**Gauss’s Law**

According to Gauss’s law, for charges in vacuum, the total electric flux over a closed surface (Gaussian surface) is equal to ( ), where q_{net} is the total charge enclosed by the Gaussian surface. Mathematically we write this as,

=

The area vector is drawn outward and normal to the surface. The left hand side of the equation is the flux through all surfaces.

**Proof of Gauss’s law (assuming Coulomb’s law)**

**Flux due to an internal charge**

Suppose a charge q is placed at a point O inside a “closed” surface (figure). Takes a point P on the surface and consider a small area ΔS on the surface around P. Let OP=r. The electric field at P due to the charge q is around P. Let OP=r. The electric field at P due to the charge q is

E=

Along the line OP. Suppose this line OP makes an angle α with the outward normal to ΔS. The flux of the electric field through ΔS is

Δɸ=Ḗ.ΔṤ = EΔṤ cos α ΔS cos α = ΔΏ

Where ΔΏ= is the solid angle subtended by ΔS at O. The flux through the entire surface is

ɸ= ΔΏ=

The sum over ΔΏ is the total solid angle subtended by the closed surface at the internal point O and hence is equal to 4π.

The total flux of the electric filed due to the internal charge q through the closed surface is therefore,

ɸ 4π=

**Application of Gauss’s law**

(A) Electric field due to a uniformly charged sphere

(B) Electric field due to linear charge distribution

(C) Electric field due to a plane sheet of charge