PHY 217
Solutions for Homework Assignment 6 from Chapter 25, Capacitors
Question 6
The capacitors in each circuit 1, 2 and 3 are connected in PARALLEL to each other.
When the switch is closed, the voltage or potential difference V between their plates must be
the same.
The charges across them however are not the same. When the switch is closed, the total
charge on the two capacitors is shared so that both capacitors have the same value of V.
Circuit 1:
With Switch Open:
V
1

q
C
1

1
q 3q
6q 3q
 ;V 2  2 
2C C
C2 C
The two capacitors are already at the same potential even before the switch is closed. After
the switch is closed they will continue to have same potential difference across them SO NO
CHARGES WILL FLOW FROM EITHER ONE OF THEM TO THE OTHER.
So the charge on the left-hand capacitor will remain the same as before the switch is closed.
Circuit 2:
Switch Open:
V
1

q
C
1
1

6q
q
2
and
3C
C
V
2
3
q
C
That means left capacitor is at a lower potential before the circuit is closed.
Switch Closed:
For both capacitors to come to the same potential charges must flow
If V is the potential now,
q
1, new
3C

q

2, new
C
q
1, new
 3 q
.............(1)
2, new
But the total charge on the two capacitors together is 9q which is shared.
q
So,
1, new

q
2, new
 9q........................(2)
Combine (1) and (2) to get
27
3
q1,new  4 q or 6 4 q and q 2,new 
Comparing this with old values of 6q
for left and right plates, ch arg e has
9
1
q or 2 q
4
4
and 3q
increased for left capacitor.
Circuit 3:
Switch Open:
Apply exactly the same arguments to find that the potential of the left plate (3q/C) is higher
than that of the right capacitor(3q/2C).
Switch Closed:
So when the switch is closed, charges must flow between the left and right capacitors to make
the potential equal.
q
1, new
2C
But
q

q
2, new
2C
1, new
+q
which gives
2, new
q
1, new
=q
2, new
= 9q
9q
 4.5q = q 2,new
2
Comparing this to old values of q = 6q and q = 3q, we see that

q
1, new
=
1
2
charge on the left capacitor has decreased and that of the right capacitor has increased.
Problem 6:
(a) Capacitance of a spherical capacitor of inner radius a (38.0mm) and outer radius
b(40.0mm) is given by the expression,
C = 4  0
ab
ba
But we know that
So C

1
4  0
 9 10 N m
9
2
3
1
9 10
9

C
38 10  40 10
2 10
3
2
3
12
 84.5 10 Farad
(b) The capacitance of a parallel plate capacitor is
C
A o
d
Let the distance between the plates be d = b-a. Then C = 0A/(b – a), or
A
C b  a 
0

84.5pF 40.0 mm  38.0 mm   191cm2 .
8.85 10
12
C2
Nm 2

Problem 12:
12. (a)
The potential difference across C1 is V1 = 10.0 V. So,
q1 = C1V1 = (10.0 F)(10.0 V) = 1.00  10–4 C.
(b) Let C = 10.0 F. We first consider the three-capacitor combination consisting of C2 and its
two closest neighbors, each of capacitance C. The equivalent capacitance of this combination
is
Ceq  C 
C2C
 1.50 C.
C  C2
Also, the voltage drop across this combination is
V
CV1
CV1

 0.40V1 .
C  Ceq C  1.50 C
Since this voltage difference is divided equally between C2 and the one connected in series
with it, the voltage difference across C2 satisfies V2 = V/2 = V1/5. Thus
 10.0V 
5
q2  C2V2  10.0  F  
  2.00 10 C.
 5 
Problem 31:
The energy stored by a capacitor is given by U  21 CV 2 , where V is the potential difference
across its plates. We convert the given value of the energy to Joules. Since 1 J  1 W  s, we
multiply by (103 W/kW)(3600 s/h) to obtain 10 kW  h  3.6 107 J . Thus,
c
b
h
7
2U 2 3.6  10 J
C 2 
 72 F.
2
V
1000 V
g
Problem 49:
Think back to the lecture where we found that the introduction of a dielectric between the
plates of a capacitor causes the capacitance to be multiplied by a factor κ and the electric
field to be reduced by the same factor κ
Here we have different dielectric materials in the top and bottom halves of the plates.
We assume there is charge q on one plate and charge –q on the other. The electric field in the
lower half of the region between the plates is
E1 
q
 1 0 A
,
where A is the plate area. The electric field in the upper half is
E2 
q
 2 0 A
.
Let d/2 be the thickness of each dielectric. Since the field is uniform in each region, the
potential difference between the plates is
V
LM
N
OP
Q
E1d E2 d
qd
1
1
qd  1   2




,
2
2
2 0 A  1  2
2 0 A  1 2
so
C
q 2 0 A  1 2

.
V
d  1  2
This expression is exactly the same as that for Ceq of two capacitors in series, one with
dielectric constant 1 and the other with dielectric constant 2. Each has plate area A and plate
separation d/2. Also we note that if 1 = 2, the expression reduces to C = 10A/d, the correct
result for a parallel-plate capacitor with plate area A, plate separation d, and dielectric
constant 1.
With A  7.89 104 m2 , d  4.62 10 3m , 1  11.0 and  2  12.0 , the capacitance is
2(8.85 1012 C2 /N  m2 )(7.89 104 m2 ) (11.0)(12.0)
C
 1.731011 F.
3
4.62 10 m
11.0 12.0