A parallel plate capacitor of capacitance $1 \mu \mathrm{~F}$ is charged to a potential difference of 20 V.…

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.
  1. $2 \times 10^{-4} \mathrm{~J} / \mathrm{m}^3$
  2. $1.8 \times 10^5 \mathrm{~J} / \mathrm{m}^3$
  3. $1.8 \times 10^3 \mathrm{~J} / \mathrm{m}^3$
  4. $2 \times 10^2 \mathrm{~J} / \mathrm{m}^3$

Solution

$\begin{aligned} \text { Energy density } & =\frac{1}{2} \varepsilon_0 E^2 \\ & =\frac{1}{2} \varepsilon_0\left(\frac{V}{d}\right)^2 \\ & =\frac{1}{2}\left(8.85 \times 10^{-12}\right)\left(\frac{20}{10^{-6}}\right)^2 \mathrm{~J} / \mathrm{m}^3 \\ & \simeq 1.8 \times 10^3 \mathrm{~J} / \mathrm{m}^3\end{aligned}$

Asked in: JEE Main 2025 (28 Jan Shift 2)

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