16.

(a) The potential difference across C

1

is V

1

= 10 V. Thus, q

1

= C

1

V

1

= (10 µF)(10 V) = 1.0

× 10

−4

C.

(b) Let C = 10 µF. We first consider the three-capacitor combination consisting of C

2

and its two

closest neighbors, each of capacitance C. The equivalent capacitance of this combination is

C

eq

= C +

C

2

C

C + C

2

= 1.5C .

Also, the voltage drop across this combination is

V =

CV

1

C + C

eq

=

CV

1

C + 1.5C

=

2

5

V

1

.

Since this voltage difference is divided equally between C

2

and the one connected in series with it,

the voltage difference across C

2

satisfies V

2

= V /2 = V

1

/5. Thus

q

2

= C

2

V

2

= (10 µF)

10 V

5



= 2.0

× 10

−5

V .