Alternate Solution:
Two small charged spheres carry charges \(q_1 = 2 \times 10^{-7}\,C\) and \(q_2 = 3 \times 10^{-7}\,C\). The distance between their centres is \(r = 30\,cm = 0.3\,m\).
According to Coulomb’s law, the electrostatic force between two point charges is given by
\(F = k \dfrac{q_1 q_2}{r^2}\)
where \(k = \dfrac{1}{4\pi\varepsilon_0} = 9 \times 10^9\,Nm^2C^{-2}\).
Substituting the given values:
\(F = \dfrac{9 \times 10^9 \times (2 \times 10^{-7}) \times (3 \times 10^{-7})}{(0.3)^2}\)
\(F = \dfrac{9 \times 10^9 \times 6 \times 10^{-14}}{0.09}\)
\(F = 6 \times 10^{-3}\,N\)
Since both spheres carry charges of the same sign, the force between them is repulsive.
Therefore, the magnitude of the electrostatic force between the two spheres is \(6 \times 10^{-3}\,N\), and the force is repulsive.