Capacitors

JEE Main level, single correct, medium

125 of 180

Question 125 of 180

Consider a parallel plate capacitor of area A (of each plate) and separation 'd' between the plates. If EE is the electric field and ε0\varepsilon_0 is the permittivity of free space between the plates, then potential energy stored in the capacitor is

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Answer: (B) 12ε0E2Ad\frac{1}{2} \varepsilon_0 \mathrm{E}^2 \mathrm{Ad}

We know energy density
ρav=12ε0E2\rho_{\mathrm{av}}=\frac{1}{2} \varepsilon_0 E^2
So potential energy =ρav×=\rho_{a v} \times Volume
⇒ P.E. =12ε0E2Ad\Rightarrow \text { P.E. }=\frac{1}{2} \varepsilon_0 E^2 A d

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