Question
(a) How strong is the attractive force between a glass rod with a 0.700 μC0.700 \textrm{ }\mu\textrm{C} charge and a silk cloth with a −0.600 μC-0.600 \textrm{ }\mu\textrm{C} charge, which are 12.0 cm apart, using the approximation that they act like point charges? (b) Discuss how the answer to this problem might be affected if the charges are distributed over some area and do not act like point charges.
Question by OpenStax is licensed under CC BY 4.0
Final Answer
  1. 0.262 N0.262 \textrm{ N}
  2. If charges are distributed, the resultant force between the objects would be the vector sum of each force on each charge due to the resultant electric field at each charge. Things would get complicated!

Solution video

OpenStax College Physics, Chapter 18, Problem 11 (Problems & Exercises)

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Calculator Screenshots

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Video Transcript
This is College Physics Answers with Shaun Dychko. We're going to calculate the size of the force between a glass rod of charge 0.7 micro Coulombs and a silk cloth of charge 0.6 micro Coulombs, negative. And the distance between them is 12 centimeters which is 12 times ten to the minus two meters. So the magnitude of the force is going to be k times q1q2 and these are magnitudes of these charges, divided by the distance between them squared. So that's Coulombs Constant times 0.7 times ten to the minus six Coulombs times 0.6 times ten to the minus six Coulombs divided by 12 times ten to the minus two meters squared, giving a force of 0.262 newtons. And this force is going to be attracted because they're of opposite charge but the question only asks us to find how strong the force is and so what is it's magnitude or absolute value in other words. Part b says, what if the charges are distributed over the objects; the glass rod and the silk cloth. Well, the resultant force then would be the vector sum of each force due to the resultant electric field, due to each other charge. And things would get quite complicated. So that's why it's nice that these are assumed to be point charges.

Comments

these questions and answers are outdated, i want my money back

Hi henrimccarti,
Thank you for your feedback. The problem is that, for chapter 18 only, the questions are numbered differently in the two editions of the textbook: College Physics (which I presume you're using), and College Physics for AP Courses. I've been meaning to sort this out, and your comment is the final piece of evidence that it's important to do so now. Question 25 in your edition should be question 39. If you enter a bit of the question text in search it won't take long to find it. In any case, please contact me directly if you wish to have a refund, and it will definitely be issued.
All the best,
Shaun

Are these calculator tips correct? I am not getting the correct answer

Hi rutleyqm,
Thank you for the question. Yes, the calculator screenshots should get you the same answer. If your calculator gives a different answer it might depend how you're entering the "times 10 to the power of" part, for which I use the "E" function. If you enter this differently, you would need brackets around the denominator.
Hope this helps,
Shaun

Hi victoria, thank you for the question. The problem asks "how strong is the attractive force" which suggests it's looking for a magnitude, which is to say the strength of the attractive force. A negative sign would indicate direction, but in this particular problem it isn't clear what the negative direction would mean - is that to the left? down?
In the video I enclosed FF between vertical lines that mean magnitude (or in a math class these lines would be called absolute value), which means I'm ignoring the direction of the force. The force is attractive but whether that's left, right, up, or down, isn't specified.
Hope this helps,
Shaun