The ratio of potential difference that must be applied across parallel and series combination of two capacitors \(\mathrm{C}_1\) and \(\mathrm{C}_2\) with their capacitance in the ratio \(1: 2\) so that energy stored in these two cases becomes same is

Given: \(\mathrm{C}_1: \mathrm{C}_2=1: 2\)
\(\begin{aligned}
\therefore \quad \mathrm{C}_2 & =2 \mathrm{C}_1 \\
\mathrm{C}_{\mathrm{P}} & =\mathrm{C}_1+\mathrm{C}_2=\mathrm{C}_1+2 \mathrm{C}_1=3 \mathrm{C}_1 \\
\mathrm{C}_{\mathrm{S}} & =\frac{\mathrm{C}_1 \mathrm{C}_2}{\mathrm{C}_1+\mathrm{C}_2}=\frac{2 \mathrm{C}_1^2}{3 \mathrm{C}_1}=\frac{2}{3} \mathrm{C}_1
\end{aligned}\)
Let \(V_P\) and \(V_S\) be the potentials applied across the parallel and series combinations respectively, then using \(\mathrm{E}=\frac{1}{2} \mathrm{CV}^2\) we can write,
\(\begin{aligned}
& \frac{1}{2} \mathrm{C}_{\mathrm{P}} \mathrm{V}_{\mathrm{P}}^2=\frac{1}{2} \mathrm{C}_{\mathrm{S}} \mathrm{V}_{\mathrm{s}}^2 \\
\therefore \quad & \frac{\mathrm{V}_{\mathrm{P}}^2}{\mathrm{~V}_{\mathrm{S}}^2}=\frac{\mathrm{C}_{\mathrm{S}}}{\mathrm{C}_{\mathrm{P}}} \\
& \frac{\mathrm{V}_{\mathrm{P}}}{\mathrm{V}_{\mathrm{S}}}=\sqrt{\frac{\frac{2}{3} \mathrm{C}_1}{3 \mathrm{C}_1}}=\frac{\sqrt{2}}{3}
\end{aligned}\)