State and prove Gauss's theorem in electrostatics.

Explanation

Gauss.s Theorem : It states that total electric flux through a closed surface is equal to `(1)/(epsilon_(0))` times the magnitude of charge enclosed by it.

Proof : Consider a point charge + q is lying at O.
Draw a sphere of radius r with centre O which acts as a gussian surface.
Consider a point P on it where electric field intensity is given by
`vec(E ) = (1)/(4 pi epsilon_(0)) (q)/(r^(2)) hat(r )`
where `hat(r )` is a unit vector which gives the direction of electric field at point P.
Now consider a small of area dS around point P and `vec(dS)` is normal to the surface of area element.
Electric flux through the area element is given by
`d phi = vec(E ). vec(dS)`
`d phi = (1)/(4 pi epsilon_(0)) (q)/(r^(2)) hat(r ) vec(dS)`
= `(1)/(4 pi epsilon_(0)) (q)/(r^(2)) . dS hat(n)`
`d phi = (1)/(4 pi epsilon_(0)) (q)/(r^(2)) dS " " [therefore hat(r ). hat(n) = 1]`
where `hat(n)` is a unit vector normal to the surface of area element and gives the direction of `vec(dS)` Now flux over the whole closed surface is given by
`int dphi = int (1)/(4 pi epsilon_(0)) (q)/(r^(2)) dS`
`phi = (1)/(4 pi epsilon_(0)) (q)/(r^(2)) int dS`
`phi = (1)/(4 pi epsilon_(0)) (q)/(r^(2)) (4pi r^(2))`
`phi = (q)/(epsilon_(0))`
which is the expression for Gauss.s theorem in electrostatic.