j
r
r
r
r
Noting that F1,3 and F3,1 are an action-and-reaction pair, as are F2,3 and F3, 2 ,
express the net force acting on the point particle whose charge is q3:
r
r
r
r
r
F3 = F1,3 + F2,3 = F3,1 F3, 2 = (0.899 N ) i ( 1.28 N ) i + (0.640 N ) j
=
(0.4 N ) i (0.64 N ) j
27
Picture the Problem Let q1 represent the charge of the point particle at the origin
and q3 the charge of the point particle initially at (8.00 cm, 0) and later positioned
at (17.8 cm, 0). The diagram shows the forces acting on the point particle whose
charge is q3 at (8.00 cm, 0). We can apply Coulombs law and the principle of
superposition of forces to find the net force acting on each of the point particles.
y
q3 = 2.00 C
F2,3
q
q1 = 5.00 C
4.00
F1,3
x, cm
8.00
r
r
r
F2 (8.00 cm,0) = F1,3 + F2,3
kq1q3 r
kqq r
r1,3 + 3 3 r2,3
3
r1,3
r2,3
q r
q r
= kq3 31 r1,3 + 3 r2,3
r2,3
r1,3
5.00 C
q
(0.0800 m ) i +
(0.0400 m )i
3
3
(0.0400 m )
(0.0800 m )
q = 3.00 C
28
Chapter 21
r
r
r
Fq = FQ on x axis, q + 2 FQ at 45, q
r
kqQ
FQ on x axis,q = 2 i
R
r
kqQ
2FQ at 45,q = 2 2 cos 45i
R
2 kqQ
=
i
2
2 R
r
Substitute for FQ on x axis,q and
r
2FQ at 45,q to obtain:
r
kqQ
2 kqQ
Fq = 2 i +
i
2
R
2 R
=
kqQ
1 + 2 i
R2
a a
2
where a =1.64 1010 m. To simplify our calculations
,a
then at ,
2
3
2 3
2
2
well set ke a = C = 8.56 109 N . We can apply Coulombs law and the
principle of superposition of forces to find the net force acting on each ion.
z
a
1
2
q1
1
2
a
x
q4 a , a , a 2
2 2 3
1
2
q2
a a 3
,
2 2 ,0
(a,0,0)
q3
r
r
r
r
F1 = F2,1 + F3,1 + F4,1
r
Find F2,1 :
r
kq q
F2,1 = 12 2 r2,1 = C i = Ci
r2,1
r
Find F3,1 :
r
kq q
F3,1 = 32 1 r3,1
r3,1
( )
a
a 3
j
0 i + 0
2
2
=C
a
1
3
= C i +
j
2
2
29
30
Chapter 21
Noting that the magnitude of point charge q4 is three times that of the other point
r
charges and that it is negative, express F4,1 :
a
a
a 2
k
j + 0
0 i + 0
r
2
2 3
3
F4,1 = 3Cr4,1 = 3C
2
2
2
a a a 2
+
+
2 2 3 3
a a a 2
k
j+
i +
1 1 2
2 2 3 3
k
= 3C
= 3C i +
j +
a
2 2 3 3
r
Substitute in the expression for F1 to
obtain:
r
1
3
F1 = Ci C i +
j
2
2
1 1 2
k
+ 3C i +
j +
2 2 3 3
= C 6k
r
r
r
F2 = F3 = F1 = C 6k
r r
r
r
F1 + F2 + F3 + F4 = 0
r
r r
r
r
F4 = F1 + F2 + F3 = 3F1
= 3C 6k
31
r
kq
E (x ) = 2 rP,0
x
N m2
8.988 109
(4.0 C)
r
C 2
E (6.0 m ) =
i = (1.0 kN/C ) i
(6.0 m )2
r
(b) Evaluate E at x = 10 m:
N m2
8.988 10 9
(4.0 C )
r
C 2
E ( 10 m ) =
i =
(10 m )2
( )
( 0.36 kN/C) i
Ex (kN/C)
250
-250
-500
-2
-1
x (m)
38
Two point charges, each +4.0 C, are on the x axis; one point charge is
at the origin and the other is at x = 8.0 m. Find the electric field on the x axis at
(a) x = 2.0 m, (b) x = 2.0 m, (c) x = 6.0 m, and (d) x = 10 m. (e) At what point on
the x axis is the electric field zero? (f) Sketch Ex versus x for 3.0 m < x < 11 m.
Picture the Problem Let q represent the point charges of +4.0 C and use
r
Coulombs law for E due to a point charge and the principle of superposition for
fields to find the electric field at the locations specified.
32
Chapter 21
Noting that q1 = q2, use Coulombs law and the principle of superposition to
express the electric field due to the given charges at point P a distance x from the
origin:
r
r
r
kq
kq 2
E (x ) = E q1 ( x ) + E q2 (x ) = 21 rq1 ,P +
r
x
(8.0 m x )2 q2 ,P
1
1
= kq1 2 rq1 ,P +
r
2 q2 , P
(8.0 m x )
x
1
= 36 kN m 2 /C 2 rq1 ,P +
r
2 q2 , P
(8.0 m x )
x
( )
( )
( 9.4 kN/C ) i
()
( )
(8.0 kN/C ) i
()
( )
( 8.0 kN/C ) i
()
()
r
1
1
i +
i =
E ( 2.0 m ) = 36 kN m 2 /C
2
2
(10 m )
(2.0 m )
r
(b) Evaluate E at x = 2.0 m:
r
1
1
E (2.0 m ) = 36 kN m 2 /C
i +
i =
2
2
(6.0 m )
(2.0 m )
r
(c) Evaluate E at x = 6.0 m:
r
1
1
E (6.0 m ) = 36 kN m 2 /C
i +
i =
2
2
(2.0 m )
(6.0 m )
r
(d) Evaluate E at x = 10 m:
r
1
1
E (10 m ) = 36 kN m 2 /C
i +
i =
2
2
(2.0 m )
(10 m )
r
E (4.0 m ) = 0
(9.4 kN/C) i
33
25
Ex ( kN m /C)
50
0
-25
-50
-75
-100
-3
-2
-1
10
11
x (m)
39
When a 2.0-nC point charge is placed at the origin, it experiences an
electric force of 8.0 104 N in the +y direction. (a) What is the electric field at
the origin? (b) What would be the electric force on a 4.0-nC point charge placed
at the origin? (c) If this force is due to the electric field of a point charge on the y
axis at y = 3.0 cm, what is the value of that charge?
Picture the Problem We can find the electric field at the origin from its
r
r
definition and the electric force on a point charge placed there using F = qE . We
can apply Coulombs law to find the value of the point charge placed at y = 3 cm.
(a) Apply the definition of electric
field to obtain:
r
r
F (0,0) 8.0 10 4 N j
E (0,0 ) =
=
q0
2.0 nC
= 4.0 10 5 N/C j
(b) The force on a point charge in an
electric field is given by:
r
r
F (0,0 ) = qE (0,0 )
= ( 4.0 nC )(400 kN/C ) j
=
( 1.6 mN ) j
kq( 4.0 nC )
j = ( 1.60 mN ) j
(0.030 m )2
( )
34
Chapter 21
q=
(1.60 mN )(0.030 m )2
(8.988 109 N m 2 /C 2 )(4.0 nC)
= 40 nC
40
The electric field near the surface of Earth points downward and has a
magnitude of 150 N/C. (a) Compare the magnitude of the upward electric force
on an electron with the magnitude of the gravitational force on the electron.
(b) What charge should be placed on a ping pong ball of mass 2.70 g so that the
electric force balances the weight of the ball near Earths surface?
Picture the Problem We can compare the electric and gravitational forces acting
on an electron by expressing their ratio. Because the ping pong ball is in
equilibrium under the influence of the electric and gravitational forces acting on
it, we can use the condition for translational equilibrium to find the charge that
would have to be placed on it in order to balance Earths gravitational force on it.
