Determine the charge on the capacitor in the following circuit:

After a long time capacitor branch will behave like open circuit so circuit will be like,

Equivalent resistance is
$R=6+\left(4\left|\right|\left(2+10\right)\right) \mathrm{\Omega } = 9 \mathrm{\Omega }$
and, the current through battery becomes $\frac{72}{9}=8 \mathrm{A}$ 
From the circuit shown, current through $10 \mathrm{\Omega }\mathrm{ }$ resistor becomes, $\frac{4}{(10+2+4)}\times 8=2 \mathrm{A}$ 

Potential available for capacitor is = potential drop across $10 \mathrm{\Omega }$
$\therefore$ Potential available for capacitor is: $10\times 2=20 \mathrm{V}$
$\therefore$ Charge on capacitor $Q=\left(10 \mathrm{\mu }\mathrm{F}\right)\left(20 \mathrm{V}\right)=200 \mathrm{\mu }\mathrm{C}$