c
$\mathrm{i}=\frac{\mathrm{E}}{\left(\mathrm{R}_{2}+\mathrm{r}\right)}$
In steady state capacitor is fully charged hence
No current will flow through line $(2)$ Hence potential difference across line $(1)$ is
$\mathrm{V}=\frac{\mathrm{E}}{\left(\mathrm{R}_{2}+\mathrm{r}\right)} \times \mathrm{R}_{2},$ the same potential difference appears across the capacitor, so charge on capacitor
$\mathrm{Q}=\mathrm{C} \times \frac{\mathrm{ER}_{2}}{\left(\mathrm{R}_{2}+\mathrm{r}\right)}$