(a) Express the magnitude of the
electric force acting on the electron:
Fe = eE
Fg = me g
Fe eE
=
Fg mg
Fe
1.602 10 19 C (150 N/C )
=
Fg
9.109 10 31 kg 9.81m/s 2
)
)(
= 2.69 10
12
or
Fe =
(2.69 10 )F
12
Fe Fg = 0
or
qE mg = 0 q =
mg
E
q=
(2.70 10
)(
kg 9.81 m/s 2
150 N/C
35
= 0.177 mC
41
[SSM] Two point charges q1 and q2 both have a charge equal to
+6.0 nC and are on the y axis at y1 = +3.0 cm and y2 = 3.0 cm respectively.
(a) What is the magnitude and direction of the electric field on the x axis at
x = 4.0 cm? (b) What is the force exerted on a third charge q0 = 2.0 nC when it is
placed on the x axis at x = 4.0 cm?
Picture the Problem The diagram shows the locations of the point charges q1 and
r
q2 and the point on the x axis at which we are to find E . From symmetry
r
considerations we can conclude that the y component of E at any point on the x
axis is zero. We can use Coulombs law for the electric field due to point charges
and the principle of superposition for fields to find the field at any point on the x
r
r
axis and F = qE to find the force on a point charge q0 placed on the x axis at
x = 4.0 cm.
y, cm
3.0 q1 = 6.0 nC
r
4.0
x , cm
r
Eq1
3.0 q2 = 6.0 nC
r
kq
E x = 2 cos i
r
r
Express E x for both charges:
r
kq
E x = 2 2 cos i
r
36
Chapter 21
Substitute for cos and r, substitute numerical values, and evaluate to obtain:
r
kq 0.040 m 2kq(0.040 m )
E (4.0 cm )x = 2 2
i=
i
r
r
r3
2 8.988 10 9 N m 2 /C 2 (6.0 nC )(0.040 m )
=
i
3 2
(0.030 m )2 + (0.040 m )2
= (34.5 kN/C )i = (35 kN/C ) i
35 kN/C @ 0
r
F = (2.0 nC )(34.5 kN/C )i
=
(69 N )i
42
A point charge of +5.0 C is located on the x axis at x = 3.0 cm, and
a second point charge of 8.0 C is located on the x axi at x = +4.0 cm. Where
should a third charge of +6.0 C be placed so that the electric field at the origin is
zero?
Picture the Problem If the electric field at x = 0 is zero, both its x and y
components must be zero. The only way this condition can be satisfied with point
charges of +5.0 C and 8.0 C on the x axis is if the point charge +6.0 C is
also on the x axis. Let the subscripts 5, 8, and 6 identify the point charges and
r
their fields. We can use Coulombs law for E due to a point charge and the
principle of superposition for fields to determine where the +6.0 C point charge
should be located so that the electric field at x = 0 is zero.
Express the electric field at x = 0 in
terms of the fields due to the point
charges of +5.0 C, 8.0 C, and
+6.0 C:
Substitute for each of the fields to
obtain:
r
r
r
r
E (0 ) = E5 C + E 8 C + E6 C
=0
kq5
kq
kq
r + 26 r6 + 28 r8 = 0
2 5
r5
r6
r8
or
kq5 kq6 kq8
i + 2 i + 2 i =0
r52
r6
r8
( )
( )
q5 q6 q8
=0
r52 r62 r28
8
5
6
2
=0
2
r
(3.0 cm ) 6 (4.0 cm )2
r6 = 2.4 cm
37
43
A 5.0-C point charge is located at x = 4.0 m, y = 2.0 m and a
12-C point charge is located at x = 1.0 m, y = 2.0 m. (a) Find the magnitude and
direction of the electric field at x = 1.0 m, y = 0. (b) Calculate the magnitude and
direction of the electric force on an electron that is placed at x = 1.0 m, y = 0.
Picture the Problem The diagram shows the electric field vectors at the point of
r
interest P due to the two point charges. We can use Coulombs law for E due to
r
point charges and the superposition principle for electric fields to find E P . We
r
r
can apply F = qE to find the force on an electron at (1.0 m, 0).
y, m
q2 = 12 C
2
1
2
1
P
x, m
E1
E2
q1 = 5.0 C
r
r
r
E P = E1 + E 2
38
Chapter 21
r
Substitute numerical values and evaluate E1 :
r
kq
8.988 10 9 N m 2 /C 2 ( 5.0 C ) ( 5.0 m ) i + (2.0 m ) j
E1 = 2 1 r1,P =
2
2
r1,P
(5.0 m )2 + (2.0 m )2
(5.0 m ) + (2.0 m )
= 1.55 10 3 N/C 0.928 i + 0.371 j
)(
r
Substitute numerical values and evaluate E 2 :
r
kq
8.988 10 9 N m 2 /C 2 (12 C ) ( 2.0 m ) i + ( 2.0 m ) j
E 2 = 2 2 r2,P =
2
2
r2,P
(2.0 m )2 + (2.0 m )2
(2.0 m ) + (2.0 m )
= 13.5 10 3 N/C 0.707 i 0.707 j
)(
r
r
r
Substitute for E1 and E 2 and simplify to find E P :
r
E P = (1.44 kN/C ) i + ( 0.575 kN/C ) j + ( 9.54 kN/C ) i + ( 9.54 kN/C ) j
= ( 8.10 kN/C ) i + ( 10.1kN/C ) j
r
The magnitude of EP is:
EP =
= 13 kN/C
r
The direction of E P is:
10.1kN/C
= 230
8.10 kN/C
E = tan 1
)[
r
r
F = qEP = 1.602 1019 C ( 8.10 kN/C) i + ( 10.1kN/C ) j
= 1.30 1015 N i + 1.62 1015 N j
) (
r
Find the magnitude of F :
F=
(1.30 10
15
39
N ) + (1.62 10 15 N )
2
= 2.1 10 15 N
r
Find the direction of F :
1.62 10 15 N
= 51
15
1.3 10 N
F = tan 1
44
Two equal positive charges q are on the y axis; one point charge is at
y = +a and the other is at y = a. (a) Show that on the x axis the x component of
the electric field is given by Ex = 2kqx/(x2 + a2)3/2. (b) Show that near the origin,
where x is much smaller than a, Ex 2kqx/a3. (c) Show that for values of x much
larger than a, Ex 2kq/x2. Explain why a person might expect this result even
without deriving it by taking the appropriate limit.
Picture the Problem The diagram shows the locations of point charges q1 and q2
r
and the point on the x axis at which we are to find E . From symmetry
r
considerations we can conclude that the y component of E at any point on the x
axis is zero. We can use Coulombs law for the electric field due to point charges
and the principle of superposition of fields to find the field at any point on the x
axis. We can establish the results called for in Parts (b) and (c) by factoring the
radicand and using the approximation 1 + 1 whenever << 1.
y
a
q1 = q
r
P
x
Eq
q2 = q
r
kq
E x = 2 2 cos i
r
40
Chapter 21
r
kq x
2kqx
2kqx
E x = 2 2 i = 3 i =
2
r r
r
x + a2
2kqx
=
i
2
2 3 2
x +a
r
The magnitude of E x is:
Ex =
Ex
32
(x
2kqx
2
+ a2 )
32
2kqx
(a )
2 32
2kqx
a3
(c) For x >> a, the charges separated by a would appear to be a single charge of
2kq
magnitude 2q. Its field would be given by Ex = 2 .
