Answer
The charge stored on capacitor 2 is $~~12~\mu C$
Work Step by Step
We can write a general expression for capacitance:
$C = \frac{q}{V}$
Thus, $C$ is the slope of the $q$ versus $V$ graph for each capacitor.
$C_1 = \frac{12.0~\mu C}{2.0~V} = 6.0~\mu F$
$C_2 = \frac{8.0~\mu C}{2.0~V} = 4.0~\mu F$
$C_3 = \frac{4.0~\mu C}{2.0~V} = 2.0~\mu F$
The equivalent capacitance of $C_2$ and $C_3$ is $4.0~\mu F+2.0~\mu F = 6.0~\mu F$
We can find the equivalent capacitance of the three capacitors:
$\frac{1}{C_{eq}} = \frac{1}{6.0~\mu F}+\frac{1}{6.0~\mu F}$
$C_{eq} = 3.0~\mu F$
We can find the charge stored by $C_{eq}$:
$q = C_{eq}~V$
$q = (3.0~\mu F)(6.0~V)$
$q = 18~\mu C$
Note that this is the charge stored in $C_1$
We can find the potential difference across $C_1$:
$V_1 = \frac{q}{C_1}$
$V_1 = \frac{18~\mu C}{6.0~\mu F}$
$V_1 = 3.0~V$
Then the potential difference $V_2$ across $C_2$ is $6.0~V-3.0~V$ which is $V_2 = 3.0~V$
We can find the charge stored on capacitor 2:
$q = C_2~V_2$
$q = (4.0~\mu F)(3.0~V)$
$q = 12~\mu C$
The charge stored on capacitor 2 is $~~12~\mu C$
