Capacitance | JEE Main 2025 Solution

JEE MAIN 2025

28-01-2025 SHIFT-2

Question

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.

Select the correct option:

$2 \times 10^{-4} \mathrm{~J} / \mathrm{m}^{3}$

$2 \times 10^{2} \mathrm{~J} / \mathrm{m}^{3}$

$1.8 \times 10^{3} \mathrm{~J} / \mathrm{m}^{3}$

$1.8 \times 10^{5} \mathrm{~J} / \mathrm{m}^{3}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

$\mathrm{C}=1 \mu \mathrm{~F} ; \mathrm{V}=20 \mathrm{~V}$
$\mathrm{d}=1 \mu \mathrm{~m}$
Energy density $==\frac{1}{2} \epsilon_{0} E^{2}$
$\mathrm{E}=\frac{\mathrm{V}}{\mathrm{d}}=20 \times 10^{6} \mathrm{v} / \mathrm{m} ; \mathrm{U}=1.77 \times 10^{3} \mathrm{~J} / \mathrm{m}^{3}$

Question Tags

JEE Main

Physics

Easy

Start Preparing for JEE with Competishun

Video Solution

BY competishun

Watch Solution

YouTube

Report Issue