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Charge on a sphere.

  1. Apr 7, 2007 #1
    1. The problem statement, all variables and given/known data
    Two identical small conducting spheres are separated by 0.60 m. The spheres carry different amounts of charges and each sphere experiences an attractive electric force of 10.8 N. The total charge on the two spheres is [itex]=-12\mu C[/itex]. What is the positive charge on one of the spheres?


    2. Relevant equations
    Coulomb's Law.


    3. The attempt at a solution
    Part of my problem is I think I am confused about how to treat the signs of the charges.

    [tex] q_1 + q_2 = -12 \times 10^{-6} [/tex]
    [tex] q_1 = -(q_2 + 12) [/tex]

    Coulomb's Law gives:

    [tex] 10.8 = k\frac{q_2(q2_+12)}{0.60^2} [/tex]

    I left out the negative sign, because of the absolute value in Coulomb's law.
    Is this correct?

    and leads me to

    [tex] 0 = q_2^2 + (12\times 10^{-6})q_2 - 4.32601 \times 10^{-10} [/tex]

    From this, I end up with two solutions that fit the conditions of the problem. Is that right? I get a positive root of 27.6 micro C, and a negative root of 15.6 micro C, which gives me the positive charge of 3.6 micro C.

    Could anyone confirm that this is the right way to handle this situation, or clear up my sign confusion for me?

    Thanks so much,
    Sheldon
     
    Last edited: Apr 7, 2007
  2. jcsd
  3. Apr 7, 2007 #2

    Hootenanny

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    Indeed, the signs seems to be your problem, the first line should read

    [tex] q_1 + q_2 = 12 \times 10^{-6} [/tex]
     
  4. Apr 7, 2007 #3
    Hi Hootenanny,

    Sorry, I made a typo when I entered the problem. The sum of the charges is -12 micro C. I'll edit the original post to fix that.

    Sheldon
     
  5. Apr 7, 2007 #4

    Hootenanny

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    Then your method looks good to me...:approve:
     
  6. Apr 7, 2007 #5
    Thanks! So there are actually two possible pairs of charges for this situation? Interesting...
     
  7. Apr 7, 2007 #6

    nrqed

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    whatch out.. there must be something wrong (I will check in a second) because it's clear that with your two solutions, the product of the charges is not the same so both can't be solutions (I mean that 27.6 times (-39.6) is obviously not the same as -15.6 times +3.6) !!!
     
  8. Apr 7, 2007 #7

    nrqed

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  9. Apr 8, 2007 #8
    Patrick,

    Thank you. That was incredibly enlightening. I really appreciate it!

    Sheldon
     
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