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 :-

(A) $\frac{1}{2} \varepsilon_{0} \mathrm{E}^{2} \mathrm{Ad}$

(B) $\frac{3}{4} \varepsilon_{0} \mathrm{E}^{2} \mathrm{Ad}$

(C) $\frac{1}{4} \varepsilon_{0} \mathrm{E}^{2} \mathrm{Ad}$

(D) $\varepsilon_{0} \mathrm{E}^{2} \mathrm{Ad}$

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · 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

ε_{0}

is the permittivity of free space between the plates, then potential energy stored in the capacitor is :-

(A) $\frac{1}{2} ε_{0} E^{2} Ad$ ✓ Correct
(B) $\frac{3}{4} ε_{0} E^{2} Ad$
(C) $\frac{1}{4} ε_{0} E^{2} Ad$
(D) $ε_{0} E^{2} Ad$

Answer: (A)

Solution

The energy density of a uniform electric field is

u = \frac{1}{2} ε_{0} E^{2}

In a parallel plate capacitor this field exists uniformly through the volume between the plates,

V = A d

. The total stored energy is this density times the volume:

U = u \cdot V = \frac{1}{2} ε_{0} E^{2} A d

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Single correct option

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