# Homework Help: Oscillation of a Charged Particle

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1. Feb 9, 2017

### peroAlex

1. The problem statement, all variables and given/known data
At our university we were given this problem: charged ball with mass of $m = 0.0001 kg$ and charge $Q = -10^{-5} C$ is placed on geometric axis of thin torus with inner radius of $r_{inner} = 0.05 m$, outer radius of $r_{outer} = 0.1 m$ and surface charge density $\sigma = 10^{-5} C$. Compute oscillation time for small deviation, this is when we only slightly flick the ball from stable state.

2. Relevant equations
First, I took a look at this article and a PDF presentation.

3. The attempt at a solution
Using the equation for electric field of a charged ring $$E_z = \frac{Qz}{4 \pi \varepsilon_0 (r^2 + z^2)^{\frac{3}{2}}}$$ I tried obtaining formula for electric field of torus (wider ring) by integration infinitesimal rings from inner to outer radius. Using Symbolab I managed to obtain following equation $$E_{torus} = \frac{Qz}{4 \pi \varepsilon_0} (\frac{r}{z^3 \sqrt{\frac{r^2}{z^2}+1}})_{r_{inner}} ^ {r_{outer}}$$.

From here on, I'm lost. Can somebody please help me or at least give me some guidance?

2. Feb 9, 2017

### BvU

Hi
means you are looking for something like the $k$ in $\vec F = -k\vec x$. So if $|\vec F|$ is a more complicated function like what you have found it here, you simply want the linear development around the equilibrium position.

3. Feb 9, 2017

### BvU

On 2nd reading I wonder if you are already aware of what I posted and are stuck in working out the electric field expression ?
Not clear what you are doing there. The exercise geves that the charge is at the surface and unfiormly distributed. Looks like the pdf assumes a thin torus (yours is fat).

4. Feb 9, 2017

### peroAlex

I believe we're on the wrong footing here.

My task considers thin torus. Due to the fact that such configuration is composed of filamentary thin rings, I tried integration individual contributions (rings) from inner to outer radius.

5. Feb 9, 2017

### BvU

Guy here (page 23) calls $a/\rho = 5$ fat. With $a/\rho = 3$, yours is 67% fatter !
Surface charge on the inside is twice as close to the axis as surface charge on the outside !

Or do I have the wrong idea of inner ($r_{\rm inner} = 0.05$ m) and outer ($r_{\rm outer} = 0.10$ m) radius ?

6. Feb 9, 2017

### vela

Staff Emeritus
Is the shape really a torus (donut-shaped)? If it is, the dimensions given seem to contradict the description that it's thin. Or is it supposed to be a flat ring of charge with the given inner and outer radii?

7. Feb 9, 2017

### BvU

A washout ?!?

This is a torus

very clear what is meant with a torus in math and physics.

This is a washer (not a very scientific name, but clear enough)

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8. Feb 13, 2017

### peroAlex

This is the exact shape I was trying to describe. I'm so sorry for ambiguity in definition, English is not m native tongue.

9. Feb 13, 2017

### BvU

Good; makes life easier. Thin disk with a hole in the center it is. I should have been more suspicious when you mentioned a 'wider ring' and called it a torus.
First expression in post #1 (field of charged ring) looks exactly as the one here , but with $\ Q\$ instead of $\ 2\pi\sigma R' dR' , \ \$ and -- apart from that -- seems OK to me. This you now want to integrate from $r_{\rm min}$ to $r_{\rm max}$ (instead of from $0$ to $r_{\rm max}$ like here) and there you lose me :

Using my common sense I got something else. Pretty unpleasant, but the first derivative at $z=0$ is what we are after and that should be fairly decent. (In fact, working out the potential instead of the E-field would have been more economical -- hindsight....)

One reason I don't trust your result is that it diverges for $z=0$ wich should not happen. Could you check ?