JEEPhysics

Consider a parallel plate capacitor of area A (of each plate) and separation '𝑑' between the plates. If 𝐸 is the electri…

Q1 FREE PREVIEW
PYQ

Consider a parallel plate capacitor of area A (of each plate) and separation '𝑑' between the plates. If 𝐸 is the electric field and \({\epsilon }_{0}\) is the permittivity of free space between the plates, then potential energy stored in the capacitor is

[JEE Main 2025, 24 Jan (Shift 1)]

a

\(\frac{1}{4}{\epsilon }_{0}{E}^{2}Ad\)

b

\({\epsilon }_{0}{E}^{2}Ad\)

c

\(\frac{3}{4}{\epsilon }_{0}{E}^{2}Ad\)

d

\(\frac{1}{2}{\epsilon }_{0}{E}^{2}Ad\)

✓ Correct answer: d)

\(\frac{1}{2}{\epsilon }_{0}{E}^{2}Ad\)

Explanation

Energy per unit volume is given by:

\(\frac{U}{V}=\frac{1}{2}{\epsilon }_{0}{E}^{2}\)

Since volume V of the capacitor is \(A\times d\), the total energy stored in the capacitor is:

\(U=\frac{1}{2}{\epsilon }_{0}{E}^{2}\times Ad\)

Practice more JEE Physics PYQs

See every question on Electrostatic Potential and Capacitance, or browse the full JEE question bank.

See all questions on Electrostatic Potential and Capacitance →