Question

prove gauss's law
verify gauss's law​

Answer 1

auss Law

"The total normal electric flux over a closed surface in an electric field is equal to  times the total charge enclosed by that surface.”

Mathematically it may be expressed as

............(1)

Where q is the net charge enclosed by the surface and εo is the permittivity (of free space) of the medium

Proof of Gauss law of electrostatics (Integral Form):

Consider a source producing the electric field E is a point charge +q situated at a point O inside a volume enclosed by an arbitrary closed surface S. let us consider a small area element dS around a point P on the surface where the electric field produced by the charge +q is E. if E is along OP and area vector dS is along the outward drawn normal to the area element dS.

The electric field strength E at the point P is given by

.............(2)

Then the electric flux over the surface, therefore

........(3)

(Solid angle)

Equation (3) becomes

or

..........(4)

Equation (4) represents Gauss law (in integral form) for electrostatics for a single point charge (in integral form).  

Gauss law in Differential Form

Using divergence Theorem ( Relates volume integral of divergence of a vector field to surface integral of the vector field)

.....(5)

Using Equation (4)

........(6)

and Let a charge q be distributed over a volume V of the closed surface S and p be the charge density; then  

or

Substituting the value of net charge in terms of charge density, equation (6) becomes

Or

....(7)

or

....(8)

Equation (7) and (8) represent Gauss Law in differential form

Differential form of Gauss law states that "the divergence of electric field E at any point in space is equal to 1/ε0 times the volume charge density,ρ, at that point".