Answer
The charge on capacitor 2 is $~~20.0~\mu C$
Work Step by Step
The equivalent capacitance of $C_2$ and the other capacitor in series is $\frac{10.0~\mu F}{2} = 5.00~\mu F$
The equivalent capacitance of these two capacitors and the other capacitor in parallel is $10.0~\mu F+ 5.00~\mu F = 15.0~\mu F$
We can find the equivalent capacitance of these capacitors with the other capacitor in series:
$\frac{1}{C_{eq}} = \frac{1}{10.0~\mu F}+\frac{1}{15.0~\mu F}$
$\frac{1}{C_{eq}} = \frac{3}{30.0~\mu F}+\frac{2}{30.0~\mu F}$
$C_{eq} = 6.00~\mu F$
We can find the charge stored on $C_{eq}$:
$q = C_{eq}~V$
$q = (6.00~\mu F)(10.0~V)$
$q = 60.0~\mu C$
$C_2$ and the capacitor in series have an equivalent capacitance of $5.00~\mu F$.
The other capacitor that is in parallel with these two capacitors has a capacitance of $10.0~\mu F$
The potential difference across both of these branches is equal, and the total charge stored is $q = 60.0~\mu C$.
Therefore, the charge on capacitor 2 is $~~20.0~\mu C$
