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