chapter 24 (PDF)

Chapter 24
1. THINK Ampere is the SI unit for current. An ampere is one coulomb per second.
EXPRESS To calculate the total charge through the circuit, we note that 1 A  1 C/s and
1 h  3600 s.
ANALYZE (a) Thus,

FG
H

84 A  h  84

C h
s

IJ FG 3600 s IJ  3.0  10
K H hK

5

C.

(b) The change in potential energy is U = q V = (3.0  105 C)(12 V) = 3.6  106 J.
LEARN Potential difference is the change of potential energy per unit charge. Unlike
electric field, potential difference is a scalar quantity.
2. The magnitude is U = eV = 1.2  109 eV = 1.2 GeV.
3. (a) The change in energy of the transferred charge is
U = q V = (30 C)(1.0 109 V) = 3.0  1010 J.
(b) If all this energy is used to accelerate a 1000-kg car from rest, then U  K  12 mv 2 ,
and we find the car’s final speed to be

v

2K
2U
2(3.0 1010 J)


 7.7 103 m/s.
m
m
1000 kg

4. (a) E  F e   3.9 1015 N  1.60 1019 C  2.4 104 N C  2.4 104 V/m.

c

hb

g

(b) V  Es  2.4  104 N C 012
. m  2.9  103 V.
5. THINK The electric field produced by an infinite sheet of charge is normal to the
sheet and is uniform.
EXPRESS The magnitude of the electric field produced by the infinite sheet of charge is
E = /20, where  is the surface charge density. Place the origin of a coordinate system
at the sheet and take the x axis to be parallel to the field and positive in the direction of
the field. Then the electric potential is

1073

1074

CHAPTER 24

z

V  Vs 

x

0

E dx  Vs  Ex,

where Vs is the potential at the sheet. The equipotential surfaces are surfaces of constant x;
that is, they are planes that are parallel to the plane of charge. If two surfaces are
separated by x then their potentials differ in magnitude by
V = Ex = (/20)x.
ANALYZE Thus, for   0.10 106 C m2 and V  50 V, we have

x 

2 0 V





c

hb g  8.8  10

2 8.85  1012 C2 N  m2 50 V
6

010
.  10 C m

2

3

m.

LEARN Equipotential surfaces are always perpendicular to the electric field lines. Figure
24-5(a) depicts the electric field lines and equipotential surfaces for a uniform electric
field.
6. (a) VB – VA = U/q = –W/(–e) = – (3.94  10–19 J)/(–1.60  10–19 C) = 2.46 V.
(b) VC – VA = VB – VA = 2.46 V.
(c) VC – VB = 0 (since C and B are on the same equipotential line).
7. We connect A to the origin with a line along the y axis, along which there is no change
 
of potential (Eq. 24-18: E  ds  0 ). Then, we connect the origin to B with a line along

z

the x axis, along which the change in potential is

V  

z

x 4

0

z

 
E  ds  4.00

4

0

F4 I
x dx  4.00 G J
H 2K
2

which yields VB – VA = –32.0 V.
8. (a) By Eq. 24-18, the change in potential is the negative of the “area” under the curve.
Thus, using the area-of-a-triangle formula, we have
V  10  

z

x 2

0

  1
E  ds  2 20
2

b gb g

which yields V = 30 V.
(b) For any region within 0  x  3m,   E  ds is positive, but for any region for which
x > 3 m it is negative. Therefore, V = Vmax occurs at x = 3 m.

1075

V  10  

z

x 3

0

  1
E  ds  3 20
2

b gb g

which yields Vmax = 40 V.
(c) In view of our result in part (b), we see that now (to find V = 0) we are looking for
some X > 3 m such that the “area” from x = 3 m to x = X is 40 V. Using the formula for a
triangle (3 < x < 4) and a rectangle (4 < x < X), we require

b gb g b

gb g

1
1 20  X  4 20  40 .
2

Therefore, X = 5.5 m.
9. (a) The work done by the electric field is
f

W   q0 E  ds 
i

q0 d
q0d (1.60 1019 C)(5.80 1012 C/m 2 )(0.0356 m)
dz


2 0  0
2 0
2(8.85 1012 C2 /N  m 2 )

 1.87 1021 J.

