Download Chapter 23 solutions to assigned problems

CHAPTER 23: Electric Potential
Solutions to Assigned Problems
2.
The work done by the electric field can be found from Eq. 23-2b.
W
Vba   ba  Wba   qVba   1.60  1019 C  55V  185V   3.84  1017 J
q

6.

The distance between the plates is found from Eq. 23-4b, using the magnitude of the electric field.
V
V
45 V
E  ba  d  ba 
 3.5  102 m
d
E 1300 V m

11. Since the field is uniform, we may apply Eq. 23-4b. Note that the electric field always points from
high potential to low potential.
(a) VBA  0 . The distance between the two points is exactly perpendicular to the field lines.
(b) VCB  VC  VB   4.20 N C  7.00 m   29.4 V
(c) VCA  VC  VA  VC  VB  VB  VA  VCB  VBA  29.4 V  0  29.4 V
16. From Example 22-6, the electric field due to a long wire is radial relative to the wire, and is of
1 
. If the charge density is positive, the field lines point radially away from the
magnitude E 
2 0 R
wire. Use Eq. 23-41 to find the potential difference, integrating along a line that is radially outward
from the wire.
Rb
Rb
 
Vb  Va    E d l   
Ra
Ra
1 
2 0 R
dR  


R
ln  Rb  Ra  
ln a
2 0
2 0 R b
29. We assume that all of the energy the proton gains in being accelerated by the
voltage is changed to potential energy just as the proton’s outer edge reaches the
outer radius of the silicon nucleus.
1 e 14e 
U initial  U final  eVinitial 

4 0
r
Vinitial 
1 14e
4 0 r

 8.99  10 N m C
9
2
2
14  1.60  1019 C 
 1.2  3.6  1015m 
r
4.2  106 V
32. Use Eq. 23-2b and Eq. 23-5.
 1
q
1  q    1 q
1  q  
VBA  VB  VA  





 4 0 d  b 4 0 b   4 0 b 4 0 d  b 

q 
1
1 1
1 
1
  

2
4 0  d  b b b d  b 
4 0
 1  1   2 q  2b  d 

 d  b b  4 0b  d  b 
q
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
66
Chapter 23
Electric Potential
41. We follow the development of Example 23-9, with Figure 23-15. The charge on a thin ring of radius
R and thickness dR will now be dq   dA   aR 2   2 RdR  . Use Eq. 23-6b to find the potential of
a continuous charge distribution.
R
aR 2  2 RdR 
1
dq
1
a
V




4 0 r
4 0 0
2 0
x2  R2

0

R0
R 3dR

x2  R2
A substitution of x 2  R 2  u 2 can be used to do the integration.
x 2  R 2  u 2  R 2  u 2  x 2 ; 2 RdR  2udu
V 
R0
a
R 3dR

2 0

x2  R2
0



a 1 2
x  R2
3

2 0

a 
2 0 

a  2
R0  2 x 2

6 0
2
1
3
2
0

2 0
3/ 2
 x  R 
R  R0
a

u
2

 x 2 udu
R 0
u


 x2 x2  R2

3/ 2
 x
 x 2 x 2  R02
2
 R02

1/ 2
1/ 2
1 Q
 V0
a
R  R0
 13 u 3  ux 2  R 0
2 0
R  R0

 R 0

1/ 2

2
3
x3 

 2 x3  , x  0

44. The potential at the surface of the sphere is V0 
V 

0
1 Q
4 0 r0
. The potential outside the sphere is
r0
, and decreases as you move away from the surface. The difference in potential
4 0 r
r
between a given location and the surface is to be a multiple of 100 V.
1 Q
 0.50  106 C 
V0 
 8.99  109 N m 2 C2 
  10, 216 V
4 0 r0
 0.44 m 

V0  V  V0  V0

r0
r
 100 V  n  r 
V0
V  100 V  n 
r0
0
V0
10, 216 V
 0.44 m  0.444 m
10,116 V
Note that to within the appropriate number of significant figures, this location is at the
surface of the sphere. That can be interpreted that we don’t know the voltage well enough
to be working with a 100-V difference.
V0
10, 216 V
(b) r10 
r0 
 0.44 m   0.49 m
V0  100 V 10 9, 216 V
(a) r1 
V  100 V 1
r0 
0
(c)
r100 
V0
V  100 V 100
0
r0 
10, 216 V
216 V
 0.44 m  
21m
45. The potential due to the dipole is given by Eq. 23-7.
8.99  109 N m 2 C2 4.8  1030 C m cos 0
1 p cos 

(a) V 
2
4 0 r 2
4.1  109 m





r
Q
Q
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
67
Physics for Scientists & Engineers with Modern Physics, 4 th Edition
 2.6  103 V
(b) V 
1
4 0
p cos 
r
2
r

 1.8  103 V
(c) V 
1
4 0
p cos 
r
2
Instructor Solutions Manual

8.99  10
9
 4.8  10
 4.1  10 m 
N m 2 C2
9
8.99  10
9
30
2
 4.8  10
1.1  10 m 
N m 2 C2
9


C m cos 45o
Q
30

C m cos135o
Q
r
2

 1.8  103 V
Q
Q
51. We use Eq. 23-9 to find the components of the electric field.
V
V
V
Ex  
 2.5 y  3.5 yz ; E y  
 2 y  2.5 x  3.5 xz ; E z  
 3.5 xy
x
y
z
E   2.5 y  3.5 yz  ˆi   2 y  2.5 x  3.5 xz  ˆj   3.5 xy  kˆ
54. Let the side length of the equilateral triangle be L. Imagine bringing the
e
e
l
electrons in from infinity one at a time. It takes no work to bring the first
electron to its final location, because there are no other charges present.
l
l
Thus W1  0 . The work done in bringing in the second electron to its
final location is equal to the charge on the electron times the potential
e
(due to the first electron) at the final location of the second electron.
 1 e
1 e2

Thus W2   e   
. The work done in bringing the third electron to its final

 4 0 l  4 0 L
location is equal to the charge on the electron times the potential (due to the first two electrons).
 1 e
1 e
1 2e 2


Thus W3   e   
. The total work done is the sum W1  W2  W3 .

