P26_085.PDF

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Chapter 26 - 26.85
85. (Fourth problem of Cluster )
(a) With the parallel pair C 2 and C 3 reduced to a single C =45 µ F capacitor, this becomes very
similar to problem 82. Using notation similar to that used in the solution to 82, we have
Q = q 1 + q
where Q = C 1 V bat = 400 µ C. Also, after switch S is closed,
V 1 = V
q 1
C 1
q
C
which yields 8 q 1 = q . Therefore,
Q = q 1 + 9
8 q 1
which gives the result q 1 = 188 µ C.
(b) We find the voltage across capacitor 1 from q 1 /C 1 (see below) and (since the capacitors are in
parallel) use the fact that V 1 = V 2 = V 3 with q = CV to obtain the charges: q 2 =71 µ Cand
q 3 = 141 µ C.
(c) See part (b).
(d) (e) and (f) The capacitors all have the same voltage. V = q 1 /C 1 =4 . 7V.
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