Answer
$0.975J$
Work Step by Step
We can find the required done in charging the capacitor as follows:
$W=U=\frac{1}{2}QV$
We plug in the known values to obtain:
$W=\frac{1}{2}(3.75\times 10^{16}\times 1.6\times 10^{-19}C)(325V)$
$W=0.975J$
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 722: 90
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