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Problem involving Coulombs Law; suspended masses

  1. Jul 25, 2009 #1
    Two point particles, each of mass m and charge q are suspended from a common point by threads of lenght L. Each thread makes an angle [tex]\theta[/tex] with the vertical. (I attached a diagram to help).

    I must show that
    [tex]q=2L\sin\theta\sqrt{\frac{mg}{k}\tan\theta}[/tex]

    I start out by writing the force sum in the coordinate system, and then I solve for q:

    [tex]0=mg+k\frac{q^2}{R^2}[/tex]

    [tex]=mg+k\frac{q^2}{4L^2\sin^2\theta}[/tex]

    [tex]q^2=-4L^2\sin^2\theta\frac{mg}{k}[/tex]

    I'm confused by the negative sign, if I ignore it I get:

    [tex]q=2L\sin\theta\sqrt{\frac{mg}{k}}[/tex]

    The [tex]\tan\theta[/tex] is missing in my solution.
     

    Attached Files:

  2. jcsd
  3. Jul 25, 2009 #2

    Doc Al

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    Staff: Mentor

    Forces are vectors--direction matters. Hint: Consider the horizontal and vertical components of the forces.

    Don't forget the tension force exerted by the thread.
     
  4. Jul 25, 2009 #3

    lol, DUH the tension! I can't believe I left that one out. I pulled it togther and obtained the given expression. Thanks for your help.
     
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