28.
(a) The potential difference across C
1
(the same as across C
2
) is given by
V
1
= V
2
=
C
3
V
C
1
+ C
2
+ C
3
=
(4.00 µF)(100 V)
10.0 µF + 5.00 µF + 4.00 µF
= 21.1 V .
Also, V
3
= V
− V
1
= V
− V
2
= 100 V
− 21.1 V = 78.9 V. Thus,
q
1
=
C
1
V
1
= (10.0 µF)(21.1 V ) = 2.11
× 10
−4
C
q
2
=
C
2
V
2
= (5.00 µF)(21.1 V ) = 1.05
× 10
−4
C
q
3
=
q
1
+ q
2
= 2.11
× 10
−4
C + 1.05
× 10
−4
C = 3.16
× 10
−4
C .
(b) The potential differences were found in the course of solving for the charges in part (a).
(c) The stored energies are as follows:
U
1
=
1
2
C
1
V
2
1
=
1
2
(10.0 µF)(21.1 V )
2
= 2.22
× 10
−3
J ,
U
2
=
1
2
C
2
V
2
2
=
1
2
(5.00 µF)(21.1 V )
2
= 1.11
× 10
−3
J ,
U
3
=
1
2
C
3
V
2
3
=
1
2
(4.00 µF)(78.9 V )
2
= 1.25
× 10
−2
J .