State and prove Gauss's law in electrostatics.

Explanation

Gauss.s Theorem. It states that the total electric flux through any closed surface is equal to `(1)/(epsi_(0))` times the total electric charge enclosed by the surface.
Mathematically,
`phi=ointvec(E).dvec(S)=(q)/(epsi_(0))`
Proof. Consider a closed spherical surface S and a point charge +q be placed at its centre O as shown in the figure. consider a small element of area dS around point P through which the electric flux `dphi` is given by
`dphi=vec(E),dvecS`

Total electric flux linked whole closed surface S will be
`phi=ointdphi=ointvecE.dvecS`
or `phi=ointEdScos0^(@)=ointEdS` . . (i)
The electric field intensity E at P due to charge q at O is given by
`E=(1)/(4piepsi_(0))(q)/(r^(2))` . . (ii)
Putting Eq. (ii) in Eq. (i), we get
`phi=oint(1)/(4piepsi_(0))(q)/(r^(2))dS`
`=(1)/(4piepsi_(0)r^(2))ointdS`
`=(1)/(4piepsi_(0)r^(2))4pir^(2)`
or `phi=(q)/(epsi_(0))`
This is Gauss theorem.
Coulomb.s law from Gauss Theorem
Consider an isolated positive charge +q at O. imagine a sphere of radius r and centre O.
Electric field `vecE` at every point on the sphere is the same and is directed outwards. direction of small area element `vec(dS)` is also along `vecE` i.e., the angle between `vec(E)`
and `vec(dS)` is `0^(@)`,
According to Gauss theorem,
`oint_(S)vecE.vec(dS)=(q)/(epsi_(0))`

or `oint_(S)EdScos0^(@)=(q)/(epsi_(0))`
or `Eoint_(S)dS=(q)/(epsi_(0))`
or `Exx4pir^(2)=(q)/(epsi_(0))`
or, `E=(q)/(4piepsi_(0)r^(2))`
If another charge `q_(0)` is placed on the sphere then force F on it is
`F=q_(0)E=(q_(0)q)/(4piepsi_(0)r^(2))`
which is coulomb.s law.