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Homework Help: Two charges add to zero

  1. Feb 5, 2009 #1
    1. The problem statement, all variables and given/known data

    I am trying to find the point at which the electric field from two charges adds to zero and came to the equation (kq)/x^2-(kQ)/(x-d)^2 = 0

    2. Relevant equations

    (kq)/x^2-(kQ)/(x-d)^2 = 0

    3. The attempt at a solution

    I know that (kq)/x^2-(kQ)/(x-d)^2 = 0 is the correct equations but algebraically do not know what to do from here.

    Thanks for the help
  2. jcsd
  3. Feb 5, 2009 #2


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    Are the charges both the same sign?
  4. Feb 5, 2009 #3
    The charge at x = 0 is negative and has magnitude q = 2 mCoulombs. The charge at x = d, (d = +12 cm), is positive and has magnitude Q = 4 mCoulombs.
  5. Feb 5, 2009 #4


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    Then your equation is already set up wrong, because the field between a positive and a negative is always non-zero at every point in between.

    So if it isn't 0 between them it must be 0 somewhere else.
  6. Feb 6, 2009 #5
    I must have typed something wrong but in the help section of the homework it gives me this:

    Ex = (1/4peo)*((q/x^2) - (Q/(x-d)^2)) (for x < 0)

    Setting this expression equal to zero yields the equation:

    x^2*(q-Q) + x*(-2qd) + qd^2 = 0

    I am just wondering mathematically how you get from the first equation to the second one because I have no idea...
  7. Feb 6, 2009 #6


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    After canceling common factors, that step can be made by cross multiplying from

    q/x2 = Q/(x-D)2

    But you say they are choosing x<0, but I would note that for x<0 the distance to the other charge is x + D not x - D. This equation would be true for x>D however.
  8. Feb 6, 2009 #7
    Ahaaa I get it now thanks a lot.

    And then how do you get to x = d*((q/(q - Q))*(1 +/- sqrt(Q/q))
  9. Feb 6, 2009 #8


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    That's the solution for a quadratic equation?

    The Quadratic Formula?
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