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Coloumbs Law when have 3 charges at unknown distances

  1. May 13, 2016 #1
    1. The problem statement, all variables and given/known data
    I have a question that I can't figure out.

    Question is
    "Two charges are fixed in location: charge q1 = +8e is located at the origin and charge q2 = -2e is located on the x-axis at x = L. At what point (other than infinitely away) can a proton (a unit positive charge e) be placed so that it has net zero force acting on it"

    2. Relevant equations
    F = (kq1q2)/r^2

    3. The attempt at a solution
    I have got that q1 is (8e x 1.602 x 10^-19C) = 1.282 x 10^-18C
    and q 2 is (-2e x 1.602 x 10^-19C) = 3.204 x 10^-19C
    k = 8.988 x 10^9Nm^2C^-2
    Am assuming that a unit positive charge is 1C (although did originally think it could possibly be 1.602 x 10^-19C, the size of one proton, would that be correct?)

    When pumping in these values into the formula, and rearranging the formula so F = 0N, this will obviously give a distance of 0 which would be incorrect, also not sure how the added proton fits in anywhere. Have spent a few hours trying to find another way but am a bit stuck. Thanks in advance for any help =)
     
  2. jcsd
  3. May 13, 2016 #2
    one should make out a force diagram and see /calculate the resultant of the two forces and make it zero.
     
  4. May 13, 2016 #3
    Thanks for your reply. How would I go about that if I dont have distances?
     
  5. May 13, 2016 #4
    Try and feel it :-
    if you put a +ve unit charge IN BETWEEN 8e and -2e, then it the 8e would cause repulsion and -2e would cause attraction, plot it and you would get that this is not possible.

    if you put a +ve unit charge on the line joining the two charges ( but not in b/w them ) then it is possible that you find two points on which the net force =0
     
  6. May 13, 2016 #5
    Arrrr yes! That would work perfectly, thanks for your time in helping =)
     
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