Problem29
Section 18.4: Coulomb's Law (Test prep for AP® courses)

Question

(a) What is the electrostatic force between two charges of 1 C each, separated by a distance of 0.5 m?
(b) How will this force change if the distance is increased to 1 m?

Question by OpenStax is licensed under CC BY 4.0

Final Answer

  1. 3.60×1010 N3.60 \times 10^{10} \textrm{ N}
  2. 8.99×109 N8.99 \times 10^9 \textrm{ N}

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Video Transcript

This is College Physics Answers with Shaun Dychko. The electrostatic force between two charges is Coulomb's constant times the first charge times the second charge divided by the distance between them squared. So that's 8.988 times ten to the nine newtons meter squared per Coulomb squared, times one Coulomb from the first charge, and times one coulomb from the second charge, divided by 0.5 meters between them squared, giving 3.60 times ten to the ten newtons of force between them. Now if the distance is increased to one meter, that represents a difference by a factor of two. So we can say that force two is k q one q two over this new distance which is two times the original distance. We square that and this makes a factor of one quarter times F one because that's what all this stuff is. This is the first force. So that's one quarter times 3.5952 times ten to the ten newtons, writing this number with more digits because we want to avoid intermediate rounding error. We get an answer for the force in part B will be 8.99 times ten to the nine newtons.

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