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I should emphasize that I view the problem as a 'research' rather than a 'textbook' level. However, as I am almost unfamiliar with the subject it may be that the answer is elementary and well-known.

I will first describe the actual experiment that was carried and then present my thoughts and questions.

The fog, or rather the water vapor is generated by some simple device, say a kettle. The fog is inside a camera with the usual indoor air conditions ($T\approx 20^{\circ}\mathrm{C}, p\approx 10^5 \times\mathrm{Pa})$. There is also a cylinder shaped electrode inside the camera capable of producing an electric field $E$ with the magnitude at most $10^6 V/m$ and with the gradient $\nabla E$ at most $0.5\times10^8 \mathrm{V/m}^2$. These values are for the immediate vicinity of the electrode and decrease as the distance to the electrode increases.

In the actual experiment the vapor is produced constantly and the fog is dense. At first, the electric field is off. When one turns it on the fog starts to dissipate quite fast. When the electric field is turned off again, the fog returns back to its high density. The video of the experiment is here https://youtu.be/qVNOZKII2SI. Unfortunately, I have not participated in the experiment and can not provide any details at the moment.

The main question is due to which effects the dissipation of the fog happens?

Below I give my thoughts on the subject. The air is neutral with the permittivity close to a unit and the direct effect of the electric field is thus negligible. The water vapor is also neutral, but the permittivity of water is large $\varepsilon \approx 80$, so the polarization effects are in play.

A dipole moment $P$ of a water drop of mass $m$ in the electric field $E$ can be estimated as $P=\varepsilon_0 E m/\rho$, with $\rho\approx 10^3 \mathrm{kg/m^3}$ — water density and $\varepsilon_0$ — permittivity of the vacuum. Note that the electric field from a cylinder shaped electrode is not homogeneous. Hence we have the following force acting on a drop $F=P\nabla E$. As estimation shows, acceleration due to this force is of order $a\sim 0.5 \times\mathrm{m/s}^2$. This estimation is for the immediate vicinity of the electrode, the force rapidly decrease with the distance to the electrode ($F\propto 1/r^3$). Note also that since the dipole moment of a drop $P$ is proportional to its mass, the acceleration does not depend on the size of a drop.

Hence one sees, that even in the region where the electric effects are strongest the effect of the electric field is negligibly small compared to the gravitation force producing acceleration $g\approx 10 \mathrm{m/s}^2$. Therefore, the direct attraction between polarized drops of water and the electrode plays no significant role in the dissipation. But what could be other effects due to electric field which lead to the dissipation?

Here I am lost and asking for your ideas.

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I don't have time to do calculations, but maybe the mechanism is as follows (if this video is not a fake): the electrode (or the leads) cause a crown discharge, charges are captured by fog droplets, and then the charged fog droplets move in the electric field.

EDIT: See also https://books.google.com/books?id=8ysXBQAAQBAJ&pg=PA127&lpg=PA127&dq=fog+dissipation+in+electric+field&source=bl&ots=SBzlg66KJS&sig=Jja-i56qSWNkAz-G787olodw7gc&hl=en&sa=X&ei=qHVgVaeoGIKooQTgtIDADQ&ved=0CCcQ6AEwATgK#v=onepage&q=fog%20dissipation%20in%20electric%20field&f=false , where this is called "the Cottrell method of dust and fog precipitation". The electric field they mention is of the same order of magnitude as what you mention.

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  • $\begingroup$ Interesting possibility. Field values seem to allow a crow discharge. I will think about this carefully, thanks for your tip. $\endgroup$ May 23 '15 at 12:49
  • $\begingroup$ It also would be very valuable if you could sketch what kind of computation should I perform to quantitatively estimate the effect. $\endgroup$ May 23 '15 at 12:55
  • $\begingroup$ @Weather Report: Looks like you should look up "electrostatic precipitation" $\endgroup$
    – akhmeteli
    May 23 '15 at 13:08
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Most likely, you're right. However, it should be noted that following your logic of drops of water should remain on the electrode. I thought so too. However, in all experiments, the fog clears almost instantly. About everything covered droplets of moisture. And electrode remained completely dry. The movement of fog was not to the electrode, and from him. With respect to A. Paley

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according with the that you describe about of the experiment, I can say the following: so that the fog droplets are precipitated to ground, its mass had been increased to form big droplets, otherwise, the droplets will remain suspended on the air . This process is called as coalescense. When their mass increase, they can be precipitated to ground by the action of the gravitational force, forming droplets. Other situations is that the ground of the chamber is to a low potential compared with the corona discharge, attracting the droplets. I recommenment to you that read the next book https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19810018106.pdf

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    $\begingroup$ The electrode is remainted dry because the high energy of the corona discharge, generating hot $\endgroup$ Jun 1 '19 at 23:09

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