Problem29
Section 18.4: Coulomb's Law

Question

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

9×103 N9 \times 10^3 \textrm{ N}

Solution video

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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.

Comments

suzannemagargee·
Problem #15 15. (a) Find the total electric field at x = 1.00 cm in Figure 18.51(b) given that q = 5.00 nC . (b) Find the total electric field at x = 11.00 cm in Figure 18.51(b). (c) If the charges are allowed to move and eventually be brought to rest by friction, what will the final charge configuration be? (That is, will there be a single charge, double charge, etc., and what will its value(s) be?)
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suzannemagargee·
Do you have the solution for Problem 15? The solution to Problem 29 is provided here. Thanks for these solutions and your help!
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ShaunDychko·
Hi suzannemagargee, Thank you for the question. The problem you're looking for is here: https://collegephysicsanswers.com/openstax-solutions/find-total-electric-field-x-100-cm-figure-1851b-given-q-500-nc-b-find-total-0 It looks like you're working from the College Physics for AP Courses textbook, so please select your chapters from the AP version of the text by choosing "College Physics for AP Courses" from the "Solutions" drop-down in the top-level menu and the problems will be numbered according to the AP version. All the best, Shaun
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ShaunDychko·
Oh, and I'm really glad the solutions are helpful! :)
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ecarey1026·
You should have divided kq^2 by 0.001^2 m, instead of 1000^2 km. The written work is correct, but I think you plugged it into your calculator incorrectly. Final answer is ~ 9x10^15 N.
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ShaunDychko·
Hello, thank you for the comment. The question mentions that the charges are separated by 1 km, which is 1000m. This means we need to divide by 1000^2, so it's worth having a second look at the video and see what you think. All the best, Shaun
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henrimccarti·
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
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ShaunDychko·
It's here: https://collegephysicsanswers.com/openstax-solutions/calculate-magnitude-electric-field-200-m-point-charge-500-mc-such-found-terminal. Found it by entering "Calculate the magnitude of the electric field 2.00 m" in [search](https://collegephysicsanswers.com/search-answers).
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