x
Factor the radicand to obtain:
For a << x, 1 +
a2
1 and:
x2
a 2
Ex = 2kqx x 2 1 + 2
x
[ ]
Ex = 2kqx x 2
3 2
3 2
2kq
x2
q2 = 5.0 C
2
P
E1
E2
x, m
21
q1 = 4.0 C
22
r
r
r
E P = E1 + E 2
r
kq
8.988 10 9 N m 2 /C 2 ( 4.0 C ) ( 5.0 m ) i + (3.0 m ) j
E1 = 2 1 r1,P =
2
2
r1.P
(5.0 m )2 + (3.0 m )2
(5.0 m ) + (3.0 m )
= ( 1.06 kN/C) 0.857 i + 0.514 j = (0.907 kN/C) i + ( 0.544 kN/C) j
r
Evaluate E 2 :
r
kq
8.988 10 9 N m 2 /C 2 (5.0 C ) ( 4.0 m ) i + ( 2.0 m ) j
E 2 = 2 2 r2,P =
2
2
r2,P
(4.0 m )2 + (2.0 m )2
(4.0 m ) + (2.0 m )
= (2.25 kN/C) 0.894 i 0.447 j = ( 2.01kN/C) i + ( 1.01kN/C) j
r
Substitute and simplify to find E P :
r
E P = (0.908 kN/C) i + ( 0.544 kN/C) j + ( 2.01kN/C) i + ( 1.01kN/C) j
= ( 1.10 kN/C) i + ( 1.55 kN/C) j
r
The magnitude of E P is:
EP =
= 1.9 kN/C
41
42
Chapter 21
r
The direction of E P is:
1.55 kN/C
= 235
1.10 kN/C
E = tan 1
)[
r
r
F = qE P = 1.602 10 19 C ( 1.10 kN/C) i + ( 1.55 kN/C) j
= 1.76 10 16 N i + 2.48 10 16 N j
) (
r
The magnitude of F is:
F=
( 1.76 10
16
) + ( 2.48 10
2
r
The direction of F is:
16
= 3.0 10 16 N
2.48 10 16 N
= 235
16
1.76 10 N
F = tan 1
E x = 2kqx x 2 + a 2
3 2
43
E x
to obtain:
x
[(
3 2
3 2
E x
d
d
2kqx x 2 + a 2
=
= 2kq
x x2 + a2
x
dx
dx
3 2
3 2
d 2
= 2kq x
x + a2
+ x2 + a2
dx
5 2
3
(2 x ) + x 2 + a 2 3 2
= 2kq x x 2 + a 2
2
= 2kq 3 x 2 x 2 + a 2
Set this derivative equal to zero for
extreme values:
Solving for x yields:
5 2
+ x2 + a2
3 2
3x 2 x 2 + a 2
x=
5 2
+ x2 + a2
3 2
=0
a
2
0.2
Ex
0.0
-0.2
-0.4
-10
-5
10
47
[SSM] Two point particles, each having a charge q, sit on the base of
an equilateral triangle that has sides of length L as shown in Figure 21-38. A third
point particle that has a charge equal to 2q sits at the apex of the triangle. Where
must a fourth point particle that has a charge equal to q be placed in order that the
electric field at the center of the triangle be zero? (The center is in the plane of
the triangle and equidistant from the three vertices.)
44
Chapter 21
Picture the Problem The electric field of the 4th charged point particle must
cancel the sum of the electric fields due to the other three charged point particles.
By symmetry, the position of the 4th charged point particle must lie on the vertical
centerline of the triangle. Using trigonometry, one can show that the center of an
equilateral triangle is a distance L 3 from each vertex, where L is the length of
the side of the triangle. Note that the x components of the fields due to the base
charged particles cancel each other, so we only need concern ourselves with the y
components of the fields due to the charged point particles at the vertices of the
triangle. Choose a coordinate system in which the origin is at the midpoint of the
base of the triangle, the +x direction is to the right, and the +y direction is upward.
y
q3 = 2 q
r
E2
L
q4 60
q1
r
E1
3
60 q 2
r
E3
i =1 to 4
=0
r r r
r
Substituting for E1 , E 2 , E3 , and E 4 yields:
k (q )
L
k (q )
k (2q ) kq
cos 60 j
j+ 2 j=0
cos 60 j +
2
2
y
L
L
3
3
y=
L
3
45
The positive solution corresponds to the 4th point particle being a distance
L 3 above the base of the triangle, where it produces the same strength and
same direction electric field caused by the three charges at the corners of the
triangle. So the charged point particle must be placed a distance L 3 below the
midpoint of the triangle.
Two point particles, each having a charge equal to q, sit on the base of
48
an equilateral triangle that has sides of length L as shown in Figure 21-38. A third
point particle that has a charge equal to 2q sits at the apex of the triangle. A fourth
point particle that has charge q is placed at the midpoint of the baseline making
the electric field at the center of the triangle equal to zero. What is the value of
q? (The center is in the plane of the triangle and equidistant from all three
vertices.)
Picture the Problem The electric field of 4th charge must cancel the sum of the
electric fields due to the other three charges. Using trigonometry, one can show
that the center of an equilateral triangle is a distance L 3 from each vertex,
where L is the length of the side of the triangle. The distance from the center point
of the triangle to the midpoint of the base is half this distance. Note that the x
components of the fields due to the base charges cancel each other, so we only
need concern ourselves with the y components of the fields due to the charges at
the vertices of the triangle. Choose a coordinate system in which the origin is at
the midpoint of the base of the triangle, the +x direction is to the right, and the +y
direction is upward.
y
q3 = 2 q
r
E2
L
q1
r
E4
60
r
E3
r
E1
L
60 q 2
q4 = q '
i =1 to 4
=0
46
Chapter 21
r r r
r
Substituting for E1 , E 2 , E3 , and E 4 yields:
k (q )
L
k (q )
k (2q )
kq'
j+
j=0
cos 60 j +
cos 60 j
2
2
2
L
L
L
3
3
2 3
q' =
1
4
(a) Because Ex is in the x direction, a positive test charge that is displaced from
(0, 0) in either the +x direction or the x direction will experience a force pointing
away from the origin and accelerate in the direction of the force. Consequently,
the equilibrium at (0,0) is unstable for a small displacement along the x axis.
If the positive test charge is displaced in the direction of increasing y (the +y
direction), the charge at y = +a will exert a greater force than the charge at
y = a, and the net force is then in the y direction; i.e., it is a restoring force.
Similarly, if the positive test charge is displaced in the direction of decreasing y
(the y direction), the charge at y = a will exert a greater force than the charge at
y = a, and the net force is then in the y direction; i.e., it is a restoring force.
Consequently, the equilibrium at (0,0) is stable for a small displacement along the
y axis.
(b) Following the same arguments as in Part (a), one finds that, for a negative test
charge, the equilibrium is stable at (0,0) for displacements along the x axis and
unstable for displacements along the y axis.
Fq at y=+ a =
q0 = 14 q
47
kqq0
kq 2
+
=0
a2
(2a )2
Remarks: In Part (c), we could just as well have expressed the net force
acting on the charge at y = a. Due to the symmetric distribution of the
charges at y = a and y = +a, summing the forces acting on q0 at the origin
does not lead to a relationship between q0 and q.
50 Two positive point charges +q are on the y axis at y = +a and y = a. A
bead of mass m and charge q slides without friction along a taut thread that runs
along the x axis. Let x be the position of the bead. (a) Show that for x << a, the
bead experiences a linear restoring force (a force that is proportional to x and
directed toward the equilibrium position at x = 0) and therefore undergoes simple
harmonic motion. (b) Find the period of the motion.
Picture the Problem In Problem 44 it is shown that the electric field on the x
axis, due to equal positive point charges located at (0, a) and (0,a), is given by
Ex = 2kqx x 2 + a 2
3 2
For x << a:
Fx = qE x =
Fx =
(x
2kq 2 x
2
+ a2
32
2kq 2 x
x2
a 2 + 1
a
32
2kq 2
x
a3
That is, the bead experiences a linear
restoring force.
Fx =
T = 2
k' =
m
k'
2kq 2
a3
48
Chapter 21
T = 2
m
ma 3
=
2
2kq 2
2kq 2
a3
1.602 10 19 C
e
=
me 9.109 10 31 kg
a=
a=
Fnet eE
=
me me
(1.602 10
C (100 N/C )
9.109 10 31 kg
19
t =
v 0.01c
=
a
a
t =
49
= 0.2 s
(d) Use a constant-acceleration
equation to express the distance
the electron travels in a given
time interval:
Substitute numerical values and
evaluate x:
x = vi t + 12 a (t )
1
2
(1.759 10
13
m/s 2 (0.1704 s )
= 0.3 m
52
The acceleration of a particle in an electric field depends on the
charge-to-mass ratio of the particle. (a) Compute q/m for a proton, and find its
acceleration in a uniform electric field that has a magnitude of 100 N/C. (b) Find
the time it takes for a proton initially at rest in such a field to reach a speed of
0.01c (where c is the speed of light). (When the speed of an electron approaches
the speed of light c, relativistic kinematics must be used to calculate its motion,
but at speeds of 0.01c or less, non-relativistic kinematics is sufficiently accurate
for most purposes.)
