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Homework Help: Identical discharging spheres

  1. Oct 6, 2009 #1
    1. The problem statement, all variables and given/known data

    Two identical small metal spheres initially carry charges q_1 and q_2. When they're x=1.0m apart, they experience a 2.5N attractive force. Then they're brought together so charge moves from one to the other until they have the same net charge. They're again placed x=1.0m apart, and now they repel with a 2.5N force.

    2. Relevant equations


    3. The attempt at a solution

    Alright, so this question seemed pretty easy but my answer isn't making any sense. First I solved for the net charge on the spheres after they had discharged,

    F=kq3^2/r^2 ... q3=sqrt(Fr^2/k) = 1.667*10^-5C

    Next I assumed that since the two spheres had balanced their charges, the original charges were 1.667*10^-5 +/- x.

    Putting this back into the Coulomb's Law equation I got

    F=k(1.667*10^-5 - x)(1.667*10^-5 + x)/r^2

    But when I solved for x and then tried to sub it back in to q1=(1.667*10^-5 - x) and q2=(1.667*10^-5 + x), both q1 and q2 were positive, which would not result in an attractive force.

    Help please. =)
  2. jcsd
  3. Oct 6, 2009 #2


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    Hi Tridius. welcome to PF.
    In the first case, F = k*q1*q2/d^2.
    In the second case, charge on each sphere is (q1 - q2)/2
    So F= k*(q1 - q2)^2/d^2
    Can you proceed now?
  4. Oct 6, 2009 #3
    Thanks a lot, this should really help. What I don't understand though is that if the charges are now (q1-q2)/2, then why is the force k(q1-q2)^2/d^2? Where did the 2 in the denominator go?
  5. Oct 6, 2009 #4


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    The force should be
    F = k(q1-q2)^2/4*d^2
  6. Oct 6, 2009 #5


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    I don't understand why each charge is (q1-q2)/2.
    If it was, then the total charge would be 2*(q1-q2)/2 = q1-q2.
    But we know the total charge is q1 + q2.
    Looks like the charge on each must be (q1 + q2)/2.
  7. Oct 6, 2009 #6


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    Since the force is attractive, the charges must be of opposite nature.
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