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 = kQ_{1}Q_{2} / r^{2}

[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 = kQ_{1}Q_{2} / r^{2}

=>**20 = kQ _{1}Q_{2} /
4................Eqn.1**

**Case 2:**

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

Thus, F = kQ_{1}Q_{2} / r^{2}

**F = kQ _{1}Q_{2} /
16........................Eqn.2**

Thus, Dividing Eqn.2 by Eqn.1:

F / 20 = (kQ_{1}Q_{2} / 16) /
(kQ_{1}Q_{2} / 4)

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

=> F / 20 = 0.25

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

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