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Calculate the number of excess electrons

  1. Mar 21, 2007 #1
    1. The problem statement, all variables and given/known data

    Two negatively charged objects repel each other with a measured force of 6.3 N when they are 0.5 cm apart. If the excess charge on one of the objects is caused by 8.3 x 10^22 extra electrons, use Coulomb's Law to calculate the number of excess electrons on the second object.



    2. Relevant equations

    F=kq1q2/r^2



    3. The attempt at a solution

    Since 1e=1.602 x 10^-19C
    8.3 X 10^22e = (1.602 x 10^-19 C)(8.3 x 10^22)
    =1.3 x 10^4 C

    Therefore, 6.3N = (9 x 10^9 N*m^2/C^2)1.3 x 10^4 C)(q2)/(0.005m)^2
    =1.3 x 10^-18 C

    Since 1C = 6.242 x 10^18e
    1.3 x 10^-18C = (1.3 x 10^-18)(6.242 x 10^18)e
    =8.1e ????


    I think I might be over complicating this one. Can anybody help me out? Please?
     
  2. jcsd
  3. Mar 21, 2007 #2

    Mentz114

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    Gold Member

    Hi, i don't understand what you've done.

    6.3N = (9 x 10^9 N*m^2/C^2)1.3 x 10^4 C)(q2)/(0.005m)^2
    =1.3 x 10^-18 C

    What is this ? 6.3N=1.3x10^-8 C ???

    If you write a=b then a should equal b. You are confused.
    Here's a tip - don't put in the numbers until you have worked out an algebraic formula for what you want. You want q1, and then to divide by the electronic charge.
     
  4. Mar 21, 2007 #3
    would this equation be all right?

    q1=F(r)^2/k(q2)
     
  5. Mar 21, 2007 #4

    Mentz114

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    Gold Member

    Yes, good, but better written like this ( because F(r) looks like a function)

    q1 = F*r^2/(k*q2)

    It's also clearer if you show the multiplication sign. Now plug in the numbers and then divide by the electronic charge.
     
  6. Mar 21, 2007 #5
    :smile: Thanks Mentz! I was over complicating it. Smooth sailing now though!
     
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