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Two suspended charged particles, find the angle from vertical.

  1. Feb 5, 2014 #1
    1. The problem statement, all variables and given/known data

    Two particles, each of mass m and having charge q, are suspended by strings of length l from a common point. Find the angle θ which each string makes with the vertical.

    2. Relevant equations

    [tex] F_e = k \frac{q^2}{r^2}, \quad F_G = -mg, \quad F_T = \text{tension on string}, \quad r = 2l\sin{\theta}[/tex]


    3. The attempt at a solution

    Since [itex]F_{net} = \vec{0}[/itex], I set [itex]F_e = -F_T \sin{\theta}[/itex] and [itex]F_G = - F_T \cos{\theta}[/itex]. After some manipulation, I get [tex] \tan{\theta} = k \frac{q^2}{r^2 m g}.[/tex]

    After I substitute [itex]r = 2l \sin{\theta}[/itex], I can't figure out how to solve for theta. I feel like I'm missing something simple. Any suggestions?
     
    Last edited: Feb 5, 2014
  2. jcsd
  3. Feb 5, 2014 #2
    Looks like it may be ugly. Are you allowed to use the small angle approximation?

    For small theta…
    Cos(theta) = 1
    Sin(theta) = Theta

    Theta is in radians
     
  4. Feb 5, 2014 #3
    Each of those can be thought of as a taylor series expansion around theta=0. If you want to me more accurate, include more terms. I think the cos term becomes something like...

    Cos(theta) = 1 - theta^2

    Might be a coefficient I'm missing
     
  5. Feb 6, 2014 #4
    I don't think I can assume small angles, nothing is mentioned about the length of the string. But I do feel like there is some simplifying assumption I'm missing.
     
  6. Feb 6, 2014 #5
    From a similar problem...

    But, if you have the charge, and need to figure out the angle - there is no closed form solution (but there is an approximate solution).

    You can use small angle as I said earlier. To get a more precise answer, you can graph both sides of the equation of theta and see where they intersect. The x (theta) coordinate of that point is your solution.
     
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