Answer
$q = 1.0\times 10^{-3}~C$
Work Step by Step
We can find the initial charge on the capacitor:
$E = \frac{q^2}{2C}$
$q^2 = 2~E~C$
$q = \sqrt{2~E~C}$
$q = \sqrt{(2)(0.50~J)(1.0\times 10^{-6}~F)}$
$q = 1.0\times 10^{-3}~C$
Fundamentals of Physics Extended (10th Edition)
by Halliday, David; Resnick, Robert; Walker, Jearl
Published by
Wiley
ISBN 10:
1-11823-072-8
ISBN 13:
978-1-11823-072-5
Chapter 27 - Circuits - Problems - Page 800: 68a
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