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Charged Aluminum Spheres

  1. Aug 30, 2008 #1
    Two small aluminum spheres, each of mass 0.0250kg, are seperated by 80cm or .8m. If the atomic mass of aluminum is 26.982 grams per mole and its atomic number is 13 then how many electrons does each sphere contain? How many electrons would have to be removed form one sphere and added to the other to cause an attractive force of 1.00 X 10^4 N? What fraction of all the electron in one of the spheres does this represent? The spheres may be treated as point charges.

    0.05kg * 1000g/1kg * 1 mole/26.982g * (6.023 X 10^23) atoms/ 1 mole * 13 electrons / 1 atom = 9.43117 X 10^25 electrons

    If an electron is removed from one sphere it become +e and if an electron is added to the other sphere it becomes -e so the number of electrons that must be subtracted and added = (Ne) and (-Ne).
    Thus
    F=k(Ne)(-Ne)/.8m^2
    but since the force = k * the magnitude of the product of the charges / r^2
    1.00 X 10^4 N= k(Ne)^2/0.64m^2

    (1.00X10^4N)*(.64m^2)/k = (Ne)^2
    (1.00X10^4N)*(.64m^2)/k = (N^2) * (1.6X10^-19)^2
    (1.00X10^4N)*(.64m^2)/[k* (1.6X10^-19)^2] = N^2
    sqrt((1.00X10^4N)*(.64m^2)/[k* (1.6X10^-19)^2]) =N
    9.90243 X 10^7 electrons = N

    The ratio is 9.90243 X 10^7 / 9.43117 X 10 ^25

    Did I do this correctly?

    Thank you.
    Stephen
     
  2. jcsd
  3. Aug 30, 2008 #2

    rl.bhat

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    sqrt((1.00X10^4N)*(.64m^2)/[k* (1.6X10^-19)^2]) =N
    9.90243 X 10^7 electrons = N


    Please check this calculation.
     
  4. Aug 30, 2008 #3
    ok I re-did the calculation finding the number of electrons that would need to be transfered and I got 7.06384 X 10^7.

    Did I find the correct number of electrons in the spheres? And if so would the ratio now be (7.06384 X 10 ^7)/(9.43117 X 10^25)?

    Thank you for your help.
    Stephen
     
  5. Aug 30, 2008 #4

    rl.bhat

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    sqrt((1.00X10^4N)*(.64m^2)/[k* (1.6X10^-19)^2]) =N
    N^2 = (1.00x10^2)^2*(0.8)^2/(9x10^9)*(1.6^-19)^2
    N = 1.00^2*(0.8)/(3x10^4)*(10^1/2)*(1.6^-19)= 5.269x10^15
     
  6. Aug 31, 2008 #5
    ok
    so after doing it one more time on my calculator I got 5.27046X10^15

    Did I calculate the amount of electrons that are in the sphere correctly? And would the ratio now be 5.27046X10^15/9.43117 X 10^25?

    Thanks for the help.
    Stephen
     
  7. Aug 31, 2008 #6
    bump.......
     
  8. Aug 31, 2008 #7
    I do have to turn this in soon so any help would be appreciated.

    thank you
    Stephen
     
  9. Aug 31, 2008 #8

    rl.bhat

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    0.05kg * 1000g/1kg * 1 mole/26.982g * (6.023 X 10^23) atoms/ 1 mole * 13 electrons / 1 atom

    =1.451x10^25
     
  10. Sep 1, 2008 #9
    I miss typed 0.05kg * 1000g/1kg * 1 mole/26.982g * (6.023 X 10^23) atoms/ 1 mole * 13 electrons / 1 atom

    it was supposed to be
    0.0250kg * 1000g/1kg * 1 mole/26.982g * (6.023 X 10^23) atoms/ 1 mole * 13 electrons / 1 atom =9.43117X10^25 electrons
     
  11. Sep 1, 2008 #10
    Not to nag, but I have to turn this in tomorrow. So if anyone could tell me if I found the correct number of electrons in each sphere, the correct number of electrons that would have to transfer, and the correct ratio, I would really appreciate it.

    Thanks.
    Stephen
     
  12. Sep 2, 2008 #11
    Any opinions????
     
  13. Sep 2, 2008 #12

    rl.bhat

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    If 0.05kg contain 1.451x10^25 electrons, 0.025kg must contain 7,25x10^24 electrons. How is that you are getting different answer?
    Now you can find out the correct ratio.
     
  14. Sep 2, 2008 #13
    again not .05 but .025

    I mis typed it in the calculations but in the problem statement it says .025kg.
     
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