A parallel plate capacitor of capacitance of $10 \mu \mathrm{~F}$ is charged to $1 \mu \mathrm{C}$.

Question

A parallel plate capacitor of capacitance of $10 \mu \mathrm{~F}$ is charged to $1 \mu \mathrm{C}$. The charging battery is removed and then the separation between the\ plates is doubled. Work done during the process is\

Professor PC Thomas Classes · Answer

When charging battery is removed, charge $q$ on capacitor remains constant. As separation between the plates is doubled, its capacity becomes half. Work done during the process.\ $\mathrm{W}=\mathrm{U}_{2}-\mathrm{U}_{1}=\frac{\mathrm{q}^{2}}{2(\mathrm{C} / 2)} \frac{\mathrm{q}^{2}}{2 \mathrm{C}}=\frac{\mathrm{q}^{2}}{2 \mathrm{C}}$\ $\mathrm{W}=\frac{\left(1 	imes 10^{-6}ight)^{2}}{2 	imes 10 	imes 10^{-6}}$\