Physics - Capacitor Question with Solution | TestHub

Given: ${\mathrm{C}}_1:{\mathrm{C}}_2=1:2$ $\begin{array}{rl}\therefore {\mathrm{C}}_2& =2{\mathrm{C}}_1\\ {\mathrm{C}}_{\mathrm{P}}& ={\mathrm{C}}_1+{\mathrm{C}}_2={\mathrm{C}}_1+2{\mathrm{C}}_1=3{\mathrm{C}}_1\\ {\mathrm{C}}_{\mathrm{S}}& =\frac{{\mathrm{C}}_1{\mathrm{C}}_2}{{\mathrm{C}}_1+{\mathrm{C}}_2}=\frac{2{\mathrm{C}}_1^2}{3{\mathrm{C}}_1}=\frac{2}{3}{\mathrm{C}}_1\end{array}$ 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, $\begin{array}{rl}& \frac{1}{2}{\mathrm{C}}_{\mathrm{P}}{\mathrm{V}}_{\mathrm{P}}^2=\frac{1}{2}{\mathrm{C}}_{\mathrm{S}}{\mathrm{V}}_{\mathrm{s}}^2\\ \therefore & \frac{{\mathrm{V}}_{\mathrm{P}}^2}{{\mathrm{V}}_{\mathrm{S}}^2}=\frac{{\mathrm{C}}_{\mathrm{S}}}{{\mathrm{C}}_{\mathrm{P}}}\\ & \frac{{\mathrm{V}}_{\mathrm{P}}}{{\mathrm{V}}_{\mathrm{S}}}=\sqrt{\frac{\frac{2}{3}{\mathrm{C}}_1}{3{\mathrm{C}}_1}}=\frac{\sqrt{2}}{3}\end{array}$