Electrostatic Energy Rate of Change in Parallel Plate Capacitor

To find the dependence of the time rate of change of electrostatic energy of the parallel plate capacitor on the plate separation $x$, we proceed step-by-step:

1. Capacitance as a function of separation $x$: For a parallel plate air capacitor with plate area $A$ and separation $x$, the capacitance $C$ is given by: $C = \frac{\varepsilon_0 A}{x}$

2. Electrostatic Energy Stored: Since the capacitor remains connected to a battery, the potential difference $V$ across the plates remains constant. The electrostatic energy $U$ stored in the capacitor is: $U = \frac{1}{2} C V^2 = \frac{1}{2} \left( \frac{\varepsilon_0 A}{x} \right) V^2 = \frac{\varepsilon_0 A V^2}{2 x}$

3. Rate of Change of Energy: The plates are pulled apart at a uniform speed $v$, which means: $\frac{dx}{dt} = v = \text{constant}$

   Differentiating $U$ with respect to time $t$ using the chain rule, we get: $\frac{dU}{dt} = \frac{d}{dx}\left( \frac{\varepsilon_0 A V^2}{2 x} \right) \cdot \frac{dx}{dt}$ $\frac{dU}{dt} = -\frac{\varepsilon_0 A V^2}{2 x^2} \cdot v$

4. Proportionality: Taking the magnitude of the rate of change of energy: $\left| \frac{dU}{dt} \right| \propto \frac{1}{x^2} = x^{-2}$

   Comparing this with $\left| \frac{dU}{dt} \right| \propto x^{\alpha}$, we get: $\alpha = -2$

Thus, the correct option is A (which corresponds to $-2$).