Q 12-02-059JEE MainJEE Main 2025 (28 Jan, Shift 2)Easy
A parallel plate capacitor of capacitance $1\ \mu\text{F}$ is charged to a potential difference of $20\ \text{V}$. The distance between the plates is $1\ \mu\text{m}$. The energy density between the plates of the capacitor is
Answer: (C) $1.8\times10^{3}\ \text{J/m}^3$
$E = \dfrac{V}{d} = \dfrac{20}{10^{-6}} = 2\times10^7\ \text{V/m}$.
$$u = \frac{1}{2}\varepsilon_0E^2 = \frac{1}{2}\times8.85\times10^{-12}\times4\times10^{14} \approx 1.8\times10^3\ \text{J/m}^3$$
Solution by Sreeraj P, M.Sc Physics