JEE Main 2026 — Capacitance Question with Solution

The electrostatic energy of a parallel plate capacitor is given by $U=\frac{1}{2}C{V}^2$.

For a parallel plate capacitor, the capacitance is $C=\frac{{\epsilon }_{0}A}{x}$, where $x$ is the separation between the plates.

Since the capacitor remains connected to the battery, the potential difference $V$ across the plates is constant.

Substituting $C$ into the energy equation gives:

$U=\frac{{\epsilon }_{0}A{V}^2}{2x}$

Differentiating $U$ with respect to time $t$ to find the time rate of change of electrostatic energy:

$\frac{dU}{dt}=\frac{d}{dt}\left(\frac{{\epsilon }_{0}A{V}^2}{2x}\right)=-\frac{{\epsilon }_{0}A{V}^2}{2x^2}\frac{dx}{dt}$

Given that the plates are pulled apart at a uniform speed $v$, we have $\frac{dx}{dt}=v$.

$\frac{dU}{dt}=-\frac{{\epsilon }_{0}A{V}^{2}v}{2x^2}$

From the above expression, it is clear that $\frac{dU}{dt}\propto x^{-2}$.

Comparing this with $x^{\alpha }$, we get $\alpha =-2$.

Answer: $-2$