State Coulomb's law and explain its scalar form.

(N/A) Coulomb's law states that the electrostatic force between two stationary point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. The force acts along the line joining the two charges.
If two point charges $q_{1}$ and $q_{2}$ are separated by a distance $r$ in vacuum,the magnitude of the electrostatic force $F$ is given by:
$F \propto \frac{|q_{1} q_{2}|}{r^{2}}$
$F = k \frac{|q_{1} q_{2}|}{r^{2}}$
where $k$ is the Coulomb constant,$k = \frac{1}{4 \pi \epsilon_{0}} \approx 8.9875 \times 10^{9} \text{ N m}^{2} \text{ C}^{-2}$.
Here,$\epsilon_{0}$ is the permittivity of free space,$\epsilon_{0} \approx 8.854 \times 10^{-12} \text{ C}^{2} \text{ N}^{-1} \text{ m}^{-2}$.
If the charges are placed in a medium with permittivity $\epsilon$,the force $F_{m}$ is:
$F_{m} = \frac{1}{4 \pi \epsilon} \frac{|q_{1} q_{2}|}{r^{2}}$
The relative permittivity (or dielectric constant) is defined as $K = \epsilon_{r} = \frac{\epsilon}{\epsilon_{0}}$.
Therefore,$\epsilon = K \epsilon_{0}$.
Substituting this into the expression for $F_{m}$:
$F_{m} = \frac{1}{4 \pi K \epsilon_{0}} \frac{|q_{1} q_{2}|}{r^{2}} = \frac{F}{K}$
Thus,the force in a medium is $1/K$ times the force in free space.