JEE Mains · Physics · STD 12 - Electric potential and capacitance
- JEE Mains 2025
- Easy
- MCQ
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
Pick an option to check your answer.
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
\(\Rightarrow \text { P.E. }=\frac{1}{2} \varepsilon_0 E^2 A d\)
\(\rho_{\mathrm{av}}=\frac{1}{2} \varepsilon_0 E^2\)
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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