Answer
$U_{max}=1.17\mu J$
Work Step by Step
We know that;
$U_{max}=\frac{Q^2}{2C}$
We plug in the known values to obtian:
$U_{max}=\frac{(2.90\times 10^{-6})^2}{2(3.60\times 10^{-6})}=1.1681\times 10^{-6}=1.17\mu J$
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 31 - Electromagnetic Oscillations and Alternating Current - Problems - Page 935: 1a
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