Question

Two point charges are brought closer together, increasing the force between them by a factor of 25. By what factor was their separation decreased?

Final Answer

Their separation was decreased by a factor of $\dfrac{1}{5}$.

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This is College Physics Answers with Shaun Dychko.
The force between two charges in the first case is

*kq1q2*over*r1*squared. And in the second case, we're told that the charges are the same but the distance between them changes and so we have a subscript two on the*r*here instead. And we're told that the force in the second case is 25 times the force in the first case. So we can say*F2*is 25 times*F1*. And we have a bunch of common factors that we can cancel divide both sides by*kq1q2*. Then we have one over*r2*squared equals 25 over*r1*squared. And we can flip both sides by raising them both to the exponent negative one or take the reciprocal of both sides in other words. And we get*r2*squared is*r1*squared over 25. Then take the square root of both sides and you get*r2*is*r1*over five. So that means the separation was decreased by a factor of one fifth.