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 744: 67e

Answer

$120\ V$

Work Step by Step

The voltage across $C_2$ can be found using the formula: $V_2 =\frac{q_2}{C_2}$ As capacitor are arranged in series, $q_1 =q_2 =480\ \mu C $ Thus: $V_2 =\frac{480\ \mu C}{4\ \mu F}$ $V_2 =120\ V$
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