# 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.

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