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JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance

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. A \(\varepsilon_0 \mathrm{E}^2 \mathrm{Ad}\)
  2. B \(\frac{1}{2} \varepsilon_0 \mathrm{E}^2 \mathrm{Ad}\)
  3. C \(\frac{1}{4} \varepsilon_0 \mathrm{E}^2 \mathrm{Ad}\)
  4. D \(\frac{3}{4} \varepsilon_0 \mathrm{E}^2 \mathrm{Ad}\)
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Answer & Solution

Correct Answer

(B) \(\frac{1}{2} \varepsilon_0 \mathrm{E}^2 \mathrm{Ad}\)

Step-by-step Solution

Detailed explanation

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