Question

Why we use gauss law to find electric field intensity in its applications?

Answer 1

Gauss's Law

Statement

The total dielectric flux passing through a closed surface in vacuum enclosing a charge is 1ϵo1ϵo times the charge enclosed by the closed surface.

Mathematically,

ϕ=1ϵo×charge enclosed (q)ϕ=1ϵo×charge enclosed (q)

ϕ=qϵoϕ=qϵo

where,ϵoϵo is permittivity of free space.

In medium,

ϕ=qϵϕ=qϵ

To verify Gauss theorem suppose a point charge placed at O in vacuum show that the electric field intensity at point 'P' that lies at distance'r' from the charge 'q' is

E=14πϵo.qr2E=14πϵo.qr2

When a spherical surface passing through point'P' is constructed that has the centre 'O' and radius 'r' at every pointon its surface electric field intensity has the same value. If ′ϕ′′ϕ′ represents the electric flux passing through the sphere and 'A' its area then

ϕ=EAϕ=EA

=14πϵo.qr24πr2=14πϵo.qr24πr2

=qϵo=qϵo

∴ϕ=qϵo∴ϕ=qϵo

Application of Gauss's Theorem