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Two suspended charged particles, find the angle from vertical.

  • Thread starter Bobbo Snap
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  • #1
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Homework Statement



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.

Homework 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]


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:

Answers and Replies

  • #2
499
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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
 
  • #3
499
2
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
 
  • #4
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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.
 
  • #5
499
2
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.
 
  • #6
I know that the answer is

tan^3(theta)/(1+tan^2(theta)) = q^2/8(pi)(permittivity of free space) mgl^2

But i have no ideal why or if mgl^2 is in the numerator or denominator. My guess is that mgl^2 is in the denominator
 
  • #7
haruspex
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I know that the answer is

tan^3(theta)/(1+tan^2(theta)) = q^2/8(pi)(permittivity of free space) mgl^2

But i have no ideal why or if mgl^2 is in the numerator or denominator. My guess is that mgl^2 is in the denominator
That is effectively the same as Bobbo Snap obtained (four years ago), but expressing it as you have it becomes a cubic in tan(θ). So in principle you could now apply the standard solution to a cubic.
 

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