Answer
$V_2 = 50.0~V$
Work Step by Step
In part (g), we found that $q_2 = 250~\mu C$
We can find $V_2$:
$V_2 = \frac{q_2}{C_2}$
$V_2 = \frac{250~\mu C}{5.00~\mu F}$
$V_2 = 50.0~V$
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