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

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

Solution Video

OpenStax College Physics Solution, Chapter 18, Problem 13 (Problems & Exercises) (0:58)

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Video Transcript

This is College Physics Answers with Shaun Dychko. The force between two charges in the first case is <i>kq1q2</i> over <i>r1</i> 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 <i>r</i> 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 <i>F2</i> is 25 times <i>F1</i>. And we have a bunch of common factors that we can cancel divide both sides by <i>kq1q2</i>. Then we have one over <i>r2</i> squared equals 25 over <i>r1</i> 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 <i>r2</i> squared is <i>r1</i> squared over 25. Then take the square root of both sides and you get <i>r2</i> is <i>r1</i> over five. So that means the separation was decreased by a factor of one fifth.