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Coulomb's Law in 1D

  1. Jun 10, 2010 #1
    1. The problem statement, all variables and given/known data

    Three charges lie along the x-axis. The positive charge q1 = 10.0 x 10^-6 C is at x = 1.00 m, and the negative charge q2 = -2.00 x 10^-6 C 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?

    2. Relevant equations

    F=kq1q2/r^2

    3. The attempt at a solution

    F23 = -F13

    F23 = k(-2.00 x 10^-6 C)q3/-x^2
    F13 = -k(10.0 x 10^-6 C)q3/(1.0 m - x)^2

    [k(-2.00 x 10^-6 C)q3/-x^2] - [k(10.0 x 10^-6 C)q3/(1.0 - x)^2] = 0
    (-2.00 x 10^-6 C)/-x^2 = (10.0 x 10^-6 C)/(1.0 -x)^2
    2/x^2 = 10/(1- x)^2
    2 - 4x + 2x^2 = 10x^2
    -8x^2 - 4x +2 = 0

    My quadratic equation doesn't work out so I'm assuming there was a terrible malfunction earlier on. Can somebody please help me out with this one?
     
  2. jcsd
  3. Jun 10, 2010 #2

    Doc Al

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    Staff: Mentor

    Hint: First figure out what region the positive charge q3 must be in. ( x < 0; 0 < x < 1; x > 1) The field from each charge may have a different sign in different regions, which affects how you'd write the equation.
     
  4. Jun 10, 2010 #3
    It's going to be left of the origin (-x). How would this affect my initial signs?
     
  5. Jun 10, 2010 #4

    Doc Al

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    Staff: Mentor

    Actually your equations are fine. (I just saw the double negative.) Why do you think your quadratic doesn't work out?
     
  6. Jun 10, 2010 #5
    Because I end up with 4 +/- sqr.[(-4_^2) - (4)(-8)(2)]/2 x 2
    = (4 - 80)/4
    = -19 which is much too high.
     
  7. Jun 10, 2010 #6
    Er, I meant low in the last reply.
     
  8. Jun 10, 2010 #7

    Doc Al

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    Staff: Mentor

    I assume you are trying to use the quadratic formula. Try again; you're making an error somewhere.

    Edit: You forgot to take the square root.
     
  9. Jun 10, 2010 #8
    Is my error in the use of the quadratic formula or in the work leading up to it?
     
  10. Jun 10, 2010 #9

    Doc Al

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    Staff: Mentor

    Your work is fine. You just forgot to take the square root. (See my last post.)
     
  11. Jun 10, 2010 #10
    Oooooooooh...duh. Thanks!!!
     
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