14.

(a) and (b) The original potential difference V

1

across C

1

is

V

1

=

C

eq

V

C

1

+ C

2

=

(3.16 µF)(100 V)

10.0 µF + 5.00 µF

= 21.1 V .

Thus ∆V

1

= 100 V

− 21.1 V = 79 V and ∆q

1

= C

1

∆V

1

= (10.0 µF)(79 V) = 7.9

× 10

−4

C.