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 $\epsilon_{o}$
