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
$F_{ext}=qE=q\frac{q}{2{\epsilon }_0}=\frac{q^2}{2A{\epsilon }_0}$
$dW=F_{ext}dx$
$W=\int F_{ext}dx$

Given that $q=\frac{{\epsilon }_{0}A{V}_0}{x}$
$W={\int }_d^{2d}\frac{{\epsilon }_{0}A{V}_0^2}{2x^2}dx=\frac{{\epsilon }_{0}A{V}_0^2}{4d}$