# Calculating mass and charge independently in a Paul Trap

1. Jul 25, 2013

### Bravus

Trapping dust particles in a Paul trap, in air, at 50 Hz (similar to Winter & Ortjohann, 1991).

Gravity compensation is achieved using a DC voltage across top and bottom electrodes.

Given the features of the trap, the charge-to-mass ratio of the trapped particle can be calculated as follows:

$$\frac{Q}{M} = \frac{z_0^3Ω^2ΔV_z}{4V_{ac}^2Δz_F}$$

where Q = charge (C), M = mass (kg), z0 is half the height of the trap (m), Ω = frequency (rad/s), Vz = applied compensation voltage (V), Vac = is the amplitude of the oscillating trapping potential (V) and zF = displacement of the particle from the trap centre (m).

OK, so that allows me to find the ratio of charge to mass, by varying the compensation voltage and measuring changes in the 'sag' from the centre of the trap. I'm not sure it allows me to separately determine the mass and charge, though...

As I'm writing this it occurs to me that graphing ΔVz vs ΔzF will give me intercepts that might be relevant...

Will keep thinking on it myself, and share any fruits if I make progress, but thought I'd turn your collective minds loose on it too...

Reference

H Winter and H W Ortjohann. Simple demonstration of storing macroscopic par-
ticles in a "Paul trap". American Journal of Physics, 59(9):807-813, May 2004.
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Last edited: Jul 25, 2013
2. Jul 25, 2013

### Bravus

Not sure why the TeX is failing me on the equation, but hopefully you can see what is top line and what is bottom line.

Edit: fixed that

Second edit: I missed something on my earlier look. There is no $c^2$ term. It should have read $V_{ac}^2$ - as it now does.

Last edited: Jul 25, 2013
3. Jul 26, 2013

### Bravus

113 views, no replies. It's OK. Here's what I just tweeted:

Work through physics problem I was stuck on. Reassured to discover it's impossible. Grinning supervisor confesses to being 'mean'.​

As far as I can tell, there's no way to do it with one particle.

The hint and key, that we got to eventually (but I had to be told), is to use two.

They'll have the same charge (sign, not necessarily magnitude), and the Coulomb force between them is independent of mass, so it allows calculating charge (or, at least, combined charge). Since we can already find Q/M, if we can find charge we can find mass.