Question

State and explain Coulomb's law in electrostatics.

Answer 1

Hint: In presence of two or more charges there is a Force of attraction between different types of charges and repulsion between the same type of charge. Then this force is directly proportional to strength of charges and inversely proportional to the square of distance between them.

Complete step-by-step answer:
Coulomb's Law:
The Electrostatic Force of interaction between two static point electric charges is directly proportional to the product of the charges, inversely proportional to the source of the distance between them and acts along the straight line joining the two charges.
If two point charges $$q_1$$ and $$q_2$$ separated by a distance r.
Let F be the electrostatic force between these two Charges. According to coulomb's law
$$F\alpha q_1{q}_2\text{ and }F\alpha {\displaystyle \frac{1}{r^2}}$$
$$F_e={\displaystyle \frac{K{q}_1{q}_2}{r^2}}$$.
Where K = coulomb's constant or electrostatic force constant.
r = Distance between both charges.
$$K={\displaystyle \frac{1}{4\pi E_0}}=9\times {10}^9{\displaystyle \frac{N{M}^2}{C^2}}$$
$$E_0=\text{ Permittivity of medium}$$
Vector Form:
$$\overrightarrow{F_{12}}=\text{Force on }{q}_1\text{ due to }{q}_2={\displaystyle \frac{K{q}_1{q}_2}{r^2}}\overrightarrow{r_{21}}$$
$$\overrightarrow{F_{21}}=\text{Force on }{q}_2\text{ due to }{q}_1={\displaystyle \frac{K{q}_1{q}_2}{r^2}}\overrightarrow{r_{12}}$$
$$\overrightarrow{r_{12}}=\text{ unit vector from }{q}_1\text{ to }{q}_2$$
$$\overrightarrow{r_{21}}=\text{ unit vector from }{q}_2\text{ to }{q}_1$$

Note: The law is based on physical observation and is not logically derivable from any other concept. Experiments till today reveal its universal Nature.
The Force is conservative.
Also coulomb's law in terms of position vector
$$\overrightarrow{F_{12}}={\displaystyle \frac{K{q}_1{q}_2}{{|\overrightarrow{r_1}-\overrightarrow{r_2}|}^3}}(\overrightarrow{r_1}-\overrightarrow{r_2})$$