Answer
$q_2 = 768~\mu C$
Work Step by Step
In part (a), we found that $q_1 = 480~\mu C$
In part (c), we found that $q_2 = 480~\mu C$
The total charge is $q = 960~\mu C$
Note that $q_1 = q-q_2$
Since the capacitors are in parallel, the potential difference across each capacitor is equal.
We can find $q_2$:
$V_2 = V_1$
$\frac{q_2}{C_2} = \frac{q_1}{C_1}$
$q_2~C_1 = C_2~q_1$
$q_2~C_1 = C_2~(q-q_2)$
$q_2~(C_1+C_2) = C_2~q$
$q_2 = \frac{C_2}{C_1+C_2}~q$
$q_2 = (\frac{8.00~\mu F}{10.0~\mu F})~(960~\mu C)$
$q_2 = 768~\mu C$
