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 _____.

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

For a parallel plate capacitor, the capacitance is $C = \dfrac{\varepsilon_0 A}{x}$, 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:

$U = \dfrac{\varepsilon_0 A V^2}{2x}$

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

$\dfrac{dU}{dt} = \dfrac{d}{dt} \left( \dfrac{\varepsilon_0 A V^2}{2x} \right) = -\dfrac{\varepsilon_0 A V^2}{2x^2} \dfrac{dx}{dt}$

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

$\dfrac{dU}{dt} = -\dfrac{\varepsilon_0 A V^2 v}{2x^2}$

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

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

Answer: $-2$