Picture the Problem We can use Newtons second law of motion to find the
acceleration of the proton in the uniform electric field and constant-acceleration
equations to find the time required for it to reach a speed of 0.01c and the
distance it travels while acquiring this speed.
(a) Use data found at the back of
your text to compute e/m for an
electron:
e
1.602 10 19 C
=
m p 1.673 10 27 kg
a=
= 9.58 10 7 C/kg
a=
Fnet eE
=
mp mp
(1.602 10
C (100 N/C )
1.673 10 27 kg
19
= 9.576 10 9 m/s 2
= 9.58 10 9 m/s 2
50
Chapter 21
t =
v 0.01c
=
a
a
t =
53
An electron has an initial velocity of 2.00 106 m/s in the +x direction.
r
It enters a region that has a uniform electric field E = (300 N/C) j . (a) Find the
acceleration of the electron. (b) How long does it take for the electron to travel
10.0 cm in the x direction in the region that has the field? (c) Through what angle,
and in what direction, is the electron deflected while traveling the 10.0 cm in the x
direction?
Picture the Problem The electric force acting on the electron is opposite the
direction of the electric field. We can apply Newtons second law to find the
electrons acceleration and use constant acceleration equations to find how long it
takes the electron to travel a given distance and its deflection during this interval
of time. Finally, we can use the pictorial representation to obtain an expression for
the angle through which the electron is deflected while traveling 10.0 cm in the x
direction.
r
E = (300 N/C) j
m, e
r
v = (2.00 10 6 m/s ) i
y
51
r
r
r Fnet eE
=
a=
me
me
r
1.602 10 19 C
(300 N/C) j
a=
9.109 10 31 kg
= 5.276 1013 m/s 2 j
)
) j
x
0.100 m
=
= 50.0 ns
v x 2.00 10 6 m/s
t =
y
x
where y is its vertical deflection.
Using a constant-acceleration
equation, relate the vertical
deflection of the electron to its
acceleration and the elapsed time:
y = 12 a y (t )
= tan 1
a y (t )2
1 a y (t )2
= tan 1
2x
x
= tan 1 2
2(10.0 cm )
= 33.4
Because the acceleration of the electron
is downward and it was moving
horizontally initially, it is deflected
downward .
54
52
Chapter 21
Picture the Problem Because the electric field is in the y direction, the force it
exerts on the electron is in the +y direction. Applying Newtons second law to the
electron will yield an expression for the acceleration of the electron in the y
direction. We can then use a constant-acceleration equation to relate its speed to
its acceleration and the distance it has traveled.
Apply
= ma y to the electron to
FE Fg = ma y
obtain:
eE mg = ma y
ay =
v y2 = v02 + 2a y y
eE
g
m
v y = 2
g y
m
)(
v y = 2
9.81 m/s 2 1.0 10 6 m
31
9.109 10 kg
= 5.8 mm/s
55
A 2.00-g charged particle is released from rest in a region that has a
r
uniform electric field E = (300 N/C) i . After traveling a distance of 0.500 m in
this region, the particle has a kinetic energy of 0.120 J. Determine the charge of
the particle.
Picture the Problem We can apply the work-kinetic energy theorem to relate the
change in the objects kinetic energy to the net force acting on it. We can express
the net force acting on the charged body in terms of its charge and the electric
field.
W = K = Fnet x
Fnet = qE
K = K f K i = qEx
53
or, because Ki = 0,
K f = qEx
q=
q=
Kf
Ex
0.120 J
(300 N/C)(0.500 m )
= 800 C
56
A charged particle leaves the origin with a speed of 3.00 106 m/s at
r
an angle of 35 above the x axis. A uniform electric field given by E = E0 j exists
throughout the region. Find E0 such that the particle will cross the x axis at
x = 1.50 cm if the particle is (a) an electron, and (b) a proton.
Picture the Problem We can use constant-acceleration equations to express the x
and y coordinates of the particle in terms of the parameter t and Newtons second
law to express the constant acceleration in terms of the electric field. Eliminating
t will yield an equation for y as a function of x, q, and m that we can solve for Ey.
x = (v cos ) t
and
y = (v sin )t 12 a y t 2
ay =
Fnet, y
m
qE0
m
y = (v sin )t
y = (tan )x
qE 0 2
t
2m
qE0
x2
2mv 2 cos 2
54
Chapter 21
E0 =
mv 2 sin 2
qx
) mq
10
) 9.109
1.602 10
31
kg
C
19
= 3.2 kN/C
27
.673 10
)11.602
10
19
kg
C
= 5.9 MN/C
57
[SSM] An electron starts at the position shown in Figure 21-39 with
an initial speed v0 = 5.00 106 m/s at 45 to the x axis. The electric field is in the
+y direction and has a magnitude of 3.50 103 N/C. The black lines in the figure
are charged metal plates. On which plate and at what location will the electron
strike?
Picture the Problem We can use constant-acceleration equations to express the x
and y coordinates of the electron in terms of the parameter t and Newtons second
law to express the constant acceleration in terms of the electric field. Eliminating
t will yield an equation for y as a function of x, q, and m. We can decide whether
the electron will strike the upper plate by finding the maximum value of its y
coordinate. Should we find that it does not strike the upper plate, we can
determine where it strikes the lower plate by setting y(x) = 0. Ignore any effects of
gravitational forces.
x = (v0 cos )t
and
y = (v0 sin )t 12 a y t 2
ay =
Fnet, y
me
eE y
me
y = (v0 sin )t
y ( x ) = (tan )x
eE y
dy
= tan
x'
dx
me v02 cos 2
eE y
2me
55
t2
eE y
x 2 (1)
= 0 for extrema
Solve for x to obtain:
x' =
mev02 sin 2
(See remark below.)
2eE y
ymax =
me v02 sin 2
2eE y
19
Because the plates are separated by 2 cm, the electron does not strike the upper
plate.
To determine where the electron will
strike the lower plate, set y = 0 in
equation (1) and solve for x to
obtain:
x=
me v02 sin 2
eE y
19
4.1cm
56
Chapter 21
58
An electron that has a kinetic energy equal to 2.00 1016 J is moving
to the right along the axis of a cathode-ray tube as shown in Figure 21-40. An
r
electric field E = 2.00 10 4 N/C j exists in the region between the deflection
r
plates; and no electric field ( E = 0 ) exists outside this region. (a) How far is the
electron from the axis of the tube when it exits the region between the plates?
(b) At what angle is the electron moving, with respect to the axis, after exiting the
region between the plates? (c) At what distance from the axis will the electron
strike the fluorescent screen?
Picture the Problem The trajectory of the electron while it is in the electric field
is parabolic (its acceleration is downward and constant) and its trajectory, once it
is out of the electric field is, if we ignore the small gravitational force acting on it,
linear. We can use constant-acceleration equations and Newtons second law to
express the electrons x and y coordinates parametrically and then eliminate the
parameter t to express y(x). We can find the angle with the horizontal at which the
electron leaves the electric field from the x and y components of its velocity and
its total vertical deflection by summing its deflections over the first 4 cm and the
final 12 cm of its flight.
x(t ) = v0t
and
y (t ) = v0, y t + 12 a y t 2
Because v0,y = 0:
x(t ) = v0t
(1)
and
y (t ) = 12 a y t 2
Using Newtons second law, relate
the acceleration of the electron to the
electric field:
Substituting for a y gives:
ay =
Fnet eE y
=
me
me
y (t ) =
y(x ) =
eE y
2me
t2
eE y
2mev02
(2)
x2 =
eE y
4K
x2
(1.602 10
19
)(
(
vy
v
= tan 1 y
v
x
v0
= tan 1
v y = v0, y + a y t
Using a constant-acceleration
equation, express vy as a function
of the electrons acceleration and
its time in the electric field:
eE y x
eE x
= tan 1 y
2
2K
me v0
= tan 1
)(
(
= tan 1
ytotal = y4 cm + y12 cm
x = v0 t
and
y = v y t
vy
y =
v0
x = (tan )x
57
58
Chapter 21
= 4.47 cm
That is, the electron will strike the
fluorescent screen 4.47 cm below the
horizontal axis.
