College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 17 - Problems - Page 655: 57


$Q = 6.12\times 10^{-4}~C$

Work Step by Step

We can find the magnitude of charge which must be removed from each plate: $Q = C~V$ $Q = (10.2\times 10^{-6}~F)(60.0~V)$ $Q = 6.12\times 10^{-4}~C$
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