Question

State and derive formula for Gauss's law....​

Answer 1

In physics, Gauss's law, also known as Gauss's flux theorem, is a law relating the distribution of electric charge to the resulting electric field. The surface under consideration may be a closed one enclosing a volume such as a spherical surface.

Answer 2

$\bf{\blue{\underline{Deriving\;Gauss's\;law\;from\;coulomb's\;law:-}}}$

Coulomb's law states that the electric field due to a stationary point change is:

$\bf{E(r)=\frac{q}{4\pi \epsilon_{0}}\frac{e_{r}}{r^{2}}}$

Where,

er is the radial unit vector,

r is a radius, |r|,

∈ 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

$\bf{E(r)=\frac{1}{4\pi \epsilon_{0}}\int\frac{\rho(s)\;(r-s)}{|r-s|^{3}}d^{3}s}$

Where rho is the charge density. It we take the divergence of both sides of this equation with respect to r, and use the known theorem;

$\bf{\Delta\;.\;\bigg(\frac{s}{|s|^{3}}\bigg)=4\pi \delta(s)}$

Where $\bf{\delta(s)}$ is the Dirac delta function, the result is,

$\bf{\Delta\;.\;E(r)=\frac{1}{\epsilon_{0}} \int \rho(s)\;\delta(r-s)d^{3}s}$

Using the "sifting property" of the Dirac delta function, we arrive at:

$\bf{\Delta\;.\;E(r)=\frac{\rho(r)}{\epsilon_{0}},\;which\;is\;differential\;form\;of\;Gauss's\;law,\;as\;desired.}$