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: \(-2\)