Answer
$q_1 = 192~\mu C$
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 $q_2 = q-q_1$
Since the capacitors are in parallel, the potential difference across each capacitor is equal.
We can find $q_1$:
$V_1 = V_2$
$\frac{q_1}{C_1} = \frac{q_2}{C_2}$
$q_1~C_2 = C_1~q_2$
$q_1~C_2 = C_1~(q-q_1)$
$q_1~(C_1+C_2) = C_1~q$
$q_1 = \frac{C_1}{C_1+C_2}~q$
$q_1 = (\frac{2.00~\mu F}{10.0~\mu F})~(960~\mu C)$
$q_1 = 192~\mu C$
