Two infinite plane parallel sheets having surface charge density +σ and –σ are kept parallel to each other at a small separation distance d. The electric field at any point in the region between the plates is

CONCEPT:

Gauss’s Law:

The total electric flux through a closed surface is 1/ε_o times the charge enclosed in the surface i.e. \({\rm{\Phi }} = \frac{q}{{{\epsilon_o}}}\)

But we know that Electrical flux through a closed surface is \(\oint \vec E \cdot \overrightarrow {ds} \)

\(\therefore\oint \vec E \cdot \overrightarrow {ds} = \frac{q}{{{\epsilon_o}}}\)

Where, E = electric field, q = charge enclosed in the surface, and ε_o = permittivity of free space.

Derivation:

The electric field at a point due to an infinite sheet of charge is

\(E = \frac{\sigma }{{2{\epsilon_0}}}\)

Where

σ = surface charge density.

Here,

E1: Electric Field due to sheet having surface charge density +σ

E2: Electric Field due to sheet having surface charge density -σ

The electric field at any point in the region between the plates is

E = E₁ + E₂

\(E = \frac{\sigma }{{2{\epsilon_0}}} - \left( {\frac{{ - \sigma }}{{2{\epsilon_0}}}} \right) = \frac{{\sigma + \sigma }}{{2{\epsilon_0}}} = \frac{{2\sigma }}{{2{\epsilon_0}}} = \frac{\sigma }{{{\epsilon_0}}}\)