Coulomb Law From Gauss Law derivation

Gauss’s law for electrostatics is used for determination of electric fields in some problems in which the objects possess spherical symmetry, cylindrical symmetry,planar symmetry or combination of these. Let us discuss the applications of gauss law of electrostatics:

1. Electric Field Due To A Point Charge Or Coulomb’s Law From Gauss Law:-

To derive Coulomb’s Law from gauss law or to find the intensity of electric field due to a point charge +q at any point in space using Gauss’s law ,draw a Gaussian sphere of radius r at the centre of which charge +q is located (Try to make the figure yourself).

All the points on this surface are equivalent and according to the symmetric consideration the electric field E has the same magnitude at every point on the surface of the sphere and it is radially outward in direction.Therefore, for a area element dS around any point P on the Gaussian surface both E and dS are directed radially outward,that is ,the angle between E and dS is zero.Therefore,

The flux passing through the area element dS ,that is,

d φ =E.dS= EdS cos 00=EdS

Hence, the total flux through the entire Gaussian sphere is obtained as,

Φ=∫EdS

Or                                φ=E∫dS

But ∫dS is the total surface area of the sphere and is equal to 4πr2,that is,

Φ=E(4πr2)                                                                  (1)

But according to Gauss’s law for electrostatics

Φ=q/ε0 (2)

Where q is the charge enclosed within the closed surface

By comparing equation (1) and (2) ,we get

E(4πr2)=q/ε0

Or                                       E=q/4πε0r2 (3)

The equation (3) is the expression for the magnitude of the intensity of electric field E at a point,distant r from the point charge +q.

In vector form,              E=1/4πε0 q/r2 =1/4πε0qr/r3

In a second point charge q0be placed at the point at which the magnitude of E is computed ,then the magnitude of the force acting on the second charge q0would be

F=q0E

By substituting value of E from equation (3),we get

F=qoq/4πε0r2 (4)

The equation (4) represents the Coulomb’s Law and it is derived from gauss law.