Answer
The charge on capacitor 4 is $~~12~\mu C$
Work Step by Step
The equivalent capacitance of $C_2, C_3,$ and $C_4$ in parallel is $4.0~\mu F+4.0~\mu F+4.0~\mu F = 12~\mu F$
The equivalent capacitance of $C_5$ and $C_6$ in parallel is $6.0~\mu F+6.0~\mu F = 12~\mu F$
We can find the equivalent capacitance of all the capacitors:
$\frac{1}{C_{eq}} = \frac{1}{6.0~\mu F}+\frac{1}{12~\mu F}+\frac{1}{12~\mu F}$
$\frac{1}{C_{eq}} = \frac{2}{12~\mu F}+\frac{1}{12~\mu F}+\frac{1}{12~\mu F}$
$C_{eq} = 3.0~\mu F$
We can find the charge stored on $C_{eq}$:
$q = C_{eq}~V$
$q = (3.0~\mu F)(12~V)$
$q = 36~\mu C$
Note that this is the total charge stored on $C_2, C_3,$ and $C_4$
Since $C_2 = C_3 = C_4$, by symmetry, each capacitor stores one third of the charge.
The charge on capacitor 4 is $~~12~\mu C$
