Answer
$U_1 = 0.0125~J$
Work Step by Step
In part (e), we found that $V_1 = 50.0~V$
We can find $U_1$:
$U_1 = \frac{1}{2}C_1~V_1^2$
$U_1 = \frac{1}{2}(10.0\times 10^{-6} F)(50.0~V)^2$
$U_1 = 0.0125~J$
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 742: 38f
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