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Charged Aluminum Spheres

  1. Aug 28, 2007 #1
    I'm having difficulty with a problem on MasteringPhysics (such wonderful software...) and as a last resort I'm posting on here. This is, I'm sure, a really simple problem but I'm getting no kind of feedback from MP and there isn't an example problem like this in the book.

    1. The problem statement, all variables and given/known data

    Two aluminum spheres of mass .025 kg are separated by 80 centimeters.

    A) How many electrons does each sphere contain?

    B)How many electrons would have to be removed from one sphere and added to the other to cause an attractive force between the spheres of magnitude 1.00 x 10^4 (roughly one ton)? Assume that the spheres may be treated as point charges.

    C)What fraction of all the electrons in one of the spheres does this represent?

    2. Relevant equations



    3. The attempt at a solution

    A) I found Part A to be 7.25 x 10^24 electrons.

    B) This is where I'm stuck.

    If you were to remove electrons from one sphere and put them on the other I understand that their charges are to be equal but opposite, as in q1 = -q2. So using Coloumb's Law (F=K*q1*q2/(r^2)) I've set the Force to 1*10^4, divided that by K=9*10^9.

    10000/(9*10^9) = q1*q2/(.8^2)

    Then multiplying that by .8^2, I have just the charges on the other side of the equation. Since the charges in the formula are absolute value I can set q1=q2 and have q1^2. Taking the square root of the entire thing I have:


    So now I can use the formula q=e(#protons-#electrons). So:

    8.4237*10^-4 = 1.6*10^-19(7.25*10^24-#electrons). Solving from this I get 7.249*10^24 electrons as my final answer.

    However, MP says I'm wrong but there isn't any feedback as to where I went wrong, and it seems straightforward enough to me that no matter how I rework it I get the same thing every time.

    Can anyone help me out here?

    C) Can't do this one until B is done.

    Any help is greatly appreciated.
  2. jcsd
  3. Aug 28, 2007 #2
    how you did 'a'?
    I am getting something different [5.58E20]
  4. Aug 28, 2007 #3
    Code (Text):
    .025 kg x [U]1000g[/U] x [U]1mol[/U] x [U]6.023*10^23 atoms in a mol[/U] x [U]13 electrons[/U] = #electrons
               1kg  26.982g               1 mol             1 atom Al
    Last edited: Aug 28, 2007
  5. Aug 28, 2007 #4
    oops, I forgot units...
    but do you get 7. something when you do this cal. on calculator?

    my cal's giving me 5.something

    and you getting
  6. Aug 28, 2007 #5

    How much does each ball weigh in your problem? It could be different, especially if you're doing this in MP.
  7. Aug 28, 2007 #6


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    Homework Helper

    multiply by the atomic number of aluminimum (ie number of protons in an atom which equals the number of electrons in an atom)
  8. Aug 28, 2007 #7


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    I'm confused by what you did in part B... the charge you calculated looks correct... you just need to divide by 1.6*10^-19 to get the number of electrons.
  9. Aug 28, 2007 #8
    You have to? I think maybe not. (We never did that)

    oo..O you don't know that part A is correct try without multiplying by #13, hopefully, you may get the right answer
    Last edited: Aug 28, 2007
  10. Aug 28, 2007 #9


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    Homework Helper

    Gaupp did... he didn't write it out..

    Remember you need the number of electrons... not the number of atoms...

    So number of atoms * (number of electrons/atom) gives the answer... I get the same answer as Gaupp.
  11. Aug 28, 2007 #10
    ooo.. yea you are right.. that gives the # of atoms, not the # of electrons
  12. Aug 28, 2007 #11

    Ok I see now. The difference between the #protons and the #electrons would be the number of electrons lost or gained. I kept solving for just the number of electrons, by moving around the #protons in the equation. DOH!

    Thanks for the help there learningphysics.
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