Physics 51
Chapter 23, 13th ed, Freedman
Homework Solutions
23.13.
IDENTIFY and SET UP: Apply conservation of energy to points A and B.
EXECUTE: K A  U A  K B  U B
U  qV , so K A  qVA  K B  qVB
KB  K A  q(VA  VB )  000250 J  (500  106 C)(200 V  800 V)  000550 J
vB  2K B /m  742 m/s
EVALUATE: It is faster at B; a negative charge gains speed when it moves to higher potential.
23.16.
IDENTIFY: Apply Ka  U a  Kb  Ub .
SET UP: Let q1  300 nC and q2  200 nC. At point a, r1a  r2a  0250 m. At point b,
r1b  0100 m and r2b  0400 m. The electron has q   e and me  9111031 kg. K a  0 since
the electron is released from rest.
EXECUTE: 
keq1 keq2
keq keq2 1

 1
 mevb2 .
r1a
r2a
r1b
r2b 2
 (300  109 C) (200  109 C) 
17
Ea  K a  U a  k (160  1019 C) 

  288  10 J.

0

250
m
0

250
m


 (300  109 C) (200 109 C)  1
1
2
17
2
Eb  Kb  U b  k (160  1019 C) 

  mevb  504  10 J  mevb

0

100
m
0

400
m
2
2


Setting Ea  Eb gives vb 
2
911  1031 kg
(504  1017 J  288  1017 J)  689  106 m/s.
EVALUATE: Va  V1a  V2a  180V. Vb  V1b  V2b  315V. Vb  Va . The negatively charged
electron gains kinetic energy when it moves to higher potential.
23.21.
IDENTIFY: For a point charge, V 
kq
. The total potential at any point is the algebraic sum of the
r
potentials of the two charges.
SET UP: (a) The positions of the two charges are shown in Figure 23.21a.
Figure 23.21a
(b) x  a : V 
kq 2kq  kq( x  a )
kq 2kq
kq(3x  a)


. 0  x  a :V 


.
x xa
x( x  a)
x ax
x( x  a)
q
 kq 2kq
kq ( x  a )
2q


. A general expression valid for any y is V  k  

x
xa
x( x  a)
 x xa
(c) The potential is zero at x  a and a /3.
(d) The graph of V versus x is sketched in Figure 23.21b.
x  0 :V 

 .

Figure 23.21 b
kq

, which is the same as the potential of a point charge
x
x2
– q. Far from the two charges they appear to be a point charge with a charge that is the algebraic sum
of their two charges.
EVALUATE: (e) For x  a : V 
23.23.
kqx
IDENTIFY and SET UP: Apply conservation of energy, Eq. (23.3). Use Eq. (23.12) to express U in
terms of V.
(a) EXECUTE: K1  qV1  K2  qV2 , q(V2  V1)  K1  K2 ; q  1602 1019 C.
K1  K 2
 156 V.
q
EVALUATE: The electron gains kinetic energy when it moves to higher potential.
K  K2
 182 V.
(b) EXECUTE: Now K1  2915 1017 J, K2  0. V2  V1  1
q
EVALUATE: The electron loses kinetic energy when it moves to lower potential.
K1  12 mev12  4099  1018 J; K 2  12 mev22  2915  1017 J. V  V2  V1 
23.27.
IDENTIFY: The potential at any point is the scalar sum of the potential due to each shell.
kq
kq
SET UP: V 
for r  R and V 
for r  R.
R
r
EXECUTE: (a) (i) r  0. This point is inside both shells so
 600  109 C 900  109 C 
q
q 
V  k  1  2   (899  109 N  m2 /C2 ) 

.
 00300 m
00500 m 
 R1 R2 

V  1798 103 V  (1618 103 V)  180 V.
(ii) r  400 cm. This point is outside shell 1 and inside shell 2.
 600  109 C 900  109 C 
q q 
V  k  1  2   (899  109 N  m2 /C2 ) 

