Gauss’s theorem states that* the total flux through a closed surface enclosing a charge is equal to $1/ε_0$ 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 $4πr^2$.

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 $(1)$, \[Φ=\frac{q}{4πε_0r^2}×4πr^2\] \[Φ=\frac{q}{ε_0}\]

This proves the Gauss’s theorem.

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