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Help with solving the problem charge problem

  1. Sep 12, 2011 #1
    1. Two small beads having a positive charges 3q and q are fixed at the opposite ends of a horizontal, insulating rod, extending from the origin to point X. This length is the distance (d). A third small charged bead (Q) is free to slide on the rod. At what position is the third bead in equilibrium?

    2. Coulomb's Law is used for this problem. F = kqq/d

    F[1][/SUB] = kQ(3q)/x[2][/SUP]
    F[2][/SUB] = kQq/(d-x)[2][/SUP]

    F[1][/SUB] = F[2][/SUB]

    3. I am attempting to solve for x.

    3 (d-x)[2][/SUP] = (x)[2][/SUP]

    I know I could make both side a square root so I have x one side, but I need help beyond this.

    I also need to show all steps. Thanks!
     
  2. jcsd
  3. Sep 12, 2011 #2

    Delphi51

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    Homework Helper

    Welcome to PF, thom.
    Looks good so far!
    Recommend you expand the left side, collect like terms to get a quadratic equation.

    I'm passing on a tip I find really helpful: don't bother with those sub, sup codes. Copy the ² symbol from the bottom of my post, or the complete set here:
    https://www.physicsforums.com/blog.php?b=346 [Broken]
     
    Last edited by a moderator: May 5, 2017
  4. Sep 12, 2011 #3
    [tex] \frac{3}{x^2}=\frac{1}{(d-x)^2} \Rightarrow 3x^2-6dx+3d^2=x^2\Rightarrow 2x^2-6dx+3d^2=0\Rightarrow x^2-3dx+3/2d^2=0[/tex]
    [tex]
    x_{1,2}=\frac{3d}{2}\pm\sqrt{\frac{9d^2}{4}-\frac{3d^2}{2}}=\frac{3d}{2}\pm\sqrt{\frac{3d^2}{4}}=\frac{3\pm\sqrt{3}}{2}d
    [/tex]
    I hope I haven't made a mistake...
     
  5. Sep 12, 2011 #4

    Delphi51

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    Homework Helper

    That checks out when using the + sign. The other solution is larger than d, so it has to be extraneous.
     
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