1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Simple application of Coulomb's Law/Equation

  1. Mar 10, 2015 #1
    1. The problem statement, all variables and given/known data

    Two pieces of copper weighing 10 grams each are 10 cm apart and 1/1000 electrons are transferred from one to the other. Find the force of electrostatic attraction between them.


    2. Relevant equations

    Molar mass Cu: 63.5 g/mol

    [tex] \frac{1}{4\pi \epsilon_0} = 9 \times 10^9 [/tex]

    ##e = 1.6 \times 10^{-19} \ C##

    Avagadro: ##6 \times 10^{23}##


    3. The attempt at a solution

    Actually, I only have one minor thing I don't understand.

    In my solution I have

    [tex] F = 9 \times 10^9 \frac{-15.12 \times +15.12}{0.1^2} [/tex]

    Whereas the solution outlined in the book only uses one charge in the numerator

    [tex] F = 9 \times 10^9 \frac{15.12}{0.1^2} [/tex]

    So my answer is out by a factor of 15.12.

    I don't get it. Wouldn't the charge on one of the copper pieces be +15.12 C and the other -15.12 C and therefore q1 x q2 would equal ##- 15.12^2##...?
     
  2. jcsd
  3. Mar 10, 2015 #2

    gneill

    User Avatar

    Staff: Mentor

    Can you show how you arrived at 15.12 C for the charge? It seems a bit small to me for 1/1000 of the electrons in a 10g chunk of copper (29 electrons per atom).
     
  4. Mar 10, 2015 #3

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    I believe each copper atom is only considered to provide one free electron.
    https://in.answers.yahoo.com/question/index?qid=20110411224645AATo5fc
     
  5. Mar 10, 2015 #4

    gneill

    User Avatar

    Staff: Mentor

  6. Mar 10, 2015 #5
    My reasoning is exactly the same as the solution as outlined in the text. No different. Number of moles of copper in a 10g piece multiplied by avagadro multiplied by 1/1000.

    I get 15.12 C the book says 15.12 C. If we're both wrong, that's interesting, but what I'd like to know is: Should it be 15.12 or -15.12^2 in the numerator of Coulomb's Equation?
     
  7. Mar 10, 2015 #6

    gneill

    User Avatar

    Staff: Mentor

    If we're questioning the book's solution then I believe that it behooves us to examine the details. But perhaps that's just my picky nature :smile:

    One could answer the question by considering the units of the equation... kQ/r2 yields a field strength (V/m or N/C) rather than a force.
     
  8. Mar 10, 2015 #7
    Sorry, I sounded a bit pushy there, I went back and edited my comment pretty much immediately after, realizing.

    Interesting! Thanks for the reply. I should have thought of doing dimensional analysis on it. I really should have. Thanks again.
     
  9. Mar 10, 2015 #8
    The text shows a worked solution, not just an answer, so I can see where my work differs from the texts, and it differs only in that it has only 15.12 in the numerator of Coulomb's equation, not -15.12^2. :)
     
  10. Mar 11, 2015 #9

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I don't understand. The book is wrong by a factor of about 15 x 109 and all we care about is a factor of 15 ? I understand kosto, but I feel with gneill a whole lot more.
    What book is this and what does teacher say ?
     
  11. Mar 11, 2015 #10
    How is the book wrong by a factor of 15 x 10^9? I don't think it is. It gives an answer of about 2 x 10^14 N for the force between the charges.

    It is Physics for Competitions by C. G Agrawal Vol II A Master Book for IIT, PMT, AIEEE, UPSEAT, CET, PET, JEE, BCECE
     
  12. Mar 11, 2015 #11

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Ah, thanks for clarifying. The book has the right numerical value but the 2 was missing.

    I was forgetting the 1 in a 1000 , and ended up comparing that wrong value to 2 x 10^14/ 15000 which is also the wrong idea of what the book might have come up with. Two wrongs never make a right and there you are :smile: talking nonsense.

    Time to slow down o:) !
    Once again. Oh boy.
    I'm almost ready to adopt a PF signature like "sorry folks!"
     
  13. Mar 11, 2015 #12
    LOL, I think I need to adopt the sorry folks signature. Though at the moment I can pass it off as being a noob.

    I think next time I should just write out everything, especially since folks here seem to want to get more deeply into explanations than in other parts of the internet. Which is excellent. I didn't want to write a longer post originally because I didn't want to waste anybody's time, just cut to chase sort of thing.

    Next time I need to include more of the working.
     
  14. Mar 11, 2015 #13
    No teacher, teaching myself right now. I imagine I'll be posting around here a lot :oldsurprised:
     
  15. Mar 11, 2015 #14

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    [/QUOTE]kudos and good luck. look forward to the next post :cool:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Simple application of Coulomb's Law/Equation
Loading...