# Dissipating charge

1. Aug 3, 2006

### Saketh

A friend of mine has given me a physics problem to solve. Here it is:
Two small equally charged spheres, each of mass $m$, are suspended from the same point by silk threads of length $\ell$. The distance between the spheres $x << \ell$. Find the rate $$\frac{dq}{dt}$$ with which the charge leaks off each sphere if their approach velocity varies as $$v = \frac{a}{\sqrt{x}}$$, where $a$ is a constant. ​
I started by writing down the forces:
$$mg\tan{\theta} = \frac{q^2}{4\pi \epsilon_0 x^2}$$
Since $x << \ell$, I concluded that $\sin{\theta} \approx \tan{\theta}$. I rewrote the above force equation with sin instead of tan.

After this, I started doing things randomly. At first, I tried using angular speed with $\ell$ as the radius of rotation, but that seemed unnecessarily complicated.

So I went back and differentiated both sides of the first equation. I don't know if this is correct, but I let $\cos{\theta}$ be 1 since $\theta \approx 0$. So, using the quotient rule, I had:
$$mg = \frac{1}{2\pi \epsilon_0}\frac{q^2(\frac{dx}{dt}) - q\frac{dq}{dt}}{x^3}$$
At this point, since $\frac{dx}{dt} = \frac{a}{\sqrt{x}}$, I started substituting things in. However, I ended up with charge as a function of time, distance as a function of time, and the time derivative of the charge function all in one equation - unsolvable.

The other way that I tried it was to start from the premise that $\ell \frac{d\theta}{dt} = \frac{a}{\sqrt{x}}$ and go from there. But I don't know if this is the correct way either.

Basically, I have no idea what I'm doing. My friend said this is an easy problem, but I am stumped. Am I approaching this incorrectly?

Thanks for the assistance.

Last edited: Aug 3, 2006
2. Aug 4, 2006

### Tide

Solve your original equation for q and approximate $\theta$ with x/l. Then differentiate q with respect to time. You should end up with dq/dt being constant within this approximation.

[Note: you may need to make minor adjustments to the preceding depending on whether your angle is the half-angle or full angle.]

3. Aug 4, 2006

### andrevdh

I don't think one should assume that the spheres are in equilibrium. That is

$$T\sin(\theta) - F_Q = m\ddot{x}$$

since

$$v_x(x) \Rightarrow v_x(t)$$

Last edited: Aug 4, 2006
4. Aug 4, 2006

### Saketh

Thanks - I have found my errors now. I had to approximate $\theta$ with $\frac{x}{2\ell}$.

For those who are interested, the answer is:
$$\frac{dq}{dt} = \frac{3}{2}a\sqrt{\frac{2\pi \epsilon_0 mg}{\ell}}$$