# Electrostatic Dust Removal

I have been recently reading a paper on dust removal and I have a few basic questions regarding it.

First off, I am citing this paper - "Dust removal system with static electricity"

Basically, an electric field is applied, like so:

Since, the particle is initially grounded, the positive charge gets neutralized.

The polarized charge q of the particle is then given by

$$q = 3 \epsilon _{o} {\epsilon_{s}-1 \over \epsilon_{s}+2} E$$

where $\epsilon _{o}$ and $\epsilon _{s}$ are permittivity of free space and relative permittivity of the particle respectively.

First Question : How did the author get this expression? there is not much explanation in the paper. I have tried to derive it on my own, without much luck.

The particle is then collected by the electrode and the trajectory is governed by the following equation.

$$m{du \over dt} = qE + mg$$

which makes sense.

Second question : What if the surface on which the dust was initially resting on was not conductive surface

Say, a situation like this -

The electric field, will then be constant $E = {V \over d}$, where $V$ is is the potential difference between two plates and the $d$ is the distance between them.

The positive charge, here will not be neutralized in this case, and the particle will remain as a induced dipole. Which I can treat as two point charges $q_{+}$ and $q_{-}$ which has the same magnitude. The total electrostatic force on the particle due to the electric field will be

$$F = q_{+}E - q_{-}E$$

but since $E$ is constant, $F= 0$, and the particle will not be collected by the top plate.

Am I thinking it right? A second opinion will help.

I have asked a similar question here : Calculating dust attraction to a charged surface

And the attraction between charge and the dipole seems to exist only because of changing E.

Third Question : The paper models the entire thing inside a vacuum chamber, will it make any difference if it is not? The same equations will apply right? except may be a slight change in $\epsilon_{o}$

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The negative side of the particle is closer to the electrode, the force depends on distance so the attraction of the negative side exceeds the repulsion of the positive side. However, you probably also have to take into account charges in the non-conducting layer. – RedGrittyBrick Nov 22 '12 at 15:16
The induced charge on the non conducting layer will not make a big difference, will it? – Ender Nov 22 '12 at 15:49
For your third question, if it's not a vacuum then you need to consider drag forces due to viscosity within the fluid medium when you determine the equation of motion. – tpg2114 Nov 22 '12 at 17:10