Dipoles
59
Two point charges, q1 = 2.0 pC and q2 = 2.0 pC, are separated by
4.0 m. (a) What is the magnitude of the dipole moment of this pair of charges?
(b) Sketch the pair and show the direction of the dipole moment.
Picture the Problem We can use its definition to find the dipole moment of this
pair of charges.
r
r
p = qL
and
p = (2.0 pC )(4.0 m )
= 8.0 10 18 C m
r
p
+q
60
A dipole of moment 0.50 enm is placed in a uniform electric field that
has a magnitude of 4.0 104 N/C. What is the magnitude of the torque on the
dipole when (a) the dipole is aligned with the electric field, (b) the dipole is
transverse to (perpendicular to) the electric field, and (c) the dipole makes an
angle of 30 with the direction of the electric field? (d) Defining the potential
energy to be zero when the dipole is transverse to the electric field, find the
potential energy of the dipole for the orientations specified in Parts (a) and (c).
Picture the Problem The torque on an electric dipole in an electric field is given
r r
r r r
by = p E and the potential energy of the dipole by U = p E .
r r r
= p E = pE sin
where is the angle between the
electric dipole moment and the electric
field.
59
(0) = pE sin 0 = 0
r r
U = p E = pE cos
Evaluate U for = 0:
General Problems
[SSM] Show that it is only possible to place one isolated proton in
61
an ordinary empty coffee cup by considering the following situation. Assume the
first proton is fixed at the bottom of the cup. Determine the distance directly
above this proton where a second proton would be in equilibrium. Compare this
distance to the depth of an ordinary coffee cup to complete the argument.
Picture the Problem Equilibrium of the second proton requires that the sum of
the electric and gravitational forces acting on it be zero. Let the upward direction
be the +y direction and apply the condition for equilibrium to the second proton.
Apply
proton:
= 0 to the second
r r
Fe + Fg = 0
or
kqp2
h2
mp g = 0 h =
kqp2
mp g
60
Chapter 21
N m2
8.988 10 9
1.602 10 19 C
2
C
h=
1.673 10 27 kg 9.81 m/s 2
(
)(
12 cm 5 in
This separation of about 5 in is greater than the height of a typical coffee cup.
Thus the first proton will repel the second one out of the cup and the maximum
number of protons in the cup is one.
62
Point charges of 5.00 C, +3.00C, and +5.00 C are located on the
x axis at x = 1.00 cm, x = 0, and x = +1.00 cm, respectively. Calculate the
electric field on the x axis at x = 3.00 cm and at x = 15.0 cm. Are there any points
on the x axis where the magnitude of the electric field is zero? If so, where are
those points?
Picture the Problem The locations of the point charges q1, q2 and q3 and the
points P1 and P2 at which we are to calculate the electric field are shown in the
r
diagram. From the diagram it is evident that E along the axis has no y
r
component. (a) We can use Coulombs law for E due to a point charge and the
r
superposition principle for electric fields to find E at points P1 and P2. (b) To
decide whether there are any points on the x axis where the magnitude of the
electric field is zero we need to consider the intervals (identified on the following
pictorial representation) I (x < 1.00 cm), II ( 1.00 cm < x < 0), III (0 < x < 1.00
cm), and IV (1.00 cm < x). Throughout interval II, the three electric fields are in
the same direction and so they cannot add up to zero. Throughout interval IV, the
electric-field strength due to the positive charge at x = 1.00 cm is larger than the
electric-field strength due to the negative charge at x = 1.00 cm. The fields due
to the positive charges at x = 0 and x = 1.00 cm both point in the +x direction, so
the resultant field throughout this interval must also point in the +x direction.
Hence we can narrow our search for points on the x axis where the magnitude of
the electric field is zero to intervals I and III. Setting the resultant electric-field
strength on the x axis equal to zero throughout these intervals and solving the
resulting quadratic equations will identify the desired points. Sketching the
electric field vectors in intervals I and III will help you get the signs of the terms
in the quadratic equations right.
q2
1.00
=5
q3
q1
=
5.0
=3
.00
.00
C
0
C
P1
1.00
3.00
P2
15.0
x, cm
l
rva
l
rva
e
Int
e
Int
l
rva
l
rva
e
Int
e
Int
IV
III
II
r
r
r
r
E P1 = E q1 + E q2 + E q3
q q
q
= k 12 + 22 + 23 i
r11 r21 r31
r
Substitute numerical values and evaluate E P1 :
r
5.00 C
3.00 C
5.00 C
+
+
E P1 = 8.988 10 9 N m 2 /C 2
i
2
2
2
(4.00 cm ) (3.00 cm ) (2.00 cm )
= 1.14 10 8 N/C i
r
r
r
r
E P2 = E q1 + E q2 + E q3
q
q
q
= k 21 + 22 + 23 i
r12 r22 r32
r
Substitute numerical values and evaluate E P2 :
r
5.00 C
3.00 C
5.00 C
E P2 = 8.988 109 N m 2 /C 2
+
+
i
2
2
2
(16.0 cm ) (15.0 cm ) (14.0 cm )
= 1.74 106 N/C i
61
62
Chapter 21
Applying Coulombs law for electric fields and the superposition of fields in
interval I yields:
5.00 C
[x ( 1.00 cm )]
3.00 C
5.00 C
=0
2
x
(x 1.00 cm )2
where k has been divided out and x is in the interval defined by x < 0. The root
of this equation that is in the interval x < 1.00 cm is x = 6.95 cm .
Applying Coulombs law for electric fields and the superposition of fields in
interval III yields:
5.00 C
[x ( 1.00 cm )]
3.00 C
5.00 C
=0
2
x
(1.00 cm x )2
The root of this equation that is in the interval 0 < x < 1.00 cm is x = 0.417 cm
63
Point charges of 5.00 C and +5.00 C are located on the x axis at
x = 1.00 cm and x = +1.00 cm, respectively. (a) Calculate the electric field
strength at x = 10.0 cm. (b) Estimate the electric field strength at x = 10.00 cm by
modeling the two charges as an electric dipole located at the origin and using
E= 2kp x
Part (a), and explain the reason for the difference between the two results.
C
.00
=5
q2
q1
=
5.0
Picture the Problem Let the point of interest (x = 10.0 cm) be identified as point
r
P. In Part (a) we can use Coulombs law for E due to a point charge and the
superposition principle to find the electric field strength at P. In Part (b) we can
use Equation 21-10 with p = 2aq to estimate the electric field strength at P.
1.00 0 1.00
P
10.0
x, cm
63
r
r
r
E P = E q1 + E q2
=
kq
1
2
1.00 cm P
i +
kq
2
2
1.00 cm P
q
q
= k 2 1
+ 2 2 i
x1.00 cm P x1.00 cm P
r
Substitute numerical values and evaluate E P :
r
N m 2 5.00 C
5.00 C
+
E P = 8.988 10 9
i = 1.83 10 6 N/C i
2
2
2
C
(11.00 cm ) (9.00 cm )
and
E P = 1.83 10 6 N/C
(b) The electric field strength due to a
dipole is given by Equation 21-10:
Because p = 2qa , where 2a is the
separation of the charges that
constitute the dipole, E is also given
by:
E=
2kp
x
E=
4kqa
x
N m2
(5.00 C )(1.00 cm )
4 8.988 10 9
C 2
EP =
= 1.80 10 6 N/C
3
10.0 cm
The exact and estimated values of E P agree to within 2%. This difference is due to
the fact that the separation of the two charges of the dipole is 20% of the distance
from the center of the dipole to point P.
A fixed point charge of +2q is connected by strings to point charges of
64
+q and +4q, as shown in Figure 21-41. Find the tensions T1 and T2.
Picture the Problem The electrostatic forces between the charges are responsible
for the tensions in the strings. Well assume that these are point charges and
apply Coulombs law and the principle of the superposition of forces to find the
tension in each string.
