Capacitors

JEE Main level, single correct, medium

124 of 180

Question 124 of 180

A parallel plate capacitor of capacitance 1μ F1 \mu \mathrm{~F} is charged to a potential difference of 20 V . The distance between plates is 1μ m1 \mu \mathrm{~m}. The energy density between plates of capacitor is.

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Answer: (C) 1.8×103 J/m31.8 \times 10^3 \mathrm{~J} / \mathrm{m}^3

 Energy density =12ε0E2=12ε0(Vd)2=12(8.85×10−12)(2010−6)2 J/m3≃1.8×103 J/m3\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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