Question
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 $\_\_\_\_$ .
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Solution
Sol. [Key Concept: With the battery fixing $V$, $U = \frac{1}{2}\frac{\varepsilon_{0}A}{x}V^{2}$, so $\frac{dU}{dt} \propto \frac{1}{x^{2}}$ and $\alpha = - 2$.]
Capacitance $= C = \frac{\varepsilon_{0}\text{ }A}{x}$
Energy stored $= U = \frac{1}{2}{Cv}^{2}$
$${U = \frac{1}{2}\frac{\varepsilon_{0}\text{ }A}{x}v^{2} }{\frac{dU}{dt} = \frac{1}{2}\varepsilon_{0}{Av}^{2}\left( - \frac{1}{x^{2}} \right)\frac{dx}{dt} }{\frac{dU}{dt} \propto \frac{1}{x^{2}} }$$
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