Question

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

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

Answer 1

Answer: (D)

$U=\frac{1}{2}C{V}^2=\frac{1}{2}\frac{{\epsilon }_{0}A}{d}{V}^2$
At any instant, let the separation between plates be$x$ .
So, $U=\frac{1}{2}\frac{{\epsilon }_{0}A}{x}{V}^2$
$\therefore \frac{dU}{dt}=\frac{1}{2}{\epsilon }_{0}A{V}^2(-1)\frac{1}{x^2}=\frac{dx}{dt}$
$=-\frac{1}{2}\frac{{\epsilon }_{0}A{V}^2}{x^2}(v)$
i.e., potential energy decreases as $\left(1/x^2\right)$.