An electron is moving round the nucleus of a hydrogen atom in a circular orbit of radius $$r.$$ The coulomb force $$F$$ between the two is
(where, $$k = \frac{1}{{4\pi {\varepsilon _0}}}$$  )

Solution :
Let charges on an electron and hydrogen nucleus be $${q_1}$$ and $${q_2}.$$ Then, Coulomb's force between them at a distance $$r$$ is, $$F = - \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{q_1}{q_2}}}{{{r^2}}}\hat r$$
Putting, $$\frac{1}{{4\pi {\varepsilon _0}}} = k\,\,\left( {{\text{given}}} \right)$$
$$F = - k\frac{{{q_1}{q_2}}}{{{r^2}}}\hat r$$
Since, the nucleus of hydrogen atom has one proton, so charge on nucleus is $$e$$ i.e. $${q_2} = + e$$
also $${q_1} = e$$  for electron
So, $$F = - k\frac{{e.e}}{{{r^2}}}\hat r = - k\frac{{{e^2}}}{{{r^2}}}\hat r$$
but $$\hat r = \frac{r}{{\left| r \right|}} = \frac{r}{r}.$$
Hence, $$F = - k\frac{{{e^2}}}{{{r^2}}}.\frac{r}{r} = - k\frac{{{e^2}}}{{{r^3}}}.r$$