.
 00400 m
00500 m 
 r R2 

V  1348 103 V  (1618 103 V)  270 V.
(iii) r  600 cm. This point is outside both shells.
899  109 N  m2 /C2
q q  k
V  k  1  2   (q1  q2 ) 
(600  109 C  (900  109 C)). V   450 V.
r  r
00600 m
r
(b) At the surface of the inner shell, r  R1  300 cm. This point is inside the larger shell,
q
q 
so V1  k  1  2   180 V. At the surface of the outer shell, r  R2  500 cm. This point is outside
 R1 R2 
the smaller shell, so
 600  109 C 900  109 C 
q q 
V  k  1  2   (899  109 N  m 2/C2 ) 

.
 00500 m
00500 m 
 r R2 

V2  1079 103 V  (1618 103 V)  539 V. The potential difference is V1  V2  719 V. The
inner shell is at higher potential. The potential difference is due entirely to the charge on the inner
shell.
EVALUATE: Inside a uniform spherical shell, the electric field is zero so the potential is constant (but
not necessarily zero).
23.29.(a) IDENTIFY and SET UP: The electric field on the ring’s axis is calculated in Example 21.9. The force on
the electron exerted by this field is given by Eq. (21.3).
EXECUTE: When the electron is on either side of the center of the ring, the ring exerts an attractive
force directed toward the center of the ring. This restoring force produces oscillatory motion of the
electron along the axis of the ring, with amplitude 30.0 cm. The force on the electron is not of the form
F = –kx so the oscillatory motion is not simple harmonic motion.
(b) IDENTIFY: Apply conservation of energy to the motion of the electron.
SET UP: Ka  U a  Kb  Ub with a at the initial position of the electron and b at the center of the
ring. From Example 23.11, V 
1
Q
4 e0
x  R2
2
, where R is the radius of the ring.
EXECUTE: xa  300 cm, xb  0.
K a  0 (released from rest), Kb  12 mv 2
Thus
1 mv 2
2
 U a  Ub
And U  qV  eV so v 
Va 
1
Q
4 e0
xa2  R 2
2e(Vb  Va )
.
m
 (8988  109 N  m 2 /C2 )
240  109 C
(0300 m) 2  (0150 m) 2
Va  643 V
Vb 
v
1
4 e0
Q
xb2  R 2
 (8988  109 N  m 2 /C2 )
240  109 C
 1438 V
0150 m
2e(Vb  Va )
2(1602  1019 C)(1438 V  643 V)

 167  107 m/s
m
9109  1031 kg
EVALUATE: The positively charged ring attracts the negatively charged electron and accelerates it.
The electron has its maximum speed at this point. When the electron moves past the center of the ring
the force on it is opposite to its motion and it slows down.
23.39.
IDENTIFY and SET UP: Use the result of Example 23.9 to relate the electric field between the plates to
the potential difference between them and their separation. The force this field exerts on the particle is
given by Eq. (21.3). Use the equation that precedes Eq. (23.17) to calculate the work.
V
360 V
EXECUTE: (a) From Example 23.9, E  ab 
 8000 V/m.
d
00450 m
(b) F  q E  (240 10 9 C)(8000 V/m)  192 105 N
(c) The electric field between the plates is shown in Figure 23.39.
Figure 22.39
The plate with positive charge (plate a) is at higher potential. The electric field is directed from high
potential toward low potential (or, E is from + charge toward  charge), so E points from a to b.
Hence the force that E exerts on the positive charge is from a to b, so it does positive work.
b
W   F  dl  Fd , where d is the separation between the plates.
a
W  Fd  (192 105 N)(00450 m)  864 107 J
(d) Va  Vb  360 V (plate a is at higher potential)
U  Ub  Ua  q(Vb  Va )  (240 109 C)(360 V)  864 107 J.
EVALUATE: We see that Wab  (Ub  U a )  U a  Ub 
23.45.
IDENTIFY and SET UP: Use Eq. (23.19) to calculate the components of E
EXECUTE: V  Axy Bx2  Cy
(a) Ex  
Ey  
V
  Ay  2Bx
x
V
  Ax  C
y
V
0
z
(b) E  0 requires that Ex  E y  Ez  0.
Ez 
Ez  0 everywhere.
E y  0 at x  C/A.
And E x is also equal to zero for this x, any value of z and y  2Bx/A  (2B/A)(C /A)  2BC /A2.
EVALUATE: V doesn’t depend on z so Ez  0 everywhere.
23.62.
IDENTIFY: Apply  Fx  0 and  Fy  0 to the sphere. The electric force on the sphere is Fe  qE.
The potential difference between the plates is V  Ed .
SET UP: The free-body diagram for the sphere is given in Figure 23.62.
EXECUTE: T cos  mg and T sin  Fe gives
Fe  mg tan  (150 103 kg)(980 m/s2 )tan(30)  00085 N.
Fe  Eq 
Fd (00085 N)(00500 m)
Vq
and V 

