# Chapter 25 - Capacitance - Problems - Page 744: 67e

$120\ V$

#### 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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