Question

Two charges are separated by 2 m and repel each other with a force of 20 N. If they are moved to a separation of 4 m, what will be the repulsive force?

Answer 1

Solution:

Let us go to the basics first.

We know that Coulomb's law states following:

F = kQ₁Q₂ / r²

[where, k = coulomb's constant; Q's are the charges; r = distance between the charges]

Case1:

Two charges are separated by 2 m (= r) and repel each other with a force of 20 N (=F).

Thus, F = kQ₁Q₂ / r²

=>20 = kQ₁Q₂ / 4................Eqn.1

Case 2:

They are moved to a separation of 4 m (= r).

Thus, F = kQ₁Q₂ / r²

F = kQ₁Q₂ / 16........................Eqn.2

Thus, Dividing Eqn.2 by Eqn.1:

F / 20 = (kQ₁Q₂ / 16) / (kQ₁Q₂ / 4)

=> F / 20 = (1/16) / (1/4)

=> F / 20 = 0.25

=> F = 0.25*20 = 5 N (Answer)