64
Chapter 21
T1 = F2 q + F4 q
T1 =
kq(2q ) kq(4q )
3kq 2
+
=
d2
d2
(2d )2
T2 = Fq + F2 q
T2 =
65
[SSM] A positive charge Q is to be divided into two positive point
charges q1 and q2. Show that, for a given separation D, the force exerted by one
charge on the other is greatest if q1 = q2 = 12 Q.
Picture the Problem We can use Coulombs law to express the force exerted on
one charge by the other and then set the derivative of this expression equal to zero
to find the distribution of the charge that maximizes this force.
F=
q2 = Q q1
F=
dF
k d
[q1 (Q q1 )]
= 2
dq1 D dq1
q1 = 12 Q q2 = Q q1 = 12 Q
kq1q2
D2
kq1 (Q q1 )
D2
k
[q1 ( 1) + Q q1 ]
D2
= 0 for extrema
65
d 2F
k d
[Q 2q1 ]
=
dq12 D 2 dq1
k
( 2)
D2
< 0 independently of q1.
=
q1 = q2 = 12 Q maximizes F .
66
A point charge Q is located on the x axis at x = 0, and a point charge
4Q is located at x = 12.0 cm. The electric force on a point charge of 2 .00C is
zero if that charge is placed at x = 4.00 cm, and is 126 N in the +x direction if
placed at x = 8.00 cm. Determine the charge Q.
Picture the Problem We can apply Coulombs law and the superposition of
forces to relate the net force acting on the charge q = 2.00 C to x. Because Q
divides out of our equation when F(x) = 0, well substitute the data given for
x = 8.00 cm.
Using Coulombs law, express the
net electric force on q as a function
of x:
F (x ) =
Q=
kqQ
kq(4Q )
+
2
x
(12.0 cm x )2
F (x )
4
kq 2 +
2
(12.0 cm x )
x
126 N
= 2.99 C
1
4
9
2
2
8.988 10 N m /C (2.00 C )
+
2
2
(8.00 cm ) (4.00 cm )
Picture the Problem Knowing the total charge of the two point particles, we can
use Coulombs law to find the two combinations of charge that will satisfy the
condition that both are positive and hence repel each other. If the particles attract
each other, then there is just one distribution of charge that will satisfy the
conditions that the force is attractive and the sum of the two charges is 200 C.
66
Chapter 21
F=
q 2 = Q q1
F=
kq1q2
r12, 2
q12 Qq1 +
Fr12, 2
k
=0
(80 N )(0.60 m )2
8.988 10 9 N m 2 /C 2
=0
q1 = 1.8 10 5 C
and
q 2 = Q q1 = 1.8 10 4 C
F =
kq1q2
r12, 2
q1 = 1.5 10 5 C
and
q 2 = Q q1 = 2.1 10 4 C
68
A point particle that has charge +q and unknown mass m is released
r
from rest in a region that has a uniform electric field E that is directed vertically
67
downward. The particle hits the ground at a speed v = 2 gh , where h is the initial
height of the particle. Find m in terms of E, q, and g.
y
0
v0 = 0
m,+q
y0 = h
r
E
Ug = 0
y1 = 2 gh
y1 = 0
Welectric = K + U g
field
= K 1 K 0 + U g,1 U g,0
or, because K1 = Ug,1 = 0,
Welectric = K 1 U g,1
field
= 12 m 2 gh
m=
mgh = mgh
qE
g
68
Chapter 21
Picture the Problem We can use Coulombs law, the definition of torque, and
the condition for rotational equilibrium to find the electrostatic force between the
two charged bodies, the torque this force produces about an axis through the
center of the rod, and the mass required to maintain equilibrium when it is located
either 25.0 cm to the right or to the left of the mid-point of the rod.
(a) Using Coulombs law, express
the electric force between the two
charges:
F=
kq1q2
d2
(8.988 10
)(
N m 2 / C 2 5.00 10 7 C
(0.100 m )2
= Fl
= (0.2247 N )(0.500 m )
= 0.1124 N m = 0.112 N m ,
counterclockwise.
(c) Apply
center of
the rod
= 0 to the rod:
mgl' = 0 m =
m=
gl'
0.1124 N m
9.81m/s 2 (0.250 m )
= 0.04583 kg = 45.8 g
(d) Apply
center of
the rod
= 0 to the rod:
+ mgl' = 0
Fl + mgl' = 0
d 2 mgl'
kq1q2'
q
'
=
+
mg
l
'
=
0
2
d2
kq1l
where q is the required charge.
5.00 10 7 C
69
70
Two 3.0-C point charges are located at x = 0, y = 2.0 m and at x = 0,
y = 2.0 m. Two other point charges, each with charge equal to Q, are located at
x = 4.0 m, y = 2.0 m and at x = 4.0 m, y = 2.0 m (Figure 21-43). The electric
field at x = 0, y = 0 due to the presence of the four charges is (4.0 103 N/C) i .
Determine Q.
Picture the Problem The field at (0, 0) produced by the two 3.0 C charges is
zero. Thus the entire field at (0, 0) is due to the two point charges whose charges
are Q. Let the numeral 1 refer to the point charge in the 1st quadrant and the
numeral 2 to the point charge in the 4th quadrant. We can use Coulombs law for
the electric field due to a point charge and the superposition of forces to express
the field at the origin and use this equation to solve for Q.
Express the electric field at the origin due to the point charges Q:
r
r
r
kQ
kQ
E (0,0 ) = E1 + E 2 = 2 r1, 0 + 2 r2, 0
r1, 0
r2, 0
kQ
( 4.0 m ) i + ( 2.0 m ) j + kQ3 ( 4.0 m ) i + (2.0 m ) j = (8.0 m3 )kQ i
r3
r
r
= E i
r = x2 + y2
Ex =
(8.0 m )kQ .
r3
(8.0 m )kQ
(x
+ y2
3 2
E x2 + y2
Q= x
k (8.0 m )
3 2
3 2
= 5.0 C
71
[SSM] Two point charges have a total charge of 200 C and are
separated by 0.600 m. (a) Find the charge of each particle if the particles repel
each other with a force of 120 N. (b) Find the force on each particle if the charge
on each particle is 100 C.
70
Chapter 21
Picture the Problem Let the numeral 1 denote one of the point charges and the
numeral 2 the other. Knowing the total charge on the two spheres, we can use
Coulombs law to find the charge on each of them. A second application of
Coulombs law when the spheres carry the same charge and are 0.600 m apart
will yield the force each exerts on the other.
F=
q 2 = Q q1
F=
kq1q2
r12, 2
kq1 (Q q1 )
r12, 2
(8.988 10
)[
N m 2 /C 2 (200 C ) q1 q12
(0.600 m )2
F=
kq1q2
r12, 2
F = (8.988 10 9 N m 2 /C 2 )
(100 C)2
(0.600 m )2
= 250 N
71
72
Two point charges have a total charge of 200 C and are separated by
0.600 m. (a) Find the charge of each particle if the particles attract each other with
a force of 120 N. (b) Find the force on each particle if the charge on each particle
is 100 C.
Picture the Problem Let the numeral 1 denote one of the point charges and the
numeral 2 the other. Knowing the total charge on the two point charges, we can
use Coulombs law to find the charge on each of them. A second application of
Coulombs law when the particles carry the same charge and are 0.600 m apart
will yield the electric force each exerts on the other.
F =
q 2 = Q q1
F =
q12 Qq1
kq1q2
r12, 2
kq1 (Q q1 )
r12, 2
r12, 2 F
k
=0
(0.600 m )2 (120 N )
8.988 10 9 N m 2 /C 2
=0
or
2
q + ( 200 C ) q1 4805 (C ) = 0
2
1
q1 = 21.7 C
and
q2 = Q q1 = 222 C
F=
kq1q2
r12, 2
72
Chapter 21
) (100 C)
F = 8.988 10 9 N m 2 /C 2
(0.600 m )2
= 250 N
73
A point charge of 3.00 C is located at the origin; a point charge of
4.00 C is located on the x axis at x = 0.200 m; a third point charge Q is located
on the x axis at x = 0.320 m. The electric force on the 4.00-C charge is 240 N in
the +x direction. (a) Determine the charge Q. (b) With this configuration of three
charges, at what location(s) is the electric field zero?
