P26 084

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84. (Third problem of Cluster)

(a) With the series pair C

2

and C

3

reduced to a single C



= 10 µF capacitor, this becomes verysimilar

to problem 82. Noting for later use that q



= q

2

= q

3

, and using notation similar to that used in

the solution to problem 82, we have

Q = q

1

+ q



where Q = C

1

V

bat

= 400 µC. Also, after switch S is closed,

V

1

=

V



q

1

C

1

=

q



C



which yields

1
4

q

1

= q



. Therefore,

Q = q

1

+



1

4

q

1



which gives the result q

1

= 320 µC.

(b) We use q

2

= q

3

=

1
4

q

1

to obtain the result 80 µC.

(c) See part (b).

(d) (e) and (f) Eq. 26-1 yields

V =

q

C

=


8.0 V

for C

1

5.3 V

for C

2

2.7 V

for C

3


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