(b) Since

V – V0 = –W/q0 = –z/20,

with V0 set to be zero on the sheet, the electric potential at P is
V 

z
(5.80 1012 C/m2 )(0.0356 m)

 1.17 102 V.
2 0
2(8.85 1012 C2 /N  m2 )

10. In the “inside” region between the plates, the individual fields (given by Eq. 24-13)
are in the same direction (  i ):


ˆ
50 109 C/m2
25 109 C/m2
Ein   

i  (4.2 103 N/C)iˆ .
12
2
2
12
2
2 
2(8.85

10
C
/N

m
)
2(8.85

10
C
/N

m
)


In the “outside” region where x > 0.5 m, the individual fields point in opposite directions:
Eout  

50 109 C/m2
25 109 C/m2
ˆi 
ˆi  (1.4 103 N/C)iˆ .
12
2
2
12
2
2
2(8.85 10 C /N  m )
2(8.85 10 C /N  m )

Therefore, by Eq. 24-18, we have
V   

0.8
0

E  ds   

 2.5 103 V.

0.5
0

Ein dx  

0.8
0.5

Eout dx    4.2 103   0.5  1.4 103   0.3

1076

CHAPTER 24

11. (a) The potential as a function of r is
r

r

qr

0

0

4 0 R

V  r   V  0    E  r dr  0  


dr  
3

qr 2
8 0 R3

(8.99 109 N  m 2 C2 )(3.50 1015 C)(0.0145 m) 2
 2.68 104 V.
3
2(0.0231 m)

(b) Since V = V(0) – V(R) = q/80R, we have
(8.99 109 N  m2 C2 )(3.50 1015 C)
V  R  

 6.81104 V.
8 0 R
2(0.0231 m)
q

12. The charge is

q  4 0 RV 

(10m) (1.0V)
8.99 10 N  m
9

2

/C

2

 1.1109 C.

13. (a) The charge on the sphere is
q  4 0 VR 

(200 V)(0.15 m)
 3.3 109 C.
9
2
2
8.99 10 N  m C

(b) The (uniform) surface charge density (charge divided by the area of the sphere) is

q
3.3109 C


 1.2 108 C/m2 .
2
2
4 R 4  0.15 m 
14. (a) The potential difference is
VA  VB 

q
4 0 rA



1 
 1
 1.0 106 C 8.99 109 N  m 2 C2  


4 0 rB
 2.0 m 1.0 m 
q

 4.5 103 V.


(b) Since V(r) depends only on the magnitude of r , the result is unchanged.
15. THINK The electric potential for a spherically symmetric charge distribution falls off
as 1/ r , where r is the radial distance from the center of the charge distribution.
EXPRESS The electric potential V at the surface of a drop of charge q and radius R is
given by V = q/40R.

1077
ANALYZE (a) With V = 500 V and q  30 1012 C, we find the radius to be

R

q
4 0V

8.99 10

9



N  m2 / C2  30 1012 C 
500 V

 5.4 104 m.

(b) After the two drops combine to form one big drop, the total volume is twice the
volume of an original drop, so the radius R' of the combined drop is given by (R')3 = 2R3
and R' = 21/3R. The charge is twice the charge of the original drop: q' = 2q. Thus,
V 

1 q
1
2q

 2 2 / 3V  2 2 / 3 (500 V)  790 V.
1/ 3
4 0 R  4 0 2 R

LEARN A positively charged configuration produces a positive electric potential, and a
negatively charged configuration produces a negative electric potential. Adding more
charge increases the electric potential.
16. In applying Eq. 24-27, we are assuming V  0 as r  . All corner particles are
equidistant from the center, and since their total charge is
2q1– 3q1+ 2 q1– q1 = 0,
then their contribution to Eq. 24-27 vanishes. The net potential is due, then, to the two
+4q2 particles, each of which is a distance of a/2 from the center:
V

4q2
1 4q2 16q2 16(8.99 109 N  m 2 C2 )(6.00 1012 C)



4 0 a / 2 4 0 a / 2 4 0 a
0.39 m
1

 2.21 V.

17. A charge –5q is a distance 2d from P, a charge –5q is a distance d from P, and two
charges +5q are each a distance d from P, so the electric potential at P is
V

q
(8.99 109 N  m 2 C2 )(5.00 1015 C)
 1 1 1 1






4 0  2d d d d  8 0 d
2(4.00 102 m)
q

 5.62 104 V.

