**Statement: **The gauss’s theorem states that the total electric flux through any closed surface is equal to \(\frac{1}{{{\varepsilon }_{0}}}\) times the net charge enclosed by the surface.

Total electric flux = \(\frac{1}{{{\varepsilon }_{0}}}\)x Net charge

\(\phi =\frac{1}{{{\varepsilon }_{0}}}~q\).

\(\oint_{S}{{\vec{E}}}~.~d\vec{S}=\frac{1}{{{\varepsilon }_{0}}}q\).

The circle on the integral means that the surface S is closed. The net charge means the algebraic sum of the charge within the surface S.

**Proof: **Consider a charge +q placed at a point O. Let P be a point at a distance r from O. The electric intensity at P is given by\(E=~\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q}{{{r}^{2}}}\) along OP

Gauss law holds good for closed surface of any shape. For the sake of simplicity, let us consider a spherical surface with O as centre and r as its radius.

This is the Gaussian surface. By symmetry, the field of the charge +q is radial. E is perpendicular to the sphere and is directed along the normal to the surface. So the angle between the direction of E and the normal to the surface of the sphere is zero (cosθ = 1). Also symmetry requires that E has the same magnitude everywhere on this sphere.

\(\vec{E}.~d~\vec{S}=EdS\cos \theta =E~dS\).

\(=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q}{{{r}^{2}}}ds\).

\(\oint_{S}{E~dS}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q}{{{r}^{2}}}~\oint_{S}{dS}\).

\(=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q}{{{r}^{2}}}4\pi {{r}^{2}}=\frac{1}{{{\varepsilon }_{0}}}q\).

Hence the theorem.