Answer
$V_2 = 60.0~V$
Work Step by Step
In part (c), we found that $q_2 = 480~\mu C$
We can find $V_2$:
$V_2 = \frac{q_2}{C_2} = \frac{480~\mu C}{8.00~\mu F} = 60.0~V$
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 744: 72d
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