Gauss Law:

The total lines of force(electric flux) passing through a surface is eqyal to 1/Eo times the total charge enclosed by that surface.
φ = 1/Eo * total charge

Application og gauss law
1) To find electric intnensity due to charged sphere
Let us consdier a charged sphere of radius E and p be a point at distance r from the centre of sphere
CASE I when P lies outside the sphere
To find the intensity at { drae a concentric sphere of radius r such that it will enclose charge on given sphere. Let it be q then according to gauss law,
Elelctrc flux, φ = 1/Eo * q
E *A = 1/Eo *q
E = 1/4πr^2 * q/Eo
E = 1/4πE0 *q/r^2

CASE II P lies on surface of sphere
IN this r = R
E = 1/4πE0 *q/R^2

CASE III P lies inside the sphere
In this case the total charge enclosed by a gaussian surface draws through P(sphere) will be zero.
φ = 1/Eo * q = 0
E *A =0
E = 0

II) Intensity due to a line charge
Let us consider a infinitely long conductor which contains the charge as linear charge density λ. Let p be a point at a distance r from the conductor where electric intensity is to be found. To find this let us draw a cylinder through P co-axial with the conductor of length L.
Then. from Gauss law, flux through this cylinder,
φ = 1/Eo * total charge
φ = 1/Eo * λ
E * A = 1/Eo λ L
E * 2πrl = 1/Eo λ l
E = λ/2πEor

III) Intensity due to charged plane sheet
Let us consider a plane charged sheet having surface charge density E. Let P be a point at distance r from plane sheet where intensity is to be found. To find the intensity, let us draw a cylinder enclosing P, perpendicular to plane sheet having surface area A. The form gauss law
φ = 1/Eo * total charge
φ = 1/Eo * 6A
E *A = 1/Eo * 6A
E = 6/Eo