Deduce Coulomb.s law from Gauss. law of electrostatics.

Explanation

Coulomb.s law from Gauss.s law Consider an isolated positive point charge q placed at centre of a spherical Gaussian surface S of radius r.

By symmetry, E has same magnitude at all the points on S. At any point on S, E and d S are directed outward and the angle between them is `0^(@)`. So electric flux through small area d S is `d phi_(E)= E.dS = EdS cos 0^(@) = E dS`
Net electric flux through closed surface S is
`phi_(E )= underset(S)oint d phi_(E )= underset(S)oint E dS = E underset(S)oint dS`
`phi_(E)= E xx 4pi r^(2)` ...(i)
From Gauss.s theorem `phi_(E)= (q)/(epsi_(0))` ...(ii)
From Eqs (i) and (ii) we get
`E xx 4pi r^(2) = (q)/(epsi_(0)) rArr E= (1)/(4pi epsi_(0)) .(q)/(r^(2))`
The force on a point charge `q_(0)`, if placed on surface S will be
`F= q_(0)E rarr F= (1)/(4pi epsi_(0)).(q q_(0))/(r^(2))` ...(iii)
Eq. (iii) proves Coulomb.s law.