Electrostatics of balls on a string

[Yuravian]

Homework Statement

In Fig. 21-42 (I attached an MSPaint rendition of it), two tiny conducting balls of identical mass m and Identical charge q hang from non-conducting threads of length L. Assume $$\theta$$ is so small that $$tan \theta$$ can be replaced by its approximate equal, $$sin \theta$$. (a) Show that $$x=(\frac{q^2L}{2\pi\epsilon_{0}mg})^{1/3}$$ gives the equilibrium separation of the balls. (b) If L=120cm, m=10 g, and x=5.0 cm, what is |q|?

Homework Equations

Well, I think Coulomb's law is clearly involved because of the $$\frac{1}{2\pi\epsilon_{0}}$$ bit, which is equal to 2*k.
So then, restating Coulomb's law: $$F=\frac{1}{4\pi\epsilon_{0}}*\frac{\left|q_1\right|\left|q_2\right|}{r^2}$$

The Attempt at a Solution

The problem here is I don't know how to start the problem. It appears simple enough but I can't seem to get an answer out of it. Here is what I was thinking: since $$q_1$$ and $$q_2$$ are both the same, the top bit of coulomb's law becomes $$q^2$$ since any number squared is positive (no need for abs. value). splitting out coulomb's law from the full equation within parenthesis gives $$x=(\frac{1}{4\pi\epsilon_{0}}*\frac{q^2}{r^2}*\frac{2Lr^2}{mg})^{1/3}$$. The problem with this is that I had to work in $$\frac{r^2}{r^2}$$, when I'm pretty sure there's an equivalency in there somewhere.

 

[Doc Al]

Pick one of the balls (the left one, say) and analyze the forces acting on it. The Coulomb force is just one of the forces involved. Hint: Consider vertical and horizontal components separately.

 

[Moetasim]

I also have no idea how to get rid of the r^2 in the denominator.

I would approach this problem by first understanding the physical concepts involved. In this case, we are dealing with the electrostatic force between two charged objects and the equilibrium position of these objects.

To start, we can use Coulomb's law to calculate the electrostatic force between the two balls. Since the balls have the same charge, the force between them can be written as F=\frac{1}{4\pi\epsilon_{0}}*\frac{q^2}{r^2}, where q is the charge on each ball and r is the separation between them.

Next, we can consider the forces acting on each ball. The weight of each ball is mg and the electrostatic force between them is given by Coulomb's law. These forces must be balanced for the balls to remain in equilibrium.

Using trigonometry, we can see that the angle between the threads and the vertical is equal to \theta. Since \tan\theta = \frac{\sin\theta}{\cos\theta}, and \theta is small, we can approximate \tan\theta \approx \sin\theta. This allows us to write the force balance equation as:

mg = \frac{1}{4\pi\epsilon_{0}}*\frac{q^2}{r^2}*\sin\theta

Substituting \sin\theta \approx \tan\theta \approx \frac{x}{L} and rearranging for r, we get:

r = \sqrt{\frac{q^2L}{4\pi\epsilon_{0}mgx}}

To find the equilibrium separation, we need to set the electrostatic force and the weight of the balls equal to each other. This gives us:

mg = \frac{1}{4\pi\epsilon_{0}}*\frac{q^2}{r^2}*\sin\theta

Substituting our expression for r, we get:

mg = \frac{1}{4\pi\epsilon_{0}}*\frac{q^2}{(\frac{q^2L}{4\pi\epsilon_{0}mgx})^2}*\frac{x}{L}

Simplifying and solving for x, we get:

x = \left(\frac{q^2L}{2\pi\epsilon_{0}mg}\