An air-filled parallel plate capacitor with plate separation 1 mm has a capacitance of 20 pF. It is charged to 4.0 μ C. Calculate the amount of work…

\[\begin{aligned} \mathrm{C}= & \frac{A \varepsilon_{\mathrm{o}}}{d} ; \quad \mathrm{C}^{\prime}=\frac{A \varepsilon_{\mathrm{o}}}{d^{\prime}} \\ & \Rightarrow \frac{\mathrm{C}^{\prime}}{\mathrm{C}}=\frac{d}{d^{\prime}} \Rightarrow \mathrm{C}^{\prime}=\frac{1}{5} \times 20 \\ & \mathrm{C}^{\prime}=4 \mathrm{pF} \end{aligned} \] Work done = change in energy \[\begin{aligned} & =\frac{1}{2} \frac{Q^{2}}{\mathrm{C}^{\prime}}-\frac{1}{2} \frac{Q^{2}}{\mathrm{C}} \\ & =\frac{Q^{2}}{2}\left[\frac{1}{4}-\frac{1}{20}\right] \times 10^{12} \end{aligned} \] On solving ; \[\begin{aligned} \text { Work done } & =\frac{8}{5} \mathrm{~J} \\ & =1.6 \mathrm{~J} \end{aligned} \]