Physics - Semiconductors Question with Solution | TestHub

The energy stored in a capacitor is given by $E=\frac{Q^2}{2C}$ Here, $C=\frac{2\pi {\in }_{0}L}{\ln \left(\frac{R_2}{R_1}\right)}$, where $\begin{array}{l}\mathrm{L}=\text{ length of the cylindrical conductor }\\ {\mathrm{R}}_1=\text{ inner radius }\\ {\mathrm{R}}_2=\text{ outer radius }\\ \therefore \mathrm{C}\propto \mathrm{L}\Rightarrow \mathrm{E}\propto \frac{{\mathrm{Q}}^2}{\mathrm{L}}\end{array}$ When both $Q$ and $L$ are doubled, by keeping other parameters fixed, the energy stored (E') becomes $\begin{array}{l}{E}^{\prime }\propto \frac{(2Q{)}^2}{2L}\\ \propto \frac{2Q^2}{L}\\ \therefore E^{\prime }=2E\end{array}$