Question

Verify Gauss's law from Coulomb's law.

Answer 1

Derivation of Gauss's Theorem by using Coulomb's law.

According to definition of electric flux,

${\varphi }_E={\oint }_S\overrightarrow{E}\overrightarrow{ds}={\oint }_{s}Edscos\theta$

The intensity of electric field $\left|\overrightarrow{E}\right|$ at same distance from charge $q$ remains constant and for spherical surface $\theta =0^o$.

Therefore ${\varphi }_E=E{\oint }_{s}dscos0=E{\oint }_{s}ds$

or ${\varphi }_E=E4{\varphi }^2....\left(1\right)$

Because ${\oint }_{s}ds=4\pi r^2$

According to Coulomb's law.intensity of electric field at distance $r$ from charge $q$.

$E=\frac{1}{4\pi {\epsilon }_0}=\frac{q}{r^2}$

$\therefore$ From equation (1).

${\varphi }_E=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}\times 4\pi r^2$

or ${\varphi }_E=\frac{q}{{\epsilon }_0}$

or ${\varphi }_E=\frac{1}{{\epsilon }_0}\left(q\right)$

or $\varphi -E=\frac{1}{{\epsilon }_0}\times$ (enclosed charge)

This is Gauss's theorem.