A capacitor of capacitance 5 μ F is connected as shown in the...

In steady state, there will be no current in the capacitor branch. Net resistance of the circuit $R=1+1+0.5=2.5 \Omega$ Current drawn from the cell, $i=\frac{V}{R}=\frac{2.5}{2.5}=1 \mathrm{~A}$ Potential drop across two parallel branches $$\begin{aligned} V=E-i r & =2.5-1 \times 0.5 \\ & =2.5-0.5 \\ & =2.0 \mathrm{~V} \end{aligned}$$ So, charge on the capacitor plates $$\begin{aligned} q=C V & =5 \times 2 \\ & =10 \mu \mathrm{C} \end{aligned}$$