Answer
The charge on capacitor 2 is $~~16.0~\mu C$
Work Step by Step
We can find the equivalent capacitance of capacitor 2 and capacitor 4 in series:
$\frac{1}{C_{24}} = \frac{1}{C_2}+\frac{1}{C_4}$
$\frac{1}{C_{24}} = \frac{1}{2.00~\mu F}+\frac{1}{4.00~\mu F}$
$\frac{1}{C_{24}} = \frac{2}{4.00~\mu F}+\frac{1}{4.00~\mu F}$
$C_{24} = \frac{4.00~\mu F}{3}$
The total potential difference across both these capacitors is $12.0~V$
We can find the charge on each of these two capacitors:
$q = C_{24}~V$
$q = (\frac{4.00~\mu F}{3})(12.0~V)$
$q = 16.0~\mu C$
The charge on capacitor 2 is $~~16.0~\mu C$
