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Coulomb's law, balancing a ball with point charges

  1. Feb 6, 2009 #1
    3 equal charges q are at the vertices of an equilateral triangle as shown, with the z-axis running through the midpoint of the triangle (such that the distance from each charge to the midpoint is d)

    [​IMG]

    a bead of charge Qb (of equal sign as the 3 charges q) is supposed to be levitated on the positive z-azis (coming out of the midpoint of the triangle). Derive an expression for the coloumb force exerted on the bead as a function of its position on the positve z-axis

    2. Relevant equations

    F = (K * |q| * |Qb| ) / r^2 is the force in the radial direction (straight line connecting the 2 charges)

    3. The attempt at a solution


    OK I'm kind of having trouble here....

    The force from each charge would simply be F = (K * |q| * |Qb| ) / (d^2 + z^2)
    (by pythagorean theorem, the square of the line connecting each charge to any point on the positive z-azis would be d^2 + z^2)
    So far so good
    because of the way the charges are positioned positioned , the x and y components of the vectors will cancel each other in between the 3 charges (I think) so we're left toworry only about the z-component of each radial vector...but how do i do calculate for this z-component?

    If i break down the radial force into 2 components, with one in the x-y plane and the other in the z-direction, then the angle etween the xy-plane component and the radial component is tan^-1 (z/d)....but now hat i have the hypotenuse ((K * |q| * |Qb| ) / (d^2 + z^2) ) and the angle tan^-1 (z/d), how do i isolate for just the z-component of the radial vector?
    z-component of the vector would be the hypotenuse times cos of the angle....but i have the angle expressed as inverse tan of (z/d) so how can i take the cos of something i only have expressed as that? I am totally lost as to how to derive this expression...
     
  2. jcsd
  3. Feb 6, 2009 #2

    rl.bhat

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    Homework Helper

    Radial force can be written as F = i*Fx + j*Fy + k*Fz
    Direction of the F is given by direction cosines. z component of the F = Fz = F*cos(gamma)= F*z/r. This is due to one charge.
     
  4. Feb 6, 2009 #3
    So how does cos(gamma) come out to z/r?
     
  5. Feb 6, 2009 #4
    OH never midn i udnerstand where it comes from....If i do it by similar triangles than
    Fz / Ftot = z / r
    Fz = Ftot * z/r

    So i guess Fz = kqQb * z / (Z^2+D^2)^2

    thanks for the help :)
     
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