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 qnet 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

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