Answer
$6.47V $
Work Step by Step
We know that
$ Q_1=CV_1$
$\implies Q_1=(11.2\times 10^{-6}F)(12.0V)$
$ Q_1=134.4\mu C $
Now $ V=\frac{Q_1-Q_2}{C_1}$
$\implies \frac{Q_2}{C_2}=\frac{Q_1-Q_2}{C_1}$
This simplifies to:
$ Q_2=\frac{Q_1}{1+\frac{C_1}{C_2}}$
We plug in the known values to obtain:
$ Q_2=\frac{134.4\mu C}{1+\frac{11.2\mu F}{9.50\mu F}}$
$ Q_2=61.7\mu C $
We can find the required voltage as
$ V=\frac{Q_2}{C_2}$
We plug in the known values to obtain:
$ V=\frac{61.7\mu C}{9.50\mu F}$
$ V=6.47V $
