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 0^{0}=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 4πr^{2},that is,

Φ=E(4πr^{2}) (1)

But according to Gauss’s law for electrostatics

Φ=q/ε_{0 }(2)_{ }

Where q is the charge enclosed within the closed surface

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

E(4πr^{2})=q/ε_{0}

Or E=q/4πε_{0}r^{2} (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=1/4πε_{0 }q/r^{2} =1/4πε_{0}qr/r^{3}

In a second point charge q_{0}be placed at the point at which the magnitude of E is computed ,then the magnitude of the force acting on the second charge q_{0}would be

F=q_{0}E

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

F=q_{o}q/4πε_{0}r^{2} (4)

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

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