Answer
The charge on capacitor 4 is $~~18.0~\mu C$
Work Step by Step
When the switch is closed, $C_2$ and $C_4$ are in parallel.
The equivalent capacitance of these two capacitors is $6.00~\mu F+6.00~\mu F = 12.0~\mu F$
We can find the equivalent capacitance of $C_1, C_2, C_3,$ and $C_4$:
$\frac{1}{C_{eq}} = \frac{1}{8.00~\mu F}+\frac{1}{12.0~\mu F}+\frac{1}{8.00~\mu F}$
$C_{eq} = 3.00~\mu F$
We can find the charge stored on $C_{eq}$:
$q = C_{eq}~V$
$q = (3.00~\mu F)(12.0~V)$
$q = 36.0~\mu C$
The total charge on $C_2$ and $C_4$ is $36.0~\mu C$
We can find the potential difference:
$V = \frac{36.0~\mu C}{12.0~\mu F} = 3.00~V$
The potential difference across $C_4$ is $3.00~V$
We can find the charge on capacitor 4:
$q = (6.00~\mu F)(3.00~V) = 18.0~\mu C$
The charge on capacitor 4 is $~~18.0~\mu C$
