Using Gausss law and a symmetry argument Derive the electric

Using Gauss\'s law and a symmetry argument: Derive the electric field for an infinite wire. Derive the electric field for a point charge. Derive the electric field for an infinite plane.

Solution

Gauss’s law states that the flux of net electric field through a closed surface equal to the net charge enclosed by the surface divided by ₀.

Consider an infinite wire having a linear charge density C/m

Now construct a cylindrical Gaussian surface around the wire of radius r and length l

Charge enclosed inside Q_in = l

cylindrical surface area A = 2rl

Let E be the electric field at any point on the surface. By symmetry, E is normal to the surface and is uniform across the surface.

E is zero on the top and bottom surfaces of the Gaussian surface

Total flux through the surface = E*2rl

                                                 = l/ ₀ , by Gauss law.

electric filed at distance r from the wire

E(r) = /2r ₀

Let Q is point charge. Construct a spherical surface of radius r around the surface.

Let E be the electric field on the surface, E is uniform across the surface and is normal to it by symmetry.

Flux through the surface

= E*4r²

   = Q /₀ , by Gauss law, Q is the only charge enclosed inside the surface.

E = Q /4r²₀

Consider an infinite plane having a surface charge density C/sq.m

Construct a cylindrical Gaussian surface enclosing an area A on the surface and passing through the plane, the cylinder being perpendicular to the plane having a length l.

Electric field is perpendicular to the surface , flux through the cylindrical surface is 0 as the electric field is parallel to the surface.

flux through is perpendicular to the left and right flat surfaces of the cylinder total area is 2A as the cylinder is covering an area A of the charged sheet

if E is the electric field the flux through the Gaussian surface is

    = E*2A

charge enclosed in the surface

Q = *A

using Gauss law

E*2A = *A/₀

E = /2₀