Two point charged bodies interact with a force equal to 1.2mN. what will the force of interaction

Two point charged bodies interact with a force equal to 1.2mN. what will the force of interaction of bodies become equal if the distance between them is increased by 2 times, and the charge of one of the bodies is increased by 5 times?

Given:

F1 = 1.2 mN – force of interaction between two charges;

q1 is the value of the first charge;

q3 = 5 * q2 – the value of the second charge was increased 5 times;

R2 = 2 * R1 – the distance between charges was doubled;

k is the Coulomb constant.

It is required to determine the force of interaction of charges F2 (mN) under the changed conditions.

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

F1 = k * q1 * q2 / R1 ^ 2.

The force of interaction in the second case will be equal to:

F2 = k * q1 * q3 / R2 ^ 2 = k * q1 * 5 * q2 / (2 * R1) ^ 2 = 5 * k * q1 * q2 / (4 * R1 ^ 2).

Then:

F1 / F2 = (k * q1 * q2 / R1 ^ 2) / (5 * k * q1 * q2 / (4 * R1 ^ 2)) = 4/5, hence:

F2 = 5 * F1 / 4 = 5 * 1.2 / 4 = 6/4 = 1.5 mN.

Answer: the force of interaction between charges will be equal to 1.5 mN (1.5 * 10 ^ -3 Newton).