# Two suspended charged particles, find the angle from vertical.

• Bobbo Snap
In summary, the two particles suspended by strings have charges q and r and an angle θ which is found by solving the equation tan(θ) = q^2/8(pi)(permittivity of free space) mgl^2.
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?

Last edited:
Looks like it may be ugly. Are you allowed to use the small angle approximation?

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

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

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.

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.

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

Dakota Christian said:
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.

## 1. How do you calculate the angle from vertical for two suspended charged particles?

To calculate the angle from vertical for two suspended charged particles, you can use the equation: θ = tan^-1(q1/q2), where q1 and q2 are the charges of the two particles.

## 2. What is the significance of finding the angle from vertical for two suspended charged particles?

Finding the angle from vertical for two suspended charged particles is important because it helps in understanding the forces acting on the particles and their equilibrium state. It can also provide information about the magnitude and direction of the electric field between the particles.

## 3. Can the angle from vertical for two suspended charged particles be negative?

Yes, the angle from vertical for two suspended charged particles can be negative. A negative angle indicates that the particles are attracted to each other and the electric field between them is directed towards the negative particle.

## 4. What factors can affect the angle from vertical for two suspended charged particles?

The angle from vertical for two suspended charged particles can be affected by the distance between the particles, their charges, and the presence of any external electric fields. It can also be influenced by the medium in which the particles are suspended.

## 5. How can the angle from vertical for two suspended charged particles be experimentally determined?

The angle from vertical for two suspended charged particles can be experimentally determined by using a torsion balance or a Coulomb's balance. These devices measure the electrical and gravitational forces acting on the particles and can calculate the angle using the equilibrium condition.

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