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Net force on a charge

  1. Nov 21, 2009 #1
    1. The problem statement, all variables and given/known data
    Three charges lie along the x -axis. The positive charge q1 = 10.0 microC is at x = 1.00 m, and the negative charge q2 = -2.00 microC is at the origin. Where must a positive charge q3 be placed on the x-axis so that the resultant force on it is zero?

    Answer: x = - 0.809 m

    2. Relevant equations
    Columb's Law
    F= k q1 q2
    - - - - - -
    r^2

    k = 8.9875 x 10^9

    3. The attempt at a solution

    Force of 1 acting on 3 = - k q1 q3 / (1 - x)^2
    Force of 2 acting on 3 = k q2 q3 / x^2

    k q2 q3 / x^2 - k q1 q3 / (1 - x)^2 = 0

    k's and q3's cancel out and I get

    q2(1 - x)^2 = q1(x^2)
    -2(1 - 2x + x^2) = 10x^2
    -2 + 4x - 2x^2 = 10x^2
    12x^2 - 4x +2 = 0

    x = .167 m

    Needing a bit of help in setting this one up perhaps. No solutions guide is available. Can someone have a more conceptual explanation on how to solve this one?
     
    Last edited: Nov 21, 2009
  2. jcsd
  3. Nov 21, 2009 #2

    berkeman

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    Staff: Mentor

    Thanks for showing your work -- makes this much easier.

    This line has a math error in it: q2(1 - x^2) = q1(x^2)

    The term on the left should be quantity squared (you pulled the squared inside the parens. So re-write as:

    [tex]q_2 (1-x)^2 = q_1 x^2[/tex]
     
  4. Nov 21, 2009 #3
    oh, that's just a typing error, it doesn't change the answer from what I had originally.

    Fixed and thanks.
     
  5. Nov 21, 2009 #4

    berkeman

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    Staff: Mentor

    Okay, then I think the issue that is left is that you double-did the negative sign for the negative charge:

    Force of 1 acting on 3 = - k q1 q3 / (1 - x)^2
    Force of 2 acting on 3 = k q2 q3 / x^2

    You should let the sign on the charges themselves dictate whether the force is in the + or - x direction.

    BTW, the book answer of -0.809m works in the equation you got to this point:

    q2(1 - x)^2 = q1(x^2)


    .
     
  6. Nov 21, 2009 #5
    double-did the negative? That doesn't make sense. I thought you had to arbitrarily add the negative sign cause that's the direction the force will be in?
     
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