How will the interaction force of two charged bodies change when the distance between them decreases by 3 times?

Given:

q1, q2 – two point charges;

r2 = r1 / 3 – the distance between charges was reduced by 3 times.

It is required to determine F2 / F1 – how the force of the Coulomb interaction between charges will change with decreasing distance.

The force of interaction between charges in the first case will be equal to:

F1 = k * q1 * q2 / r1 ^ 2, where k is an electrical constant.

In the second case, the force of the Coulomb interaction will be equal to:

F2 = k * q1 * q2 / r2 ^ 2 = k * q1 * q2 / (r1 / 3) ^ 2 = 9 * k * q1 * q2 / r1 ^ 2.

Then:

F2 / F1 = (9 * k * q1 * q2 / r1 ^ 2) / (k * q1 * q2 / r1 ^ 2) = 9, that is, it will increase 9 times.

Answer: when the distance between charges decreases by 3 times, the force of interaction will increase by 9 times.