Calculate the charge on capacitor A (in $\mu$C) in the circuit shown in figure in steady state.

The current distribution is shown in figure. As we know that in steady state no current will flow in the two branches consisting of capacitors.

From figure the currents in the two loops are given as

$i_1=\frac{24}{4+2}=4A and i_2=\frac{18}{3+6}=2A$

In the figure we consider a potential x at the right plate of the 2 $\mu F$ capacitor and zero at the left plate of this capacitor 

 Also, the values of charges on each plate is shown in figure.

$q_A=q_B$

$\begin{array}{l}\Rightarrow 2x=3(18-x+12)\\ \Rightarrow 5x=90\\ \Rightarrow x=18\mathrm{V}\end{array}$

Charge on 2 $\mu$F capacitor is given as $q_{2\mu F}=2\times 18=36\mu \mathrm{C}$
Charge on 3 $\mu$F capacitor is given as $q_{3\mu F}=3\times (18-6)=36\mu \mathrm{C}$