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

For a parallel-plate air capacitor of plate area $$A$$ and separation $$x$$, the capacitance is given by
$$C = \frac{\varepsilon_0 A}{x}\quad -(1)$$

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

Electrostatic energy stored in a capacitor at constant voltage is
$$U = \tfrac12 C V^{2}\quad -(2)$$

Substituting $$C$$ from $$(1)$$ into $$(2)$$:
$$U = \tfrac12 \varepsilon_{0} A \frac{V^{2}}{x}\quad -(3)$$

Differentiate $$(3)$$ with respect to time to find the rate of change of energy:
$$\frac{dU}{dt} = \tfrac12 \varepsilon_{0} A V^{2}\,\frac{d}{dt}\!\left(\frac{1}{x}\right)$$
Using $$\dfrac{d}{dt}\!\left(\frac{1}{x}\right)= -\dfrac1{x^{2}}\dfrac{dx}{dt}$$, we get
$$\frac{dU}{dt}= -\tfrac12 \varepsilon_{0} A V^{2}\,\frac{1}{x^{2}}\frac{dx}{dt}\quad -(4)$$

The plates are being pulled apart at a uniform speed $$v$$, so $$\dfrac{dx}{dt}=v$$ (a positive constant). Substituting this into $$(4)$$:
$$\frac{dU}{dt}= -\tfrac12 \varepsilon_{0} A V^{2}\,v\,\frac{1}{x^{2}}$$

The magnitude of the time rate of change of electrostatic energy is therefore proportional to $$x^{-2}$$. Hence
$$\left|\frac{dU}{dt}\right|\propto x^{-2}$$ and the required exponent is $$\alpha=-2$$.

Option A which is: $$-2$$