Answer
$V = 45~V$
Work Step by Step
We can find the required potential:
$U = \frac{1}{2}CV^2$
$V^2 = \frac{2U}{C}$
$V = \sqrt{\frac{2U}{C}}$
$V = \sqrt{\frac{(2)(1.0~J)}{1.0\times 10^{-3}~F}}$
$V = 45~V$
Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)
by Knight, Randall D.
Published by
Pearson
ISBN 10:
0133942651
ISBN 13:
978-0-13394-265-1
Chapter 26 - Potential and Field - Exercises and Problems - Page 738: 31
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