Question

A parallel plate capacitor is connected to a battery. The plate are pulled apart with a uniform speed. If ${}^{\prime }{x}^{\prime }$ is the separation between the plates, then the time rate of change of electrostatic energy of the capacitor is proportional to :

• A. $x^2$
• B. $x$
• C. $x^{-1}$
• D. $x^{-2}$

Answer 1

The correct option is D $x^{-2}$

Parallel plate capacitance at instant time t , $C=\frac{{ϵ}_{0}A}{x}$

Also potential $\left(V\right)$ will remain same (constant) because battery is attached.
Energy $U=\frac{1}{2}C{V}^2=\frac{1}{2}\times \frac{{ϵ}_{0}A}{x}\times V^2$

$U=\frac{1}{2}{ϵ}_{0}A{V}^2\left(\frac{1}{x}\right)$

Thus, $U\propto \frac{1}{x}$

$\frac{dU}{dt}=\frac{1}{2}{ϵ}_{0}A{V}^2(-\frac{1}{x^2}\frac{dx}{dt})$

$\frac{dU}{dt}\propto \frac{1}{x^2}$

Thus rate of change of U is proportional to $x^{-2}$.