Question

Two charged particles are placed 1.0 cm apart. What is the minimum possible magnitude of the electric force acting on each charge?

Answer 1

We know that minimum charge on a body (q) = charge on an electron
⇒ q = 1.6 × 10⁻¹⁹ C
Given:
Separation between the charges, r = 1 cm = 10^-2 m
By Coulomb's Law, force,
$F=\frac{1}{4\mathrm{\pi }{\epsilon }_0}\frac{q_1{q}_2}{r^2}$
$$\begin{gathered} \Rightarrow F=\frac{9\times {10}^9\times {\left(1.6\times {10}^{-19}\right)}^2}{{\left({10}^{-2}\right)}^2} \\ \Rightarrow F=2.3\times {10}^{-24}\mathrm{N} \end{gathered}$$