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Homework Help: Distance between point charges in equilibrium (simple)

  1. Jan 5, 2013 #1
    1. The problem statement, all variables and given/known data

    Two fixed charges, +1.07 uC and -3.28 uC, are 61.8 cm apart. Where may a third charge be located so that no net force acts on it?

    2. Relevant equations

    Coulombs law

    3. The attempt at a solution

    q1 = 1.07 uC
    q2 = -3.28 uC
    q3 = q

    q1 ----- q2 ----- q

    q1 and q2 are separated by 0.618 m and q2 and q are separated by x

    Since the charges are in equilibrium,

    0 = k (1.07E-6)(q) / (0.618+x)^2 + k(1.07E-6)(q) / (x)^2

    factoring out and eliminating k and q,

    (1.07E-6)/(0.618+x)^2 + (3.28E-6)/(x)^2 = 0

    Solving for x gives me an imaginary distance. I feel like this is a very simple problem but I can't spot my error. What am I doing incorrectly? Thank you for your time.
  2. jcsd
  3. Jan 5, 2013 #2


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    Hello Mitocarta, welcome to PF!

    You've placed q closer to the larger magnitude charge. Is that going to work?
  4. Jan 5, 2013 #3
    Thank you for replying. When I switch the charges,

    (3.28E-6)/(0.618+x)^2 + (1.07E-6)/(x)^2 = 0

    I still get an imaginary answer.

    Edit: Is there a way to input math in an easier to read format on this forum?
  5. Jan 5, 2013 #4


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    The two forces act in opposite directions on q. The net force will be zero if the magnitudes of the forces are equal to each other.
  6. Jan 5, 2013 #5
    Oh I was overlooking that! Thank you so much, got the correct answer.
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