Answer
$57.14\ V$
Work Step by Step
The potential difference between the plates of a capacitor is:
$V = \frac{q}{C} $
$V = \frac{200\times 10^{-12} c}{3.5\times 10^{-12}F}$
$V=57.14$ V
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 25 - Capacitance - Problems - Page 739: 1c
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