Two suspended charged particles, find the angle from vertical.

[Bobbo Snap]

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

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

The Attempt at a Solution

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

After I substitute $r = 2l \sin{\theta}$, I can't figure out how to solve for theta. I feel like I'm missing something simple. Any suggestions?

 

[flatmaster]

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

 

[flatmaster]

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

 

[Bobbo Snap]

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.

 

[flatmaster]

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.

 

[Dakota Christian]

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

 

[haruspex]

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.