Answer
$q_2=1.5\times10^{-4}C$
Work Step by Step
Recall the following equation:
$q_2=C_2V$
We plug in the known values to obtain:
$q_2=3\times10^{-6}\times50=150\times10^{-6}$
$q_2=1.5\times10^{-4}C$
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 741: 21c
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