Two identical stationary point charges are at a distance of 1 m from each other. What should be the distance between
Two identical stationary point charges are at a distance of 1 m from each other. What should be the distance between the charges so that the interaction force between the charges decreases by 4 times?
Given:
R1 = 1 meter – the distance between two identical point charges.
It is required to determine R2 (meter) – the distance between the charges, at which the force of interaction between the charges will decrease by 4 times.
Let q be the same point charges. Then the force of interaction in the first case will be equal to:
F1 = k * q ^ 2 / R1 ^ 2, where k is the Coulomb constant.
In the second case, the force of interaction will be equal to:
F2 = k * q ^ 2 / R2 ^ 2.
By the condition of the problem, F1 / F2 = 4, then:
F1 / F2 = 4;
(k * q ^ 2 / R1 ^ 2) / (k * q ^ 2 / R2 ^ 2) = 4;
R2 ^ 2 / R1 ^ 2 = 4;
R2 / R1 = 2;
R2 = 2 * R1 = 2 * 1 = 2 meters.
Answer: the distance between charges should be 2 meters.