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Charged Spheres

  1. Jan 13, 2008 #1
    [SOLVED] Charged Spheres

    1. The problem statement, all variables and given/known data
    Two small aluminum spheres, each of mass 0.0250 kilograms, are separated by 80.0 centimeters. 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 1.00x10^4 N (roughly one ton)? Assume that the spheres may be treated as point charges.


    2. Relevant equations
    F = kq1q2/r^2
    q = ne


    3. The attempt at a solution
    In a previous question, it was worked out that each sphere has 7.25x10^24 electrons. It's confirmed that this answer is correct.

    F = kq1q2/r^2.

    We're taking an amount from the first, and giving it to the second. I declared variable 'x' to be this amount. Since each sphere has the same number of electrons, we can use a common variable 'n', instead of n1 and n2.

    F = kq1q2/r^2
    F = k(ne)(ne)/r^2
    F = k((n-x)e)((n+x)e)/r^2, introducing the variable x, taking x electrons from one, and giving them to the other, then solving for F = 10^4.

    Fr^2/k = e^2(n^2 - x^2)
    Fr^2/(ke^2) = n^2 - x^2
    (Fr^2/(ke^2)) - n^2 = -x^2
    -(Fr^2/(ke^2)) - n^2 = x^2
    Sqrt(-(Fr^2/(ke^2)) - n^2) = x

    Sqrt(-((10000)(0.8)^2/((8.988*10^9)(1.60217646*10^-19)^2)) - (7.25*10^24)^2) = x
    x = Not a real value.
    I took the absolute value of the huge number above before square rooting, and it came out to 7.25*10^24... which would mean that x = n, therefore to give the said force, 100% of the electrons would need to be transfered.

    I tried this answer, and was told I was wrong.
     
  2. jcsd
  3. Jan 13, 2008 #2

    Kurdt

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    One can just assume that [itex]q_1=n e = -q_2 [/itex].

    Remember that the charge on an electron is negative.
     
  4. Jan 13, 2008 #3
    Not too too sure that I follow what you mean, but I'll try flipping the e value.

    Sqrt(-(Fr^2/(ke^2)) - n^2) = x
    Sqrt(-((10000)(0.8)^2/((8.988*10^9)(-1.60217646*10^-19)^2)) - (7.25*10^24)^2) = x

    Still a non-real result (Root of a negative). Taking absolute value just to see what I get.
    Got same answer, 7.25*10^24.
    I understand getting the wrong answer, but I do find it weird that x = n, exactly...
     
  5. Jan 13, 2008 #4

    Kurdt

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    What I was getting at was if you work from the following you should get a sensible answer.

    [tex] F=k\frac{-n^2e^2}{r^2} [/tex]
     
  6. Jan 13, 2008 #5
    F = -kn^2*e^2 / r^2
    Sqrt(-Fr^2/(ke^2)) = n
    Sqrt(-(10000)(0.8)^2/((8.988*10^9)(-1.60217646*10^-19)^2)) = n
    n = 5.26682*10^15

    Trying...
    Correct answer.

    Just curious though, what did I do wrong? Why did my (n-x)(n+x) theory not work?
    Oooh.... it would take 5.26682*10^15 electrons to cause force F... blah.
    Thank-you for your time :)

    /solved.
     
  7. Jan 13, 2008 #6

    Kurdt

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    The n-x and n+x approach didn't work because these are just the numbers of electrons on each sphere. What you want is the number that is contributing to the force.
     
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