JEE Mains · Physics · STD 12 - Electric potential and capacitance
- JEE Mains 2025
- Easy
- MCQ
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 _________.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(1.8 \times 10^3 \mathrm{~J} / \mathrm{m}^3\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} \text { Energy density } & =\frac{1}{2} \varepsilon_0 E^2 \\ & =\frac{1}{2} \varepsilon_0\left(\frac{V}{d}\right)^2 \\ & =\frac{1}{2}\left(8.85 \times 10^{-12}\right)\left(\frac{20}{10^{-6}}\right)^2 \mathrm{~J} / \mathrm{m}^3 \\ & \simeq 1.8 \times 10^3 \mathrm{~J} / \mathrm{m}^3\end{aligned}\)
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