Answer
$C=3.6\mu F$
Work Step by Step
The charge on a capacitor is equal to $$Q=CV$$ Solving for $C$ yields $$C=\frac{Q}{V}$$ Substituting known values of $Q=32 \mu C$ and $V=9.0V$ yields a capacitance of $$C=\frac{32 \mu C}{9.0V}=3.6\mu F$$
Physics Technology Update (4th Edition)
by Walker, James S.
Published by
Pearson
ISBN 10:
0-32190-308-0
ISBN 13:
978-0-32190-308-2
Chapter 20 - Electric Potential and Electric Potential Energy - Problems and Conceptual Exercises - Page 720: 52
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