Obtain Gauss's law from the flux associated with a sphere of radius r and charge 'q' at centre.

Explanation

Let us consider the total fiux through a sphere of radius r, which encloses a point charge q at its centre.

Divide the sphere into small area elements, as shown in figure.
The flux through an area element `DeltavecS` is,
`Deltaphi = vecE.DeltavecS = vecE.hatr.DeltaS`
where we have used Coulomb.s law for the electric field due to a single charge q.
The unit vector `hatr` is along the radius vector from the centre to the area element.
The area element `DeltavecS` and `hatr` have the same direction,
`therefore Deltaphi =q/(4pi epsilon_(0)r^(2))DeltaS = (kq)/r^(2) DeltaS (therefore k = 1/(4piepsilon_(0)))`
The total flux through the sphere is obtained by adding up flux through all the different area elements.
`therefore phi = sum_(DeltaS)(kq)/r^(2).DeltaS`
`therefore phi = (kq)/r^(2) sum_(DeltaS).DeltaS = q/(4pi epsilon_(0)r^(2)) S (therefore sumDeltaS=S)`
`therefore phi = q/(4pi epsilon_(0)r^(2)) xx 4pir^(2) = q/epsilon_(0)` (`therefore S = 4pir^(2)` for sphere)
Gauss.s law: Electric flux through a close surface S.
`phi = sum q/epsilon_(0)`
`sumq` = total charge enclosed by S.
The electric flux associated with any close surface is equal to the ratio of net charg, enclosed by surface to `epsilon_(0)`
`therefore phi = int_(S) vecE.dvecS = (sumq)/epsilon_(0)`