Gauss theorem

Gauss theorem :- According to it the electric flux passing through a close surface is 1/ε0 multiplier of charge present on that surface.
Let Q charge present on a close surface then it's electric flux
ɸ = Q/ε0
It is only applicable for close surface.

Verification of Gauss theorem :- Let an unit positive charge +q is present at any point and we will find electric field at 'r' distance from it. So imagine a spherical close surface (Gaussian surface) of radius 'r' around that charge

Assume an unit area dA at surface of Gaussian surface
so electric field due to +q at dA
E = kQ/r^2 ....(1)
according to definition of electric flux
dɸ  = EdA Cosθ
here E ∥ dA (∥ means parallel)
so θ = 0 , Cos0 = 1
dɸ = E.dA    ....(2)
putting value of E from equation (1) into (2)
dɸ = (kQ/r^2)dA   ....(3)
Integrating both sides of equation (3)
∫dɸ = ∫(kQ/r^2)dA
since ∫dɸ = ɸ
ɸ  = kQ/r^2∫dA
since ∫dA = A
here kQ / r^2 is constant with respect to A.
so ɸ = kQ/r^2.A
here A is surface area of spherical surface = 4πr^2
& k = 1/(4πε0)
so ɸ = 1/(4πε0)[Q/r^2]4πr^2
ɸ  = Q/ε0   ....(4) Hence Proved.
It is the statement of Gauss Law.