Physics 210 Mastering Physics Solutions to Week 5 Assignment 23.28.IDENTIFY and SET UP: Expressions for the electric potential inside and outside a solid conducting sphere are
derived in Example 23.8.
kq k (350  109 C)

 656 V.
EXECUTE: (a) This is outside the sphere, so V 
r
0480 m
k (350  109 C)
 131 V.
0240 m
(c) This is inside the sphere. The potential has the same value as at the surface, 131 V.
EVALUATE: All points of a conductor are at the same potential.
23.32.IDENTIFY: The voltmeter reads the potential difference between the two points where the probes are placed.
Therefore we must relate the potential difference to the distances of these points from the center of the cylinder.
For points outside the cylinder, its electric field behaves like that of a line of charge.
(b) This is at the surface of the sphere, so V 
SET UP: Using V 

ln ( rb /ra ) and solving for rb , we have rb  ra e 2 e0 V/ .
2 e0


1
(175 V)

9
2 2
2  900  10 N  m /C 
 0648, which gives
EXECUTE: The exponent is 
150  109 C/m
rb = (2.50 cm) e0.648 = 4.78 cm.
The distance above the surface is 4.78 cm  2.50 cm  2.28 cm.
EVALUATE: Since a voltmeter measures potential difference, we are actually given V , even though that is
not stated explicitly in the problem. We must also be careful when using the formula for the potential difference
because each r is the distance from the center of the cylinder, not from the surface.
23.70.IDENTIFY: Divide the rod into infinitesimal segments with charge dq. The potential dV due to the segment is
1 dq
dV 
. Integrate over the rod to find the total potential.
4 e0 r
SET UP: dq   dl , with   Q/ a and dl  ad .
EXECUTE: dV 
1  dl
1 Q dl
1 Q d
1  Q d
1 Q
dq




. V
.

0
a
4 e0 r
4 e0 a
4 e0  a a 4 e0  a
4 e0
4 e0 a
1
EVALUATE: All the charge of the ring is the same distance a from the center of curvature.
23.73.IDENTIFY: The sphere no longer behaves as a point charge because we are inside of it. We know how the electric
field varies with distance from the center of the sphere and want to use this to find the potential difference
between the center and surface, which requires integration.
kQ 
r2 
SET UP: Use the result of Problem 23.72. For r  R, V 
 3  2  .
2R 
R 
kQ
3kQ
EXECUTE: At the center of the sphere, r  0 and V1 
.
. At the surface of the sphere, r  R and V2 
R
2R
The potential difference is V1 V2 
kQ (899  109 N  m 2 /C2 )(400  106 C)

 360  105 V.
2R
2(00500 m)
EVALUATE: To check our answer, we could actually do the integration. We can use the fact that E 
R
kQ
0
R3 0
V1  V2   Edr 

R
rdr 
kQ  R 2  kQ
.


R3  2  2 R
24.1.IDENTIFY: The capacitance depends on the geometry (area and plate separation) of the plates.
Q
Q
SET UP: For a parallel-plate capacitor, Vab  Ed , E 
, and C 
.
e0 A
Vab
EXECUTE: (a) Vab  Ed  (400  106 V/m)(250  103 m)  100  104 V.
kQr
R3
so
(b) Solving for the area gives
800  109 C
Q

 226  103 m 2  226 cm 2 .
A
Ee0 (400  106 V/m)[8854  1012 C2 /(N  m 2 )]
Q 800  109 C

 800  1012 F  800 pF.
Vab 100  104 V
EVALUATE: The capacitance is reasonable for laboratory capacitors, but the area is rather large.
IDENTIFY: Three of the capacitors are in series, and this combination is in parallel with the other two capacitors.
SET UP: For capacitors in series the voltages add and the charges are the same;
1
1
1


 . For capacitors in parallel the voltages are the same and the charges add; Ceq  C1  C2  L.
Ceq C1 C2
(c) C 
24.21.
C
Q
.
V
EXECUTE: (a) The equivalent capacitance of the 18.0 nF, 30.0 nF and 10.0 nF capacitors in series is 5.29 nF. When
these capacitors are replaced by their equivalent we get the network sketched in Figure 24.21. The equivalent
capacitance of these three capacitors in parallel is 19.3 nF, and this is the equivalent capacitance of the original
network.
Figure 24.21
(b) Qtot  CeqV  (193 nF)(25 V)  482 nC.
(c) The potential across each capacitor in the parallel network of Figure 24.21 is 25 V.
Q65  C65V65  (65 nF)(25 V)  162 nC
d) 25 V.
EVALUATE: As with most circuits, we must go through a series of steps to simplify it as we solve for the
unknowns.