In the circuit shown in the figure $K_1$ is open. The charge on capacitor $C$ in steady state is $q_1$. Now key is closed and at steady state charge on $C$ is $q_2$. The ratio of charges $q_1/q_2$ is
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When key is open
$\mathrm{V}_{\mathrm{C}}=\mathrm{E} \Rightarrow \mathrm{q}_{1}=\mathrm{CE}$
When key is closed
$\mathrm{V}_{\mathrm{C}}^{1}=\frac{\mathrm{E}}{2+3} \times 3=\frac{3}{5} \mathrm{E}$
$\therefore $ $\mathrm{q}_{2}=\mathrm{CV}_{\mathrm{C}}^{1}=\mathrm{C} \times \frac{3}{5} \mathrm{E}=\frac{3}{5} \mathrm{q}_{1} \Rightarrow \frac{\mathrm{q}_{1}}{\mathrm{q}_{2}}=\frac{5}{3}$
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