Answer
$U_{max}=4.5$ nJ
Work Step by Step
We know that
$U_{max}=\frac{1}{2}\frac{Q^2}{C}$
We plug in the known values to obtain:
$U_{max}=\frac{1}{2}\frac{(3\times 10^{-9})^2}{1\times 10^{-9}}=4.5\times 10^{-9}=4.5nJ$
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 936: 15c
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