The zero of the electric potential was taken to be at infinity.
18. When the charge q2 is infinitely far away, the potential at the origin is due only to the
charge q1 :
q1
V1 =
= 5.76 × 107 V.
4 0 d

1078

CHAPTER 24

Thus, q1/d = 6.41 × 1017 C/m. Next, we note that when q2 is located at x = 0.080 m, the
net potential vanishes (V1 + V2 = 0). Therefore,
0

kq2
kq
 1
0.08 m d

Thus, we find q2 = (q1 / d )(0.08 m) = –5.13 × 1018 C = –32 e.
19. First, we observe that V (x) cannot be equal to zero for x > d. In fact V (x) is always
negative for x > d. Now we consider the two remaining regions on the x axis: x < 0 and
0 < x < d.
(a) For 0 < x < d we have d1 = x and d2 = d – x. Let

V ( x)  k

FG q
Hd

1
1



IJ
K

q2
q

d2
4  0

FG 1  3 IJ  0
H x d  xK

and solve: x = d/4. With d = 24.0 cm, we have x = 6.00 cm.
(b) Similarly, for x < 0 the separation between q1 and a point on the x axis whose
coordinate is x is given by d1 = –x; while the corresponding separation for q2 is d2 = d – x.
We set
q
q
q
1
3
V ( x)  k 1  2 

0
d1 d 2
4  0  x d  x

FG
H

IJ
K

FG
H

IJ
K

to obtain x = –d/2. With d = 24.0 cm, we have x = –12.0 cm.
20. Since according to the problem statement there is a point in between the two charges
on the x axis where the net electric field is zero, the fields at that point due to q1 and q2
must be directed opposite to each other. This means that q1 and q2 must have the same
sign (i.e., either both are positive or both negative). Thus, the potentials due to either of
them must be of the same sign. Therefore, the net electric potential cannot possibly be
zero anywhere except at infinity.
21. We use Eq. 24-20:
9
2
2
30
p  8.99 10 N  m C  1.47  3.34 10 C  m 
V

 1.63 105 V.
2
9
4 0 r 2
52.0 10 m 

1

22. From Eq. 24-30 and Eq. 24-14, we have (for i = 0º)

1079

 p cos  p cos i
Wa  qV  e 

2
4 0 r 2
 4 0 r

 ep cos 
cos   1

2 
 4 0 r

with r = 20 × 109 m. For  = 180º the graph indicates Wa = 4.0 × 1030 J, from which
we can determine p. The magnitude of the dipole moment is therefore 5.6  1037 C  m .
23. (a) From Eq. 24-35, we find the potential to be
V 2

 L / 2  ( L2 / 4)  d 2
ln 
4 0
d



 2(8.99 10 N  m C )(3.68 10
9

2

2




12

 (0.06 m / 2)  (0.06 m) 2 / 4  (0.08 m) 2
C/m) ln 
0.08 m






 2.43102 V.

(b) The potential at P is V = 0 due to superposition.
24. The potential is
VP 

dq
1
Q
(8.99 109 N  m 2 C2 )(25.6 1012 C)

dq



4 0  rod R 4 0 R  rod
4 0 R
3.71102 m
1

 6.20 V.

We note that the result is exactly what one would expect for a point-charge –Q at a
distance R. This “coincidence” is due, in part, to the fact that V is a scalar quantity.
25. (a) All the charge is the same distance R from C, so the electric potential at C is
5Q1
1  Q1 6Q1 
5(8.99 109 N  m2 C2 )(4.20 1012 C)
V

 2.30 V,
 
 
4 0  R
R 
4 0 R
8.20 102 m

where the zero was taken to be at infinity.
(b) All the charge is the same distance from P. That distance is R 2  D2 , so the electric
potential at P is
Q1
6Q1 
5Q1
1 
V


