In the given figure each plate of capacitance $C$ has partial value of charge
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$\mathrm{i}=\frac{\mathrm{E}}{\left(\mathrm{R}_{2}+\mathrm{r}\right)}$

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

$\mathrm{Q}=\mathrm{C} \times \frac{\mathrm{ER}_{2}}{\left(\mathrm{R}_{2}+\mathrm{r}\right)}$

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