Obtain Gauss law from Coulomb 's law .

Explanation

Gauss law states that if a charge Q is enclosed by an arbitrary closed surface then the total electric flux `Phi_(E)` through the closed surface is
`Phi_(E)oint vecE.dvecA= (Q_("end"))/(epsilon_(0))`
A positive point charge Q is surrounded by an imaginary sphere of radius r as shown in figure. We can calculate the total electric flux through the closed surface of the sphere using the equation .

`Phi_(E)=ointvecE .dvecA= oint EdAcos theta `
The electric field of the point charge is directed radially outward at all point on the surface of the sphere . Therefore the direction of the area element `dvecA` is along the electric field `vecE` and `theta= 0^(@)` .
`Phi_(E)=oint EdA ` since cos `0^(@)=1 `
E is uniform on the surface of the sphere
`Phi_(E) = Eoint dA` Substituting for
`oint dA = 4pir^(2)` and E `= (1)/(4piepsilon_(0))Q ` in equation 3 we get
`Phi_(E)= (1)/(4piepsilon_(0))(q)/(r^(2))xx4pir^(2)= 4pi(1)/(4piepsilon_(0))= (Q)/(epsilon_(0))`
The equation (4) is called as Gauss.s law . The remarkable point about this result is that the equation ( 4) is equally true for any arbitrary shaped which encloses the charge Q.