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Coulombs Law Question

  1. Oct 16, 2011 #1
    1. The problem statement, all variables and given/known data
    I don't understand how to apply coulombs law when there are 3 balls, giving each other charge. I know that the answer for part a) is 0.02N and part b) is 0.005 N but have no idea how to get these answers.

    2. Relevant equations
    Two identical plastic balls A and B with a metallic coating are given identical but opposite charges and placed a fixed distance apart. There is found to be an attractive force of 0.04N between them. A third identical, but initially uncharged ball, C is now brought into contact with ball A and then removed.
    a) what will be the force between A and B now
    b) if ball c, still with the charge it obtained from ball a is now brought into contact with ball b and then removed, what will the force be between a and b now?

    3. The attempt at a solution
    For part a) I tried substituting -1 and +1 for the charge but I still don't understand how the ball's A and C have transferred charge.
    And for part b) I need to understand how to get part a) first
  2. jcsd
  3. Oct 16, 2011 #2
    Think of charge distribution.

    When you first touch ball C to ball A, some of the charge from ball A will be distributed through C.

    How much do you think that would be proportionally if ball C is initially uncharged?
  4. Oct 17, 2011 #3
    erm...hmm when you first touch ball A to C would the total charge from A half-out with C? so now both ball is the same charge? I'm not sure if electricity works in the same way as heating and cooling :confused:
  5. Oct 17, 2011 #4
  6. Oct 17, 2011 #5
    Yes, the charges would evenly distribute themselves over the connection made by touching.
  7. Oct 17, 2011 #6
    so now that both balls are the same charge, the charge of the ball A is halfed? so thats why the 0.04 N --> 0.02 N for part a) of the question
  8. Oct 17, 2011 #7


    so when q1 is halved, F is halved

    say A had and original charge of q0

    A and C now have a charge of 1/2 q0

    C touches B the total charge on both is q0 + 1/2 q0 as equal surface area
    => evenly distributed again, then sub into your formula
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