Question

A charge of 1.0 C is placed at the top of your college building and another equal charge at the top of your house. Take the separation between the two charges to be 2.0 km. Find the force exerted by the charges on each other. How many times your weight is this force?

Answer 1

Given:
q₁ = q₂ = q = 1.0 C
Distance between the charges, r = 2 km = 2 × 10³ m
By Coulomb's Law, electrostatic force,
$F=\frac{1}{4\mathrm{\pi }{\epsilon }_0}\frac{q_1{q}_2}{r^2}$
$$\begin{gathered} F=9\times {10}^9\times \frac{1\times 1}{{\left(2\times {10}^3\right)}^2} \\ =2.25\times {10}^3\mathrm{N} \end{gathered}$$

Let my mass, m, be 50 kg.
Weight of my body, W = mg
⇒ W = 50 × 10 N = 500 N
Now,
$\frac{\text{W}\mathrm{eight}\mathrm{of}\text{my}\mathrm{body}}{\text{F}\mathrm{orce}\mathrm{between}\text{the}\mathrm{charges}}=\frac{500}{2.25\times {10}^3}=\frac{1}{4.5}$
So, the force between the charges is 4.5 times the weight of my body.