 4 0 l 4 0 l  4 0 l
W  W1  W2  W3  0 
1 e2
4 0 l

1 2e 2
4 0 l

1 3e 2
4 0 l



3 8.99  109 N m 2 C2 1.60  1019 C
1.0  10
10
m


2
1eV

  43eV
19 
 1.60  10 J 
 6.9  1018 J  6.9  1018 J 
59. Following the same method as presented in Section 23-8, we get the following results.
(a) 1 charge:
No work is required to move a single charge into a position, so U1  0.
2 charges:
U2 
3 charges:
U3 
4 charges:
This represents the interaction between Q1 and Q2 .
1 Q1Q2
4 0 r12
This now adds the interactions between Q1 & Q3 and Q2 & Q3 .
1  Q1Q2
QQ Q Q 
 1 3  2 3

4 0  r12
r13
r23 
This now adds the interaction between Q1 & Q4 , Q2 & Q4 , and Q3 & Q4 .
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
68
Chapter 23
Electric Potential
U4 
1  Q1Q2
QQ QQ Q Q Q Q Q Q 
 1 3 1 4  2 3 2 4  3 4

4 0  r12
r13
r14
r23
r24
r34 
Q1
r14
Q4
r12
r13
r24
Q2
r23
r34
Q3
This now adds the interaction between Q1 & Q5 , Q2 & Q5 , Q3 & Q5 , and Q4 & Q5 .
(b) 5 charges:
U5 
1  Q1Q2
QQ QQ QQ Q Q Q Q Q Q Q Q Q Q Q Q 
 1 3 1 4  1 5  2 3 2 4  2 5  3 4  3 5 4 5

4 0  r12
r13
r14
r15
r23
r24
r25
r34
r35
r45 
Q1
r14
Q4
r12
r13
r24
r15
r34
r45
Q3
Q2
r23
r25
r35
Q5
62. We find the energy by bringing in a small amount of charge at a time, similar to the method given
in Section 23-8. Consider the sphere partially charged, with charge q < Q. The potential at the
1 q
, and the work to add a charge dq to that sphere will increase the
surface of the sphere is V 
4 0 r0
potential energy by dU  Vdq. Integrate over the entire charge to find the total potential energy.
Q
U   dU  
0
1
q
4 0 r0
dq 
1 Q2
8 0 r0
72. (a) All eight charges are the same distance from the center of the cube. Use Eq. 23-5 for the
potential of a point charge.
Vcenter  8
1
4 0
Q

16
1 Q
 9.24
1 Q
4 0 l
3
3 4 0 l
l
2
(b) For the seven charges that produce the potential at a corner, three are a distance l away from
that corner, three are a distance
from that corner.
2l away from that corner, and one is a distance
3l away
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
69
Physics for Scientists & Engineers with Modern Physics, 4 th Edition
Vcorner  3
1 Q
4 0 l
3
1
Q
4 0
1

2l


Q
4 0
Instructor Solutions Manual
 3
3l
3

2
1  1 Q
1 Q
 5.70

4 0 l
3  4 0 l
(c) The total potential energy of the system is half the energy found by multiplying each charge
times the potential at a corner. The factor of half comes from the fact that if you took each
charge times the potential at a corner, you would be counting each pair of charges twice.


3
U  12 8 QVcorner   4  3 
1  1 Q2
1 Q2

22.8

4 0 l
3  4 0 l

2
78. Since the E-field points downward, the surface of the Earth is a lower potential than points above the
surface. Call the surface of the Earth 0 volts. Then a height of 2.00 m has a potential of 300 V. We
also call the surface of the Earth the 0 location for gravitational PE. Write conservation of energy
relating the charged spheres at 2.00 m (where their speed is 0) and at ground level (where their
electrical and gravitational potential energies are 0).


Einitial  Efinal  mgh  qV  12 mv 2  v 
2  gh 
qV 

m 

4.5  104 C   300 V  

2
v  2  9.80 m s   2.00 m  
  6.3241m s
 0.340 kg 



4.5  104 C   300 V  

2
v  2  9.80 m s   2.00 m  
  6.1972 m s
 0.340 kg 


v  v  6.3241m s  6.1972 m s  0.13m s
80. Use Eq. 23-7 for the dipole potential, and use Eq. 23-9 to determine the electric field.
x
V 
4 0
Ex  

V
x
r
2

x
p

4 0
 y2

x y
2
2
2


2
2
p  x y
4 0 


3/ 2
1/ 2


p
x

4 0 x 2  y 2
 x 23 x 2  y 2
x
2
 y2


1/ 2

3/ 2
2x 
3
p  2x2  y2 



5/ 2
 4 0   x 2  y 2  
p  2 cos 2   sin 2  
4 0 
Ey  
Notice the
p cos 
1
V
y
1
r3



r3

px  3 2
 2 x  y2
4 0 

5 / 2
2 y 

p 
 p 3cos  sin 
3 xy




5/ 2
3


2
2
4 0   x  y   4 0 
r


dependence in both components, which is indicative of a dipole field.
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
70