Question

State and explain coulombs law in electrostatics.

Answer 1

Statement and explanation of Coulombs law in electrostatics

Coulomb's law in electrostatic states that a charge at rest $q_1$ applies a force $F$ on the other charge $q_2$​, also at rest, which are separated by a distance $r$ such that the force is directly proportional to the product of both charges and inversely proportional to the square of the distance between them

1. $F\propto q_1{q}_2$
2. $F\propto \frac{1}{r^2}$

Thus we get ​$F\propto \frac{q_1{q}_2}{r^2}$

$\Rightarrow F=k\frac{q_1{q}_2}{r^2}$

where $k=\frac{1}{4\pi {\epsilon }_0}$

${\epsilon }_0=$Permittivity of medium

In vector form

$\overrightarrow{F_{12}}=$Force on $q_1$ due to $q_2$$=k\frac{q_1{q}_2}{r^2}\overrightarrow{r_{21}}$

$\overrightarrow{F_{21}}=$Force on $q_2$ due to $q_1$ $=k\frac{q_1{q}_2}{r^2}\overrightarrow{r_{12}}$

$\overrightarrow{r_{21}}=$Unit vector from $q_1$ to $q_2$

$\overrightarrow{r_{12}}=$Unit vector from $q_2$ to $q_1$