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JEE Main 202524 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 24 January
Shift: Morning Shift
Subject: Physics
Chapter: Capacitance

JEE Main 2025 · 24 January · Morning Shift · Physics · Capacitance

Consider a parallel plate capacitor of area A (of each plate) and...

Consider a parallel plate capacitor of area A (of each plate) and separation 'd' between the plates. If

E

is the electric field and

\varepsilon_0

is the permittivity of free space between the plates, then potential energy stored in the capacitor is

Options

A $\varepsilon_0 \mathrm{E}^2 \mathrm{Ad}$
B $\frac{1}{2} \varepsilon_0 \mathrm{E}^2 \mathrm{Ad}$
C $\frac{1}{4} \varepsilon_0 \mathrm{E}^2 \mathrm{Ad}$
D $\frac{3}{4} \varepsilon_0 \mathrm{E}^2 \mathrm{Ad}$

Correct answer

B. $\frac{1}{2} \varepsilon_0 \mathrm{E}^2 \mathrm{Ad}$

Step-by-step solution

We know energy density

\rho_{\mathrm{av}}=\frac{1}{2} \varepsilon_0 E^2

So potential energy

=\rho_{a v} \times

Volume

\Rightarrow \text{ P.E. }=\frac{1}{2} \varepsilon_0 E^2 A d

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