Change the chapter
If two equal charges each of 1 C each are separated in air by a distance of 1 km, what is the magnitude of the force acting between them? You will see that even at a distance as large as 1 km, the repulsive force is substantial because 1 C is a very significant amount of charge.
Question by OpenStax is licensed under CC BY 4.0.
Final Answer

$9 \times 10^3 \textrm{ N}$

Solution Video

OpenStax College Physics Solution, Chapter 18, Problem 15 (Problems & Exercises) (0:30)

Sign up to view this solution video!

View sample solution

Calculator Screenshots

OpenStax College Physics, Chapter 18, Problem 29 (PE) calculator screenshot 1
Video Transcript

This is College Physics Answers with Shaun Dychko. Two charges of one Coulomb each separated by kilometer will have a force of nine times ten to the three Newtons of repulsive force between them. And so we have Coulombs Constant times one Coulomb times it selves. Again because each charge is the same, both one Coulomb. And divided by one kilometer converted into meters by multiplying by 1000 meters per kilometer and we square that denominator. We end up with nine times ten to the three.


Submitted by henrimccarti on Mon, 02/10/2020 - 11:13

the question from the textbook is "Calculate the magnitude of the electric field 2.00 m from a point charge of 5.00 mC (such as found on the terminal of a Van de Graaff).", these questions/answers are outdated