Question

deduce coulomb's law from gauss's law​

Answer 1

Let's consider a binary charge system separated by distance r .

Now considering one of the charges.

A spherical Gaussian Surface is imagined with the charge at the centre and the radius of that surface be r.

Let the electric field intensity through an elemental surface area be E

$\int \vec E . \: d\vec s = \dfrac{q}{ \epsilon_{0}}$

Since the angle between field vector and area vector is zero, we can say :

$= > \int E \times \: d s \times \cos(0 \degree) = \dfrac{q}{ \epsilon_{0}}$

$= > E \times \: \int d s = \dfrac{q}{ \epsilon_{0}}$

$= > E \times \: 4\pi {r}^{2} = \dfrac{q}{ \epsilon_{0}}$

$= > E = \dfrac{q}{4\pi \epsilon_{0} {r}^{2} }$

Now the force experienced by another charge at a position of r will be F

$= >F = E \times Q$

$= >F = \dfrac{qQ}{4\pi \epsilon_{0} {r}^{2} }$

$= >F = \dfrac{1}{4\pi \epsilon_{0}} \dfrac{qQ}{ {r}^{2} }$

This formula is as per Coulomb's Law