JEE Main 2025 Jan 24 Shift 1, Physics Q1: Energy Stored In A Capacitor

JEE Main2025Jan 24, Shift 1Physics

Consider a parallel plate capacitor of area $A$ (of each plate) and separation ' $d$ ' between the plates. If $E$ is the electric field and $ε_{0}$ is the permittivity of free space between the plates, then potential energy stored in the capacitor is :-

A
$\frac{1}{2} ε_{0} E^{2} A d$
B
$\frac{3}{4} ε_{0} E^{2} A d$
C
$\frac{1}{4} ε_{0} E^{2} A d$
D
$ε_{0} E^{2} A d$

Solution

The energy density (energy per unit volume) of an electrostatic field is

$u = \frac{1}{2} ε_{0} E^{2}$

The field is uniform between the plates and confined to the volume $V = A d$ , so the total stored energy is

$U = u \cdot V = \frac{1}{2} ε_{0} E^{2} (A d)$

$∴$ Option (1).

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JEE Main 2025 Jan 24 Shift 1, Physics Q1: Energy Stored In A Capacitor

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