Chapter 23 Problem 31
†
C
1
C
2
C
3
Given
C
1
C
2
= 0
= 0 .
.
02
01
µF
µF
C
3
= 0 .
02 µF
V = 100 V
Solution a) Find the equivalent capacitance of the combination.
C
1
C'
C
2 and C
3 are in parallel, their equivalent capacitance is
C
0
= C
2
+ C
3
= 0 .
01 µF + 0 .
02 µF = 0 .
03 µF
This leaves us with a combination of C
1 and C
0 in series. This gives us an equivalent capacitance of
1
C e
1
=
C
1
1
+
C 0
C e
=
1
C
1
1
+
1
C
0
=
1
0 .
02 µF
1
+
1
0 .
03 µF
= 0 .
012 µF
Note: Although the textbook asks to solve for the charge on each capacitor, I find it easier to solve for the voltages first. Therefore, I am doing part c) first followed by part b).
c) Find the voltage across each capacitor.
From the second diagram the voltage across C
1 and C
0 must total 100 V .
V
1
+ V
0
= 100 V
†
Problem from Essential University Physics, Wolfson

From the definition of capacitance
C =
Q
V
Therefore, then V =
Q
C
Q
1
C
1
Q
0
+
C 0
= 100 V
When the voltage is first applied, the charge drawn off of C
1 charge on each of these capacitors will be the same.
will have to accumulate on C
0
, therefore, the
Q
C
1
+
Q
C 0
= 100 V
Solving for Q gives
Q =
100 V
1
C
1
+
1
C
0
=
100 V
1
0 .
02 µF
+
1
0 .
03 µF
= 1 .
2 µC
The voltage across C
1 is then
1 .
2 µC
V
1
=
0 .
02 µF
= 60 V
Therefore, the voltage across C
0 is
V
0
= 100 V − V
1
= 100 V − 60 V = 40 V
Since C
2 and C
3 are in parallel, the voltage across each of them must be the same. Therefore,
V
2
= V
3
= 40 V b) Find the charge on each capacitor.
From part b the charge on C
1 is 1 .
2 µC . The charge on C
2 is
C
2
=
Q
2
V
2
Solving for Q
2 gives
Q
2
= C
2
V
2
= (0 .
01 µF )(40 V ) = 0 .
4 µC
By the same process the charge on C
3 is
Q
3
= C
3
V
3
= (0 .
02 µF )(40 V ) = 0 .
8 µC
Notice that the charge flowing off of C
1 is equal to the total charge on C
2 and C
3
.