Physics · Capacitors and R-C Circuits

JEE Main 2025 — 28 January, Evening Shift — Question 48

A parallel plate capacitor of capacitance $1 \mu \mathrm{~F}$ is charged to a potential difference of 20 V . The distance between plates is $1 \mu \mathrm{~m}$ . The energy density between plates of capacitor is :

Option A:
$1.8 \times 10^{3} \mathrm{~J} / \mathrm{m}^{3}$
Correct
Option B:
$2 \times 10^{-4} \mathrm{~J} / \mathrm{m}^{3}$
Option C:
$2 \times 10^{2} \mathrm{~J} / \mathrm{m}^{3}$
Option D:
$1.8 \times 10^{5} \mathrm{~J} / \mathrm{m}^{3}$

Answer: A

Step-by-step solution

$\mathrm{C}=1 \mu \mathrm{~F}$

$\mathrm{V}=20 \mathrm{~V}$

$\mathrm{d}=1 \mu \mathrm{~m}$

Energy density $==\frac{1}{2} \in_{0} E^{2}$

$E=\frac{V}{d}=20 \times 10^{6} v / m$

$\mathrm{U}=1.77 \times 10^{3} \mathrm{~J} / \mathrm{m}^{3}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 28 January, Evening Shift
Subject: Physics
Chapter: Capacitors and R-C Circuits
Topic: Force Between Plates and Potential Energy Stored