Question

1) A parallel plate capacitor has plate area A and separation d. It is charged to a potential difference V₀ . The charging battery is disconnected and the plates are pulled apart to three times the initial separation. The work required to separate the plate is $\frac{e_{0}A{V}_0^2}{Nd}$. Then find the value of N.

Answer 1

Dear Student,
$$\begin{gathered} HereBatteryisdisconnectedmeanspotentialdifferencewill \\ notchangehence. \\ Soenergyincapacitorwithoutpullingcapacitor \\ =\frac{1}{2}C{V_0}^2=\frac{{\epsilon }_{0}A{V_0}^2}{d} \\ Afterpullingcapacitonce=\frac{{\epsilon }_{0}A}{3d} \\ andenergy=\frac{{\epsilon }_{0}A{V_0}^2}{3d} \\ henceChangeinpotentialenergy=\frac{{\epsilon }_{0}A{V_0}^2}{3d}-\frac{{\epsilon }_{0}A{V_0}^2}{d}=-\frac{2{\epsilon }_{0}A{V_0}^2}{3d}=-workdone \\ hence \\ N=\frac{2}{3} \end{gathered}$$
Regards