Question

At what distance should two charges, each equal to 1 C, be placed so that the force between them equals your weight?

Answer 1

Given: $q_1=q_2=1\mathrm{C}$

By Coulomb's law, the force of attraction between the two charges is given by
$$\begin{gathered} F=\frac{1}{4\mathrm{\pi }{\in }_0}\frac{q_1{q}_2}{r^2} \\ =\frac{9\times {10}^9\times 1\times 1}{r^2} \end{gathered}$$
However, the force of attraction is equal to the weight (F = mg).
$$\begin{gathered} \therefore mg=\frac{9\times {10}^9}{r^2} \\ \Rightarrow r^2=\frac{9\times {10}^9}{m\times 10}=\frac{9\times {10}^8}{m}(\mathrm{Taking}g=10\mathrm{m}/{\mathrm{s}}^2) \\ \Rightarrow r^2=\frac{9\times {10}^8}{m} \\ \Rightarrow r=\frac{3\times {10}^4}{\sqrt{m}} \end{gathered}$$

Assuming that m = 81 kg, we have:
$$\begin{gathered} r=\frac{3\times {10}^4}{\sqrt{81}} \\ =\frac{3}{9}\times {10}^4\mathrm{m} \\ =3333.3\mathrm{m} \end{gathered}$$

∴ The distance r is 3333.3 m.