When the distance between the two charges changed, the force of their interaction

When the distance between the two charges changed, the force of their interaction decreased by 4 times. how has the distance between charges changed?

Given:

q1, q2 – values of point electric charges;

k – electrical constant;

F1 = 4 * F2 – the force of interaction between charges has decreased 16 times.

It is required to determine r2 / r1 – how the distance between charges has changed.

The distance between charges in the first case was equal to:

r1 = (F1 / (k * q * q2) ^ 0.5 = (4 * F2 / (k * q * q2) ^ 0.5.

The distance between charges in the second case was equal to:

r2 = (F2 / (k * q * q2) ^ 0.5.

Then:

r2 / r1 = ((F2 / (k * q * q2)) ^ 0.5) / ((4 * F2 / (k * q * q2) ^ 0.5) = (1/4) ^ 0.5 = 1/2,

that is, it will decrease by 2 times.

Answer: the distance between charges will decrease by 2 times.