Answer
The charge on capacitor 1 is $~~8.40~\mu C$
Work Step by Step
We can find the equivalent capacitance of capacitor 1 and capacitor 2 in parallel:
$C_{12} = C_1+C_2 = 3.00~\mu F$
We can find the equivalent capacitance of capacitor 3 and capacitor 4 in parallel:
$C_{34} = C_3+C_4 = 7.00~\mu F$
We can find the equivalent capacitance of the four capacitors:
$\frac{1}{C_{eq}} = \frac{1}{3.00~\mu F}+\frac{1}{7.00~\mu F}$
$\frac{1}{C_{eq}} = \frac{7}{21.0~\mu F}+\frac{3}{21.0~\mu F}$
$C_{eq} = \frac{21.0~\mu F}{10}$
$C_{eq} = 2.10~\mu F$
We can find the charge $q$:
$q = C_{eq}~V = (2.10~\mu F)(12.0~V) = 25.2~\mu C$
Note that the sum of the charge on capacitor 1 and capacitor 2 is equal to this charge $q$
We can find the potential difference across $C_{12}$:
$V_{12} = \frac{q}{C_{12}} = \frac{25.2~\mu C}{3.00~\mu F} = 8.40~V$
Note that this is the potential difference across capacitor 1 and capacitor 2
We can find the charge on capacitor 1:
$q_1 = (1.00~\mu F)(8.40~V) = 8.40~\mu C$
The charge on capacitor 1 is $~~8.40~\mu C$
