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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