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, Chapter 18, Problem 25 (Test Prep for AP® Courses)

OpenStax College Physics, Chapter 18, Problem 25 (AP) video thumbnail

In order to watch this solution you need to have a subscription.

Start free trial Log in
vote with a rating of votes with an average rating of .
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 F one 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 r two squared. We're told that r two is half of r one. So the distance between them is halved. Then we replace r one over two in place of r two in this force two formula, and this makes r one 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 k Q one Q two over r one squared and this is F one. So F two is four times F one. So that's why the answer is A.