Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 25 - Capacitance - Problems - Page 742: 38c

Answer

$U_3 = 0.0188~J$

Work Step by Step

In part (b), we found that $V_3 = 50.0~V$ We can find $U_3$: $U_3 = \frac{1}{2}C_3~V_3^2$ $U_3 = \frac{1}{2}(15.0\times 10^{-6} F)(50.0~V)^2$ $U_3 = 0.0188~J$
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