Consider a parallel plate capacitor of area A (of each plate) and separation ' d ' between the plates. If E is the electric field and ε₀ is the permittivity of free space between the plates, then potential energy stored in the capacitor is (a) ε₀ E² Ad (b) 3/4ε₀ E² Ad (c) 1/4ε₀ E² Ad (d) 1/2ε₀ E² Ad

Answer: (d). U = 1/2CV² with C = (ε₀ A)/(d) and V = Ed. U = 1/2·(ε₀ A)/(d)· E² d² = 1/2ε₀ E² Ad So the answer is (d) 1/2ε₀ E² Ad.

Consider a parallel plate capacitor of area A (of each plate) and… | JEE Main 2025 (24 Jan, Shift 1)

Q7JEE Main 2025 · 24 JAN · SHIFT 1Easy

Energy Stored & Force between Plates

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

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Answer: (d)

$U = \frac{1}{2}CV^2$ with $C = \frac{\varepsilon_0 A}{d}$ and $V = Ed$ .
$U = \frac{1}{2}\cdot\frac{\varepsilon_0 A}{d}\cdot E^2 d^2 = \frac{1}{2}\varepsilon_0 E^2 Ad$
So the answer is (d) $\frac{1}{2}\varepsilon_0 E^2 Ad$ .

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