Physics · Capacitors and R-C Circuits

JEE Main 2025 — 24 January, Morning Shift — Question 47

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 :-

Option A:
$\frac{1}{2} \varepsilon_{0} E^{2} \mathrm{Ad}$
Correct
Option B:
$\frac{3}{4} \varepsilon_{0} \mathrm{E}^{2} \mathrm{Ad}$
Option C:
$\frac{1}{4} \varepsilon_{0} E^{2} \mathrm{Ad}$
Option D:
$\varepsilon_{0} E^{2} A d$

Answer: A

Step-by-step solution

$\frac{U}{V}=\frac{1}{2} \epsilon_{0} E^{2}$

$\mathrm{U}=\frac{1}{2} \epsilon_{0} \mathrm{E}^{2} \mathrm{~V}$

$=\frac{1}{2} \in_{0} \mathrm{E}^{2}(\mathrm{Ad})$

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Exam: JEE Main 2025
Paper: 24 January, Morning Shift
Subject: Physics
Chapter: Capacitors and R-C Circuits
Topic: Force Between Plates and Potential Energy Stored

A parallel plate capacitor was made with two rectangular plates, each with a length of l=3 cm and breath of b=1 cm . The distance between…

Answer: A

Step-by-step solution

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