=4
=
3.0
.00
0
C
Picture the Problem (a) We can use Coulombs law for point charges and the
superposition of forces to express the net electric force acting on q2 and then solve
this equation to determine the charge Q. (b) To identify the location(s) at which
the electric field is zero, well need to systematically examine the resultant
electric fields in the intervals I, II, III, and IV identified below on the pictorial
representation. Drawing the electric field vectors in each of these intervals will
help you get the signs of the terms in the field equations right.
q2
q1
Interval I
Q
0.320
0
Interval II
0.200
Interval III
x,m
Interval IV
r
r
r
F2 = F1, 2 + FQ , 2
=
( )
kq1q2 kQq2
i + 2 i
r12, 2
rQ , 2
q
Q
= kq2 21 2 i = F2 i
r1, 2 rQ , 2
Solving for Q yields:
q
F
Q = rQ2, 2 21 2
r1, 2 kq2
240 N
2 3.00 C
Q = (0.120 m )
= 97.2 C
2
9
2
2
(0.200 m ) (8.988 10 N m /C )(4.00 C )
73
(b) Applying Coulombs law for electric fields and the superposition of fields in
interval I yields:
3.00 C
(x 0)
4.00 C
(x 0.200 m )
97.2 C
(x 0.320 m )2
=0
where k has been divided out and x is in the interval defined by x < 0. The roots of
this equation are at x = 0.170 m and x =0.220 m. Neither of these are in interval I.
Applying Coulombs law for electric fields and the superposition of fields in
interval II yields:
3.00 C
(x 0)
4.00 C
(x 0.200 m )
97.2 C
(x 0.320 m )2
=0
3.00 C
(x 0)
4.00 C
(x 0.200 m )
97.2 C
(x 0.320 m )2
=0
The roots of this equation are at x = 0.0649 m and x = 0.0454 m. Neither of these
are in interval III.
Applying Coulombs law for electric fields and the superposition of fields in
interval IV yields:
3.00 C
(x 0)
4.00 C
(x 0.200 m )
97.2 C
(x 0.320 m )2
=0
The roots of this equation are at x = 0.170 m and x = 0.220. Neither of these are in
interval IV.
Summarizing, the electric field is zero
at two locations in interval II:
Two point particles, each of mass m and charge q, are suspended from
74
a common point by threads of length L. Each thread makes an angle with the
vertical as shown in Figure 21-44. (a) Show that q = 2 L sin (mg k ) tan
where k is the Coulomb constant. (b) Find q if m = 10.0 g, L = 50.0 cm, and
= 10.0.
74
Chapter 21
r
T
r
Fe
r
r
Fg = m g
Fx = FE T sin =
kq 2
T sin = 0
r2
and
Fy = T cos mg = 0
tan =
kq 2
mg tan
q = r
2
mgr
k
r = 2L sin
q = 2 L sin
mg tan
k
= 0.241 C
r
T
r
Fe
r
r
Fg = m g
Fx =Fe T sin =
kq 2
T sin = 0
r2
and
Fy =T cos mg = 0
Eliminate T between these equations
to obtain:
tan =
kq 2
mgr 2
r = 2L sin
tan =
kq 2
4mgL2 sin 2
or
sin 2 tan =
kq 2
4mgL2
(8.988 10
N m 2 /C 2 (0.75 C )
= 5.73 10 3
2
2
4(0.010 kg ) 9.81 m/s (1.5 m )
9
sin tan
and
3 5.73 10 3
(1)
75
76
Chapter 21
= 0.179 rad = 10
(8.988 10
N m 2 /C 2 (0.50 C )(1.0 C )
= 5.09 10 3 3
2
2
4(0.010 kg ) 9.81 m/s (1.5 m )
9
76
Four point charges of equal magnitude are arranged at the corners of a
square of side L as shown in Figure 21-45. (a) Find the magnitude and direction
of the force exerted on the charge in the lower left corner by the other three
charges. (b) Show that the electric field at the midpoint of one of the sides of the
square is directed along that side toward the negative charge and has a magnitude
8q
1
.
E given by E = k 2 1
L 5 5
Picture the Problem Let the origin be
at the lower left-hand corner and
designate the point charges as shown in
the diagram. We can apply Coulombs
law for point charges to find the forces
exerted on q1 by q2, q3, and q4 and
superimpose these forces to find the net
force exerted on q1. In Part (b), well
use Coulombs law for the electric field
due to a point charge and the
superposition of fields to find the
electric field at point P(0, L/2).
y
L
q2 = q
q3 = +q
E2
E1
P (0, L / 2)
E3
L 2
E4
F2,1
q1 = +q
F3,1
q4 = q
F4,1
r
r
r
r
F1 = F2,1 + F3,1 + F4,1
r
kq q
kq q r
F2,1 = 22 1 r2,1 = 23 1 r2,1
r2,1
r2,1
=
k ( q )q
kq 2
L j = 2 j
3
L
L
r
kq q
kq q r
F4,1 = 42 1 r4,1 = 43 1 r4,1
r4,1
r4,1
=
2
k ( q )q
= kq i
L
i
L3
L2
r
kq q
kq q r
F3,1 = 32 1 r3,1 = 33 1 r3,1
r3,1
r3,1
kq 2
L i L j
23 2 L3
kq 2
= 3 2 2 i + j
2 L
=
r
Use Coulombs law to express E1 :
kq 2
1
1
i+ j
2
L 2 2
r
r
r
r
r
E P = E1 + E 2 + E 3 + E 4
r kq
kq L
E1 = 2 1 r1, P = 3
r1, P
r1, P 2
=
r
Use Coulombs law to express E 2 :
r kq 2
kq 2
kq 2
F1 = 2 j 3 2 2 i + j + 2 i
L
L
2 L
2
2
kq
kq
= 2 i + j 3 2 2 i + j
L
2 L
kq L 4kq
j = 2 j
3
L 2 L
2
r
kq
k ( q ) L
E 2 = 2 2 r2, P = 3 j
r2, P
r2, P 2
=
kq L 4kq
j = 2 j
3
L 2 L
2
(1)
77
78
Chapter 21
r
Use Coulombs law to express E 3 :
r
kq
kq
L
E 3 = 2 3 r3, P = 3 Li j
2
r3, P
r3, P
=
r
Use Coulombs law to express E 4 :
8kq 1
i j
53 2 L2
2
r
kq
k ( q )
L
E 4 = 2 4 r4, P = 3 Li + j
2
r4, P
r4, P
=
8kq 1
i
2
53 2 L2
j + 8kq i 1
32 2
2
5 L
j = k 8q 1 1 j
L2 5 5
77
[SSM] Figure 21-46 shows a dumbbell consisting of two identical
small particles, each of mass m, attached to the ends of a thin (massless) rod of
length a that is pivoted at its center. The particles carry charges of +q and q, and
r
the dumbbell is located in a uniform electric field E . Show that for small values
of the angle between the direction of the dipole and the direction of the electric
field, the system displays a rotational form of simple harmonic motion, and obtain
an expression for the period of that motion.
Picture the Problem We can apply Newtons second law in rotational form to
obtain the differential equation of motion of the dipole and then use the small
angle approximation sin to show that the dipole experiences a linear
restoring torque and, hence, will experience simple harmonic motion.
Apply
= I to the dipole:
d 2
dt 2
where is negative because it acts in
such a direction as to decrease .
pE sin = I
pE = I
I = 12 ma 2
d 2
dt 2
79
p = qa
1
2
ma 2
d 2
= qaE
dt 2
or
d 2
2qE
=
2
dt
ma
the differential equation for a simple
harmonic oscillator with angular
frequency = 2qE ma .
2
T=
T = 2
ma
2qE
K + U = 0
or, because Ki = 0,
(1)
K + U = 0
where K is the kinetic energy when it is
aligned with the field.
