Deduce 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 <a title="Gauss’s law for electrostatics" href="https://winnerscience.com/electromagnetic-field-theory/gauss-law-electrostatics-derivation/">Gauss’s law for electrostatics</a>

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=q_oq/4πε₀r² (4)

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