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2. Electric Potential and Capacitance
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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
A
$x^{-2}$
B
$x$
C
$x^{-1}$
D
$x^{2}$
Solution
(a) $U = \frac{1}{2}C{V^2} = \frac{1}{2}\left( {\frac{{{\varepsilon _0}A}}{x}} \right)\,{V^2}$
$\frac{{dU}}{{dt}} = \frac{1}{2}{\varepsilon _0}A{V^2}\left( { – \frac{1}{{{x^2}}}\frac{{dx}}{{dt}}} \right)$ $==>$ $\frac{{dU}}{{dt}} \propto {x^{ – 2}}$
Standard 12
Physics
