Calculate the amount of charge on capacitor of  $4\mu F$. The internal resistance of battery is  $1\Omega$:

Equivalent circuit 

Capacitor blocks the DC current. So current will flow in lower loop only. So, $V_{PQ}=\frac{5}{4+1}\times 4=4V$

Charge on $4\mu F=V\text{'}C=\left(\frac{4\times 4}{4+4}\right)\times 4\Rightarrow 2\times 4=8\mu C$