JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Capacitance

A parallel plate capacitor of capacitance

1 μ F

is charged to a potential difference of 20 V . The distance between plates is

1 μ m

. The energy density between plates of capacitor is :

Answer: (A)

Solution

The energy density stored in the electric field between the plates of a parallel plate capacitor is

u = \frac{1}{2} ϵ_{0} E^{2}

where

E

is the field between the plates.

Given

V = 20 V

and

d = 1 μ m = 1 0^{- 6} m

E = \frac{V}{d} = \frac{20}{1 0 ^{- 6}} = 2 \times 1 0^{7} V/m

Substituting into the energy density formula with

ϵ_{0} = 8.85 \times 1 0^{- 12} F/m

u = \frac{1}{2} \times 8.85 \times 1 0^{- 12} \times (2 \times 1 0^{7})^{2} = \frac{1}{2} \times 8.85 \times 1 0^{- 12} \times 4 \times 1 0^{14}

u \approx 1.8 \times 1 0^{3} J/m^{3}

JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Single correct option