Answer
The charge on capacitor 3 is $~~9.00~\mu C$
Work Step by Step
We can find the equivalent capacitance of capacitor 1 and capacitor 3 in series:
$\frac{1}{C_{13}} = \frac{1}{C_1}+\frac{1}{C_3}$
$\frac{1}{C_{13}} = \frac{1}{1.00~\mu F}+\frac{1}{3.00~\mu F}$
$\frac{1}{C_{13}} = \frac{3}{3.00~\mu F}+\frac{1}{3.00~\mu F}$
$C_{13} = \frac{3.00~\mu F}{4}$
The total potential difference across both these capacitors is $12.0~V$
We can find the charge on each of these two capacitors:
$q = C_{13}~V$
$q = (\frac{3.00~\mu F}{4})(12.0~V)$
$q = 9.00~\mu C$
The charge on capacitor 3 is $~~9.00~\mu C$.