80
Chapter 21
U = U f U i
K + qaE (cos 60 1) = 0
q=
K
aE (1 cos 60)
q=
5.00 10 3 J
(0.300 m )(600 N/C)(1 cos 60)
= pE cos f + pE cos i
= qaE (cos 60 1)
= 56 C
79
[SSM] An electron (charge e, mass m) and a positron (charge +e,
mass m) revolve around their common center of mass under the influence of their
attractive coulomb force. Find the speed v of each particle in terms of e, m, k, and
their separation distance L.
Picture the Problem The forces the electron and the proton exert on each other
constitute an action-and-reaction pair. Because the magnitudes of their charges
are equal and their masses are the same, we find the speed of each particle by
finding the speed of either one. Well apply Coulombs force law for point
charges and Newtons second law to relate v to e, m, k, and their separation
distance L.
ke 2
v2
ke 2
=m1
= 2mv 2
2
L
L
2 L
v=
ke 2
2mL
80
L
r
T
L sin
r r
Fe = qE
(mg + qE ) L sin = I P
mL2
d 2
dt 2
d 2
= (mg + qE ) L
dt 2
or
d 2 mg + qE
+
=0
dt 2
mL
This is the equation of simple
harmonic motion with:
2 =
T' =
m, q
r
r
Fg = m g
mg + qE
mL
T = 2
= 2
L
g
mL
mg + qE
d 2
,
dt 2
81
82
Chapter 21
T
=
T'
mL
mg + qE
= 1+
E=
L
g
mg + qE
mg
qE
mg
mg T
1
q T'
Noting that the period of the simple pendulum in the absence of the electric field
is 2.006 s, substitute numerical values and evaluate E:
E=
1.2 s
kqQ
mg = 0 y0 =
y 02
F=
kqQ
( y 0 + y )
kqQ
y 02
kqQ
mg
83
=
4
3
y 0 + 2 y 0 y
y 03
y
y 04 1 + 2
y0
Substitute for
kqQ
and simplify to
y02
kqQ
= mg
y02
F =
2mg
y
y0
obtain:
Apply Newtons second law to
the displaced particle to obtain:
2mg
d 2 y
=
y
2
dt
y0
or
d 2 y 2 g
+
y = 0
dt 2
y0
the differential equation of simple
harmonic motion with =
2 g y0 .
82 Two neutral
r molecules on the x axis attract each other. Each molecule
has a dipole moment p , and these dipole moments are on the +x axis and are
separated by a distance d. Derive an expression for the force of attraction in terms
of p and d.
Picture the Problem We can relate the force of attraction that each molecule
exerts on the other to the potential energy function of either molecule
using F = dU dx. We can relate U to the electric field at either molecule due to
the presence of the other through U = pE. Finally, the electric field at either
molecule is given by E = 2kp x 3 .
Express the force of attraction
between the dipoles in terms of the
spatial derivative of the potential
energy function of p1:
F =
U 1 = p1 E1
where E1 is the field at p1 due to p2.
dU 1
dx
(1)
84
Chapter 21
2kp2
x3
where x is the separation of the dipoles.
U1 =
F =
F=
E1 =
2kp1 p 2
x3
d 2kp1 p2 6kp1 p2
=
dx
x 3
x4
6kp 2
d4
r2, P
r
r1, P
q1 = Q
q2 = Q
12 a
m,q
x
1
2
r
r
r
E y = E1 + E 2
kq1
kq
r1, P + 2 2 r2, P
2
r1, P
r2, P
r1, P
r1, P =
given by:
r1, P
12 ai + yj
r1, P
and
r2, P =
r2, P
r2, P
1
2
ai + yj
r2, P
2
Substituting for q1, q2, r1, P , r2, P and r1, P = r2,P = y 2 + ( 12 a )
12
yields:
r
kQ 1 ai + yj kQ 12 ai + yj kQ 1 kQ
Ey = 2 2
+
=
2 ai + yj + 3
r2 r
r3
r1, P
r1, P
r2, P
2, P
2, P
1, P
[y
kQ
2
+ ( a)
1
2
2 32
(yj )+
[y
kQ
2
+ ( a)
1
2
2 32
(yj ) =
[y
2kQy
2
r
r
F = qE y =
+ a
1
4
[y
2 32
2kqQy
2
+ 14 a 2
( ai + yj )
1
2
32
r
where q must be positive if F always
points away from the origin.
Wnet y = K = K f K i
or, because Ki = 0,
Wdone by Fy = 12 Mvf2
vf =
2Wdone by Fy
Wdone by Fy =
y max
[y
2kqQy
2
+ ( 12 a )
2 3 2
dy
85
86
Chapter 21
dFy
y max =
dy
for vf to obtain:
[y
+ ( 12 a )
2 32
[y
d
y
dy y 2 + ( 12 a )2
+ ( a)
32
2 32
1
2
dy
=0
4kqQ
M
2 2
dy
y
2
y + ( a)
Substitute for
[y
+ ( a)
1
2
2 32
dy
a
2 2
vf =
ymax
2 2
4kqQ
m
vf =
vf =
1
2
32
dy
0.367
a
4kqQ 0.367
kqQ
= 1.21
M a
aM
87
r
Wnet = Fe dr = K
r0
ke(79e )
or, because Fe dr =
dr and
r2
r0
r0
Ki = 0,
dr 1
= 2 mp vf2
2
r
r0
79ke 2
Evaluating the integral yields:
79ke
1
= 12 mp vf2
79ke 2 =
r
r
r0
0
vf =
158ke 2
158k
=e
mp r0
mp r0
vf = 1.602 10 19 C
N m2
158 8.988 10 9
C 2
= 1.48 10 7 m/s
1.673 10 27 kg 1.00 10 13 m
) (
)(
Comment: What????
88
Chapter 21
Wnet =
rmin
r
dr = K
or, because
rmin
rmin
r r
k (2e )(79e )
F
d
r
=
e
r 2 dr and
Kf = 0,
158ke
rmin
dr
= Ki
r2
rmin
158ke
1
= Ki
158ke 2 =
r
rmin
rmin =
158ke 2
Ki
N m2
1.602 10 19 C
158 8.988 10 9
2
C
=
1.602 10 19 J
5.0 Mev
eV
rmin
= 4.6 10 14 m
r
Fe
m, Ne
r
mg
r
Fd
Fe mg Fd = ma y
or, because ay = 0,
Fe mg Fd, terminal = 0
qE Vg 6rvt = 0
or, because q = Ne,
NeE 43 r 3 g 6rvt = 0
Ne =
4
3
r 3 g + 6rv t
E
(1)
4
3
)(
)(
Ne =
7.18 10 15 N + 2.16 10 14 N
6.00 10 4 N / C
= 4.8 10 19 C
(b) Divide the result in (a) by e to
obtain:
N=
4.80 10 19 C
= 3
1.602 10 19 C
89
90
Chapter 21
Fd, terminal Fe mg = 0
or
6rv t NeE 43 r 3 g = 0
vt =
NeE + 43 r 3 g
6r
3
N
kg
m
vt =
6 1.8 10 5 Pa s 5.50 10 7 m
)(
= 0.19 mm/s
r
Fe
m, Ne
r
mg
r
Fd
Fe mg Fd = ma y
or, because ay = 0,
Fe mg Fd, terminal = 0
qE mg 6rvu = 0
or, because q = Ne,
NeE mg 6rvu = 0
vu =
91
NeE mg
6r
(1)
Fd, terminal Fe mg = 0
or
6rvd NeE mg = 0
vd =
NeE + mg
6r
(2)
u = v u + vd =
Measuring both vu and vd has the advantage that you dont need to know the mass
of the microsphere.
(b) Letting u represent the change
in the terminal speed of the
microsphere due to a gain (or loss)
of one electron we have:
Noting that v will be the same
whether the microsphere is moving
upward or downward, express its
terminal speed when it is moving
upward with N electronic charges on
it:
Express its terminal speed upward
when it has N + 1 electronic charges:
u = v N +1 v N
vN =
NeE mg
6r
vN +1 =
(N + 1)eE mg
6r
92
Chapter 21
u =
=
u =
(N + 1)eE mg NeE mg
6r
6r
eE
6r
(1.602 10
6 (1.8 10
= 52 m/s
19
)(
C 6.00 10 4 N/C
5
Pa m 5.50 10 7 m
)(