Question
For questions 25–27, suppose that the electrostatics force between two charges is F. What will be the force if the distance between them is halved?
1. 4F
2. 2F
3. F/4
4. F/2
Question by OpenStax is licensed under CC BY 4.0.
Final Answer

(a)

Solution Video

# OpenStax College Physics for AP® Courses Solution, Chapter 18, Problem 25 (Test Prep for AP® Courses) (1:06)

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Video Transcript

This is College Physics Answers with Shaun Dychko. We have two charges and the force between them is F. Now in the first scenario, we have <i>F one</i> is Coulomb's constant times the product of the two charges divided by the original distance between them squared. Then in the second scenario we have the same numerator but we have <i>r two</i> squared. We're told that <i>r two</i> is half of <i>r one</i>. So the distance between them is halved. Then we replace <i>r one</i> over two in place of <i>r two</i> in this force two formula, and this makes <i>r one</i> over two all squared, and this denominator here will become four and we multiply top and bottom by four to get rid of the four that's in the denominator of the denominator, which is a bit confusing. So put it in the numerator and we end up with four times <i>k Q one Q two</i> over <i>r one</i> squared and this is <i>F one</i>. So <i>F two</i> is four times <i>F one</i>. So that's why the answer is A.