At steady state, no current flows through the capacitor. Thus a capacitor behaves like a open switch. 

Equilibrium circuit is shown in the above figure.

Effective resistance of the equivalent circuit  R_eff​=R₂​+r

Potential drop across resistor R₁​,  $\mathrm{V}=\frac{{\mathrm{R}}_2\mathrm{E}}{{\mathrm{R}}_{\mathrm{eff}}}$​

∴ $V=\left(\frac{R_2}{R_2+r}\right)E$

As R_1​ and C are connected in parallel to each other. Thus potential drop across C is same as that across R₁​.

Thus charge on the capacitor plates  Q=CV

∴$Q= \left(\frac{R_{2}C}{R_2+r}\right)E$