 2

2
4 0  R  D 2
R2  D2 
4 R 2  D
0



5(8.99 10 N  m C )(4.20 10
9

2

2

12

C)

(8.20 102 m) 2  (6.71102 m) 2

 1.78 V.

1080

CHAPTER 24

26. The derivation is shown in the book (Eq. 24-33 through Eq. 24-35) except for the
change in the lower limit of integration (which is now x = D instead of x = 0). The result
is therefore (cf. Eq. 24-35)
V=

 L + L 2 + d2 

2.0 10 6  4  17 
4
ln
=

ln 
 = 2.18  10 V.
4o  D + D2 + d2
4 0
 1 2 

27. Letting d denote 0.010 m, we have
V

Q1
3Q1
3Q1
Q1
(8.99 109 N  m 2 C2 )(30 109 C)




4 0 d  0 d  0 d  0 d
2(0.01 m)

 1.3104 V.

28. Consider an infinitesimal segment of the rod, located between x and x + dx. It has
length dx and contains charge dq =  dx, where  = Q/L is the linear charge density of the
rod. Its distance from P1 is d + x and the potential it creates at P1 is
dV 

dq
1  dx

.
4 0 d  x 4 0 d  x
1

To find the total potential at P1, we integrate over the length of the rod and obtain:

V



4 0



L

0

L
dx

Q
 L

ln(d  x) 
ln 1  
d  x 4 0
4 0 L  d 
0

(8.99 109 N  m 2 C2 )(56.110 15 C) 
0.12 m 
ln 1 

0.12 m
 0.025 m 

 7.39 103 V.
29. Since the charge distribution on the arc is equidistant from the point where V is
evaluated, its contribution is identical to that of a point charge at that distance. We
assume V  0 as r   and apply Eq. 24-27:

V

1 Q1
1 4Q1
1 2Q1
1 Q1



4 0 R 4  2 R 4  R
4 0 R

(8.99 109 N  m 2 C2 )(7.211012 C)
2.00 m
2
 3.24 10 V.


30. The dipole potential is given by Eq. 24-30 (with  = 90º in this case)

1081
V

p cos  p cos 90

0
4 0 r 2
4 0 r 2

since cos(90º) = 0 . The potential due to the short arc is q1 / 4 0 r1 and that caused by the
long arc is q2 / 4 0 r2 . Since q1 = +2 C, r1 = 4.0 cm, q2 = 3 C, and r2 = 6.0 cm, the
potentials of the arcs cancel. The result is zero.
31. THINK Since the disk is uniformly charged, when the full disk is present each
quadrant contributes equally to the electric potential at P.
EXPRESS Electrical potential is a scalar quantity. The potential at P due to a single
quadrant is one-fourth the potential due to the entire disk. We first find an expression for
the potential at P due to the entire disk. To do so, consider a ring of charge with radius r
and (infinitesimal) width dr. Its area is 2r dr and it contains charge dq = 2r dr. All the
charge in it is at a distance r 2  D2 from P, so the potential it produces at P is
dV 

1 2p rdr
 rdr

.
2
2
4p 0 r  D
2 0 r 2  D 2

ANALYZE Integrating over r, the total potential at P is


V
2 0



R
0



r 2  D2
2
2
2 0
r D
rdr

R
0



  2
R  D2  D  .


2 0

Therefore, the potential Vsq at P due to a single quadrant is
V   2
(7.731015 C/m2 ) 
2

Vsq  
R D D 
(0.640 m)2  (0.259 m)2  0.259 m 

12
2
2
 8(8.85 10 C /N  m ) 

4 8 0 
 4.71105 V.

LEARN Consider the limit D
Vsq 

R. The potential becomes

  2

R  D2  D  


8 0
8 0

  1 R2

 D 1 
2
2
D





  D



qsq
 R 2  R 2 / 4



8 0 2 D 4 0 D 4 0 D

where qsq   R 2 / 4 is the charge on the quadrant. In this limit, we see that the potential
resembles that due to a point charge qsq .
32. Equation 24-32 applies with dq =  dx = bx dx (along 0  x  0.20 m).