The force of electrical interaction between two charged bodies is 12mN. If the charge of one body is increased by 2 times

The force of electrical interaction between two charged bodies is 12mN. If the charge of one body is increased by 2 times, and the charge of the other body is reduced by 3 times and the distance between the bodies is reduced by 2 times, then the force of interaction between the bodies will become equal.

Given:

F1 = 12 mN – force of interaction between two charges;

q3 = 2 * q1 – the charge of one body was doubled;

q4 = q2 / 3 – the charge of another body was reduced by 3 times;

r2 = r1 / 2 – the distance between charges was reduced by 2 times.

It is required to determine F2 (mN) – the force of interaction between charges after changes.

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

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

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

F2 = k * q3 * q4 / r2 ^ 2 = k * 2 * q1 * q2 / (3 * r2 ^ 2) = 8 * k * q1 * q2 / (3 * r1 ^ 2).

Then:

F1 / F2 = (k * q1 * q2 / r1 ^ 2) / (8 * k * q1 * q2 / (3 * r1 ^ 2));

F1 / F2 = 3/8;

F2 = 8 * F1 / 3 = 8 * 12/3 = 8 * 4 = 32 mN.

Answer: the force of interaction between the charges will be equal to 32 mN.