Question

A parallel plate capacitor with plate area A and initial separation d is connected to battery of EMF \(V_0\). Work done by external force to increase separation of plates from d to 2d slowly is

[Battery remains connected]

• A. $\frac{{\epsilon }_{0}A{V}_0^2}{4d}$
• B. $\frac{{\epsilon }_{0}A{V}_0^2}{d}$
• C. $\frac{{\epsilon }_{0}A{V}_0^2}{2d}$
• D. $\frac{{\epsilon }_{0}A{V}_0^2}{3d}$

Answer 1

The correct option is A $\frac{{\epsilon }_{0}A{V}_0^2}{4d}$

Let plate separation be x
\(F_{ext} = qE = q \frac{q}{2A \varepsilon_0} = \frac{q^2}{2 A \varepsilon_0}\)
\(dW = F_{ext}~dx\)
\(W = \int F_{ext}~dx\)

Given that \(q = CV=\frac{\varepsilon_0 A V_0}{x}\)

Now integrating from d to 2d we get,
\(W = \int^{2d}_{d} \frac{\varepsilon_0 AV_0^2}{2x^2} dx = \varepsilon_0 AV_0^2\int^{2d}_{d} \frac{1}{2x^2} dx=\varepsilon_0 AV_0^2[\frac{-1}{2x}]_d^{2d}=\frac{\varepsilon_0 AV^2_0}{4d}\)