Question

Two equal charges are placed at a separation of 1.0 m. What should be the magnitude of the charges, so that the force between them equals the weight of a 50 kg person?

Answer 1

Let the magnitude of each charge be q.
Separation between them, r = 1 m
Force between them, F = 50 × 9.8 = 490 N
By Coulomb's Law force,
$F=\frac{1}{4\mathrm{\pi }{\epsilon }_0}\frac{q_1{q}_2}{r^2}$
$$\begin{gathered} \Rightarrow 490=9\times {10}^9\times \frac{q^2}{1^2} \\ \Rightarrow q^2=54.4\times {10}^{-9} \\ \Rightarrow q=\sqrt{54.4\times {10}^{-9}}=23.323\times {10}^{-5}\mathrm{C} \\ \mathrm{Or}q=2.3\times {10}^{-4}\mathrm{C} \end{gathered}$$