Answer to Question #130521 in Physics for Anurag parihar

Question

Putting two equal and equal charges at a distance of 3 meters, the force of repulsion of a 1.6 N between them, then what will be the value of each charge

Answer 1

According to Coulomb's law, we have

$$F=\frac{1}{4\pi\epsilon_0}\frac{q\cdot q}{r^2}=\frac{1}{4\pi\epsilon_0}\frac{q^2}{r^2},\\\space\\ q=2r\sqrt{\pi\epsilon_0F},\\ q=2\cdot3\sqrt{3.14\cdot8.85\cdot10^{-12}\cdot1.6}=40\cdot10^{-6}\text{ C}.$$

The charges are 40 μC.