Question

At what separation should two equal charges, 1.0 C each, be placed, so that the force between them equals the weight of a 50 kg person?

Answer 1

Given:
Magnitude of charges, q₁ = q₂ = 1 C
Electrostatic force between them, F = Weight of a 50 kg person = mg = 50 × 9.8 = 490 N
Let the required distance be r.
By Coulomb's Law, electrostatic force,
$$\begin{gathered} F=\frac{1}{4\mathrm{\pi }{\epsilon }_0}\frac{q_1{q}_2}{r^2} \\ \Rightarrow 490=\frac{9\times {10}^9\times 1\times 1}{r^2} \\ \Rightarrow r^2=\frac{9\times {10}^9}{490} \end{gathered}$$

$\Rightarrow r=\sqrt{\frac{9}{49}\times {10}^8}=\frac{3}{7}\times {10}^4\mathrm{m}=4.3\times {10}^3\mathrm{m}$