State and prove Gauss Theorem .

 

Deriving Gauss's law from Coulomb's law

Gauss's law can be derived from Coulomb's law, which states that the electric field due to a stationary point charge is:

mathbf{E}(mathbf{r}) = frac{q}{4pi epsilon_0} frac{mathbf{e_r}}{r^2}

where

er is the radial unit vector,
r is the radius, |r|,
epsilon_0 is the electric constant,
q is the charge of the particle, which is assumed to be located at the origin.

Using the expression from Coulomb's law, we get the total field at r by using an integral to sum the field at r due to the infinitesimal charge at each other point s in space, to give

mathbf{E}(mathbf{r}) = frac{1}{4piepsilon_0} int frac{rho(mathbf{s})(mathbf{r}-mathbf{s})}{|mathbf{r}-mathbf{s}|^3} ,  d^3 mathbf{s}

where ρ is the charge density. If we take the divergence of both sides of this equation with respect to r, and use the known theorem[5]

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thnxx......:)

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The total electric flux passing through a close surface area into free space is equal to 1/epsilon times the royal charge enclosed by the surface

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The total electric flux over the closed surface in free space is equal to the 1/epsilon times the total charg enclosed by that surface.
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State and prove gauss's theorem
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Total electric flux passing through a close surface is given as 1/Eo times charge enclosed inside the surface
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The total electric flux over the close surface (S) in vacuum is equal to 1/epsilon times to the total charge contain inside the close surface (S)
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