P26_020.PDF

20. (a) In this situation, capacitors 1 and 3 are in series, which means their charges are necessarily the

same:

q 1 = q 3 = C 1 C 3 V

C 1 + C 3 = (1 . 0 µ F)(3 . 0 µ F)(12V)

1 . 0 µ F+3 . 0 µ F

=9 . 0 µ C .

Also, capacitors 2 and 4 are in series:

q 2 = q 4 = C 2 C 4 V

C 2 + C 4 = (2 . 0 µ F)(4 . 0 µ F)(12V)

2 . 0 µ F+4 . 0 µ F

=16 µ C .

(b) With switch 2 also closed, the potential diﬀerence V 1 across C 1 must equal the potential diﬀerence

across C 2 and is

V 1 =

C 3 + C 4

C 1 + C 2 + C 3 + C 4 V =

(3 . 0 µ F+4 . 0 µ F)(12V)

1 . 0 µ F+2 . 0 µ F+3 . 0 µ F+4 . 0 µ F =8 . 4V .

Thus, q 1 = C 1 V 1 =(1 . 0 µ F)(8 . 4V) = 8 . 4 µ C, q 2 = C 2 V 1 =(2 . 0 µ F)(8 . 4V) = 17 µ C, q 3 =

C 3 ( V

V 1 )=(3 . 0 µ F)(12V

−

8 . 4V)=11 µ C, and q 4 = C 4 ( V

−

V 1 )=(4 . 0 µ F)(12V

−

8 . 4V)=14 µ C.

−