P26_083.PDF

83. (Second problem of Cluster )

(a) The change (from the previous problem) is that the initial charge (before switch S is closed) is

Q + Q where Q is as before but Q = C 2 (10 V) = 600 µ C. We assume the polarities of these

capacitor charges are the same. With this modiﬁcation, we follow steps similar to those in the

previous solution:

Q + Q = q 1 + q 2

= q 1 + 3

4 q 1

which yields q 1 = 571 µ C.

(b) The relation 4 q 1 = q 2 gives the result q 2 = 429 µ C.

(d) Similarly, V 2 = q 2 /C 2 =14 . 3V.

(e) The initial energy now includes 2 C 2 (20 V) 2 in addition to the 2 C 1 V bat computed in the previous

case. Thus, the totalinitialenergy is 8 . 00

10 − 3 J. And the ﬁnal storedenergy is 2 C 1 V 1 + 2 C 2 V 2 =

7 . 14

10 − 3 J. The decrease is therefore 8 . 6

10 − 4 J, as it was in the previous problem.