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(A) The electrostatic energy $U$ of a capacitor connected to a battery is $U = \frac{1}{2}CV^2$.Here $C = \frac{\epsilon_0 A}{x}$, so $U = \frac{1}{2}

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-$2$

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(A) The electrostatic energy $U$ of a capacitor connected to a battery is $U = \frac{1}{2}CV^2$.Here $C = \frac{\epsilon_0 A}{x}$, so $U = \frac{1}{2} \left( \frac{\epsilon_0 A}{x} \right) V^2$.Since the battery is connected, the potential difference $V$ remains constant.Therefore, $U \propto \frac{1}{x} = x^{-1}$.The time rate of change of energy is $\frac{dU}{dt} = \frac{dU}{dx} \cdot \frac{dx}{dt}$.Given that the plates are pulled apart at a uniform speed $v$, we have $\frac{dx}{dt} = v$ (a constant).Now, differentiating $U$ with respect to $x$: $\frac{dU}{dx} = \frac{d}{dx} \left( \frac{\epsilon_0 A V^2}{2} \cdot x^{-1} \right) = -\frac{\epsilon_0 A V^2}{2} \cdot x^{-2}$.Thus, $\frac{dU}{dt} = \left( -\frac{\epsilon_0 A V^2}{2} \cdot x^{-2} \right) \cdot v$.Since $\epsilon_0, A, V,$ and $v$ are constants, we get $\frac{dU}{dt} \propto x^{-2}$.Comparing this with $x^\alpha$, we find $\alpha = -2$.

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