Write Coulomb’s law and explain its scalar form.

The electric force between two point charges varied inversely as the square of the distance between the charges and was directly proportional to the product of the magnitude of the two charges and acted along the line joining the two charges.

Thus, if two point charges $q_{1}, q_{2}$ are separated by a distance $r$ in vacuum, the magnitude of the force (F) between them is given by,

$\mathrm{F} \propto \frac{q_{1} q_{2}}{r^{2}}$

$\therefore \mathrm{F}=k \frac{q_{1} q_{2}}{r^{2}} \quad \ldots$ $(1)$

where, $k$ is the proportionality constant also called Coulombian constant.

Experimentally it is found that $k=8.9875 \times 10^{9} \mathrm{Nm}^{2} \mathrm{C}^{-2}$. In general, it is taken as $k=9 \times$ $10^{9} \mathrm{Nm}^{2} \mathrm{C}^{-2}$.

$K=\frac{1}{4 \pi \epsilon_{0}}$

where, $\epsilon_{0}=$ permittivity in free space

$=8.854185 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}$

$\approx 8.9 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}$

Hence, electric force is free space,

$\mathrm{F}=\frac{1}{4 \pi \epsilon_{0}} \frac{q_{1} q_{2}}{r^{2}}$

$\ldots$ $(2)$

If the charges are at distance ' $r$ ' in any medium instead of free space, then electric force in medium,

$\mathrm{F}_{m}=\frac{1}{4 \pi \varepsilon} \frac{q_{1} q_{2}}{r^{2}}$

where, $\varepsilon=$ permittivity in medium.

Now, relative permittivity $=\epsilon_{r}=\frac{\epsilon}{\epsilon_{0}}=k=$ dielectric constant

By taking ratio of equation (3) to (2),

$\frac{\mathrm{F}_{\mathrm{m}}}{\mathrm{F}}=\frac{\epsilon_{0}}{\epsilon}=\frac{1}{k}$

$\therefore \quad \mathrm{F}_{\mathrm{m}}=\frac{\mathrm{F}}{k}$

Thus, electric force between two charges in medium is $\frac{1}{k}$ times the electric force in free space.