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Question

(a) $U = \frac{1}{2}C{V^2} = \frac{1}{2}\left( {\frac{{{\varepsilon _0}A}}{x}} \right)\,{V^2}$
 $\frac{{dU}}{{dt}} = \frac{1}{2}{\varepsilon _0}

Vedclass · Accepted Answer

$x^{-2}$

Vedclass · Answer

(a) $U = \frac{1}{2}C{V^2} = \frac{1}{2}\left( {\frac{{{\varepsilon _0}A}}{x}} \right)\,{V^2}$
 $\frac{{dU}}{{dt}} = \frac{1}{2}{\varepsilon _0}A{V^2}\left( { - \frac{1}{{{x^2}}}\frac{{dx}}{{dt}}} \right)$ $==>$ $\frac{{dU}}{{dt}} \propto {x^{ - 2}}$

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 capacitor is proportional to

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