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(A) The electrostatic energy $U$ stored in a parallel plate capacitor connected to a battery of voltage $V$ is given by $U = \frac{1}{2}CV^2$.Since th

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$x^{-2}$

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(A) The electrostatic energy $U$ stored in a parallel plate capacitor connected to a battery of voltage $V$ is given by $U = \frac{1}{2}CV^2$.Since the capacitance $C = \frac{\varepsilon_0 A}{x}$,we have $U = \frac{1}{2} \left( \frac{\varepsilon_0 A}{x} \right) V^2$.To find the time rate of change of energy,we differentiate $U$ with respect to time $t$:$\frac{dU}{dt} = \frac{d}{dt} \left( \frac{\varepsilon_0 A V^2}{2x} \right) = \frac{\varepsilon_0 A V^2}{2} \frac{d}{dt} (x^{-1})$.Using the chain rule,$\frac{dU}{dt} = \frac{\varepsilon_0 A V^2}{2} (-x^{-2}) \frac{dx}{dt}$.Since the plates are pulled apart at a uniform speed,$\frac{dx}{dt} = v$ (a constant).Therefore,$\frac{dU}{dt} = -\frac{\varepsilon_0 A V^2 v}{2} x^{-2}$.This shows that $\frac{dU}{dt} \propto x^{-2}$.

$A$ parallel plate capacitor is connected to a battery. The plates are pulled apart with a uniform speed. If $x$ is the separation between the plates,the time rate of change of electrostatic energy of the capacitor is proportional to

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