JEE Main 2026 Apr 4 Shift 1, Physics Q15: Energy Stored In A Capacitor

JEE Main2026Apr 4, Shift 1Physics

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 $x^{α}$ , where $α$ is _______.

A
$- 2$
B
$1$
C
$- 1$
D
$2$

Solution

The capacitor stays connected to the battery, so $V$ is constant. Capacitance $C = \frac{ε _{0} A}{x}$ and stored energy

$U = \frac{1}{2} C V^{2} = \frac{ε _{0} A V ^{2}}{2 x} .$

$\frac{d U}{d t} = - \frac{ε _{0} A V ^{2}}{2 x ^{2}} \cdot \frac{d x}{d t} = - \frac{ε _{0} A V ^{2} v}{2 x ^{2}} .$

Hence $\frac{d U}{d t} \propto \frac{1}{x ^{2}} = x^{- 2}$ , so $α = - 2.$

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JEE Main 2026 Apr 4 Shift 1, Physics Q15: Energy Stored In A Capacitor

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