# Gauss’s Theorem

​Statement:
The total flux through a closed surface enclosing a charge is equal to 1/ε₀ times the magnitude of the charge enclosed.

The surface enclosing the charge is known as Gaussian surface and it can be of any shape.

​Suppose a charge +q is placed at a point O. Draw a sphere of radius r such that O is its centre. The total flux due to the charge crosses through the surface of the sphere where area is 4πr².

​If E is the electric intensity at any point on the surface of the sphere, then total flux Φ is, $Φ=EA$ $Φ=E×4πr^2 \text{ ___(1)}$ If the charge is placed in air, then the electric intensity at a point which is r distance apart from it is given by, $E=\frac{q}{4πε_0r^2}$ Putting the value of E in (1), $Φ=\frac{q}{4πε_0r^2}×4πr^2$ $Φ=\frac{q}{ε_0}$