Question

Prove gauss's law and coulomb's law of electrostatics are equivalent

Answer 1

Answer:

See below

Explanation:

Gauss's law in electrostatics states that the total electrical flux through a closed surface is numerically equal to 1/ε₀ times the charge (q) enclosed by that surface. That is ,

$\implies\displaystyle \phi_E =\oint \vec{E}.\vec{ds} =\dfrac{q}{\epsilon_0}$

Gauss's law from Coulomb's Law :-

From Gauss's law,

$\implies \displaystyle \phi_E =\oint \vec{E}.\vec{ds}$

$\implies \displaystyle \phi_E = \oint E \ ds \ \cos\theta$

Electric field produced by point charge q at a distance of r from is is given by,

$\implies \displaystyle \vec{E}=\dfrac{1}{4\pi\epsilon_0}\dfrac{q}{r^2}$

On substituting, we have;

$\implies \displaystyle \phi_E = \oint \dfrac{1}{4\pi\epsilon_0} \dfrac{q}{r^2} ds\cos\theta$

$\implies \displaystyle \phi_E = \dfrac{q}{4\pi \epsilon_0}\oint \dfrac{ds\cos\theta}{r^2}$

$\displaystyle \implies \phi_E = \dfrac{q}{4\pi\epsilon_0}\oint d\omega$

Total solid angle substended by a closed surface at a point enclosed by it is numerically equal to 4π .

That is ,

$\displaystyle \implies \omega =\int d\omega = \int \dfrac{ds\cos\theta}{r^2}= \boxed{4\pi}$

From this , we have;

$\displaystyle \implies \phi_E = \dfrac{q}{4\pi\epsilon_0}\times 4\pi$

$\displaystyle \implies \underline{\underline{\green{ \phi_E = \dfrac{q}{\epsilon_0}}}}$

Hence Gauss's law and Coulomb's Law are equivalent.

and we are done!