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 .