Answer
$q = 21.6~\mu C$
Work Step by Step
We can find the maximum charge on the capacitor:
$q = C~\mathscr{E}$
$q = (1.8\times 10^{-6}~F)(12.0~V)$
$q = 21.6\times 10^{-6}~C$
$q = 21.6~\mu C$
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 799: 58b
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