Consider a parallel plate capacitor of area A (of each plate) and separation 'd' between the plates. If $E$…

Consider a parallel plate capacitor of area A (of each plate) and separation 'd' between the plates. If $E$ is the electric field and $\varepsilon_0$ is the permittivity of free space between the plates, then potential energy stored in the capacitor is
  1. $\varepsilon_0 \mathrm{E}^2 \mathrm{Ad}$
  2. $\frac{1}{2} \varepsilon_0 \mathrm{E}^2 \mathrm{Ad}$
  3. $\frac{1}{4} \varepsilon_0 \mathrm{E}^2 \mathrm{Ad}$
  4. $\frac{3}{4} \varepsilon_0 \mathrm{E}^2 \mathrm{Ad}$

Solution

We know energy density
$\rho_{\mathrm{av}}=\frac{1}{2} \varepsilon_0 E^2$
So potential energy $=\rho_{a v} \times$ Volume
$\Rightarrow \text { P.E. }=\frac{1}{2} \varepsilon_0 E^2 A d$

Asked in: JEE Main 2025 (24 Jan Shift 1)

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