Question

A parallel plate capacitor has a uniform electric field E in the space between the plates. If the distance between the plates is d and the area of each plate is A, the energy stored in the capacitor is:
1. \(\frac{1}{2}{\varepsilon _0}{E^2}\)
2. E²Ad / ε₀
3. \(\frac{1}{2}{\varepsilon _0}{E^2}Ad\)
4. Ε₀EAd

Answer 1

Correct Answer - Option 3 : \(\frac{1}{2}{\varepsilon _0}{E^2}Ad\)

CONCEPT:

• Capacitor: It is a device that stores electrical energy in an electric field. It is a passive electronic component with two terminals.
  • The effect of a capacitor is known as capacitance.

The capacitance of the parallel plate capacitor :

C = ϵ₀A/d......(1)

where, ϵ₀ = permittivity of air, A = area of crossection, d = distance between plates.

The potential difference between the plates :

E = Vd.......(2)

Energy stored in capacitor:

Total energy(U) = 1/2 CV² 

where, C = capacitance, V = potential diffrence.

EXPLANATION:

From the above discussion its clear that

\(\begin{array}{l} U = \frac{1}{2}C{V^2} = \frac{1}{2}\left( {\frac{{{\varepsilon _0}A}}{d}} \right)\left( {E{d^2}} \right)\\ = \frac{1}{2}{\varepsilon _0}{E^2}Ad \end{array}\)

The correct option is 3.