Coulomb measured the force between two point charges and found that it 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
$$
F \equiv k \frac{\mid q_{1} q_{2}}{r^{2}}
$$
Putting this value of \(k\) in Eq. (1.1), we see that for \(q_{1}=q_{2}=1 c_{1} r=1 \mathrm{~m}, F=9 \times 10^{9}\) \(N\). That is, \(1 C\) is the charge that when placed at a distance of \(1 \mathrm{~m}\) from another charge of the same magnitude in vacuum experiences an electrical force of repulsion of magnitude \(9 \times 10^{9} \mathrm{~N}\). One coulomb is evidently too big unit to be used. In practice, in electrostatics, one uses smaller units like \(1 \mathrm{mC}\) or \(1 \mu \mathrm{C}\).
The constant \(k\) in Eq. \((1.1)\) is usually put as \(k=1 / 4 \pi \varepsilon_{0}\) for later convenience, so that Coulomb's law is written as
$$
F=\frac{1}{4 \pi \varepsilon_{0}} \frac{\left|q_{1} q_{2}\right|}{r^{2}} \quad \text { sossese } \quad(1.2)
$$
\(\varepsilon_{0}\) is called the permittivity of free space. The value of \(\varepsilon_{0}\) in SI units is \(\varepsilon_{0}=8.854 \times\) \(10^{-1} 2 C^{2} N^{-1} m^{-2}\).
Coulomb's force law between two point charges \(q_{1}\) and \(q_{2}\) located at \(r_{1}\) and \(r_{2}\) is then expressed as
$$
\mathbf{F}_{21}=\frac{1}{4 \pi \varepsilon_{a}} \frac{q_{1} q_{2}}{r_{21}^{2}} \hat{\mathbf{r}}_{21}
$$
The force \(F_{12}\) on charge \(q_{1}\) due to charge \(q_{2}\), is
$$
\mathbf{F}_{12}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q q_{2}}{r_{12}^{2}} \hat{\mathbf{r}}_{12}=-\mathbf{F}_{21}
$$