 478 V.
q
d
890 106 C
EVALUATE: E  V/d  956 V/m. E   /e0 and   Ee0  846 109 C/m2.
Figure 23.62
23.79.
IDENTIFY: Slice the rod into thin slices and use Eq. (23.14) to calculate the potential due to each
slice. Integrate over the length of the rod to find the total potential at each point.
(a) SET UP: An infinitesimal slice of the rod and its distance from point P are shown in Figure
23.79a.
Figure 23.79a
Use coordinates with the origin at the left-hand end of the rod and one axis along the rod. Call the axes
x and y so as not to confuse them with the distance x given in the problem.
EXECUTE: Slice the charged rod up into thin slices of width dx. Each slice has charge
dQ  Q(dx/a) and a distance r  x  a  x from point P. The potential at P due to the small slice dQ
is
1  dQ 
1 Q  dx 



.
4 e0  r  4 e0 a  x  a  x 
Compute the total V at P due to the entire rod by integrating dV over the length of the rod
( x  0 to x  a):
dV 
V   dV 
Q
dx
a
Q
Q
 xa
.
x 

[ ln( x a  x)]0 
ln 
4 e0a 0 ( x  a  x) 4 e0a
4 e0a 
EVALUATE: As x  , V 
a
Q
 x
ln    0.
4 e0a  x 
(b) SET UP: An infinitesimal slice of the rod and its distance from point R are shown in Figure
23.79b.
Figure 23.79b
dQ  (Q/a)dx as in part (a)
Each slice dQ is a distance r  y 2  (a  x) 2 from point R.
EXECUTE: The potential dV at R due to the small slice dQ is
dV 
1  dQ 
1 Q


4 e0  r  4 e0 a
dx
y  (a  x)2
2
.
V   dV 
Q
4 e0a
dx
a
0
y  (a  x)2
2
.
In the integral make the change of variable u  a  x;du  dx
V 
Q
du
0
4 e0a  a
y2  u2
=-


0
ln u  y 2  u 2  .
 a
4 e0a 
Q
 
2
2
ln y  ln a  y 2  a 2   Q ln  a  a  y


4 e0a 
y
 4 e0a  
 
(The expression for the integral was found in Appendix B.)
 y
Q
ln    0.
EVALUATE: As y  , V 
4 e0a  y 
V 

Q
(c) SET UP: part (a): V 


 .


Q
Q
 xa
 a
ln 
ln 1  .

4 e0a  x  4 e0a  x 
From Appendix B, ln(1  u)  u  u 2/2 ... , so ln(1  a/x)  a/x  a2/2x2 and this becomes a/x when x
is large.
Q a
Q
. For large x, V becomes the potential of a point charge.
EXECUTE: Thus V 
 
4 e0a  x  4 e0 x
part (b): V 
  a  a2  y2
ln 
4 e0a  
y
 
Q
From Appendix B,
2


   Q ln  a  1  a
  4 e0a  y
y2



.


1  a 2 /y 2  (1  a 2 /y 2 )1/2  1  a 2 /2 y 2  …
Thus a/y  1  a 2 /y 2  1  a/y  a 2 /2 y 2  … 1  a/y. And then using ln(1  u)  u gives
a
Q
.
 
4 e0 a
4 e0 a  y  4 e0 y
EVALUATE: For large y, V becomes the potential of a point charge.
V
Q
ln(1  a/y ) 
Q