Let me show a figure first about the shape of flame for a candle without any external electric field in the dark night:

enter image description here

Now consider the following physical configuration: If you put this lighted candle into the parallel plate capacitor, in which the electric field produced by it can be turned up or turn down arbitrarily.

So my question is: what's the shape of the flame of the candle when we keep the electric field stationary and how is it changed as we increased the field or decrease the field yielded by the parallel plate electrical capacitor?


According to http://archive.iypt.org/iypt_book/2011_3_Bouncing_flame_Iran_RMN_HA_RA_v3.pdf , in a typical scenario, electrons and positive ions in the flame plasma move to opposite plates of the capacitor, electrons leave the flame area and form negative ions with oxygen atoms, the flame gets positively charged and attracted to the negatively charged plate, as you can see in numerous videos. In some videos, though, the flame is "divided" in two parts attracted to opposite plates, so in such cases the negative ions probably remain within the flame.

  • $\begingroup$ what are positive ions and negative ions in flame and what is flame ? $\endgroup$ – user3728644 Aug 4 '19 at 12:28

To understand what will happen to the flame, we should first know what a flame is actually made up of, so sit tight!. A flame is basically a partial plasma state, in plasma electrons are ripped off the atoms and +ve and -ve charges are separated, same things happen in a flame but in much less amount. In a flame reactions occur on molecular level, wax vapors, carbon, hydrogen and oxygen slam into one another and form new molecules, in some ( it's probabilistic ) cases electrons are ripped off due to extreme thermal energies, these electrons then become free and roam here and there in the flame. So when you put a flame inside a capacitor plate electric field between the plates attracts those funky electrons and the flame gets tilted towards the $+ve$ side, now you may ask "Hey there are $+ve$ charges in there too so shouldn't it tilt towards $-ve$ side too ?", I think probability of $+ve$ charge tilt is low as nucleus is very stable and can't move here and there, so net effect is that some electrons in the flame(which are ionized, not excited ones.) are attracted towards the $+ve$ plate and the flame tilts, now if the voltage between the electrons (present in the flame) and the $+ve$ plate of the capacitor is enough to cause air breakdown, then there will be a spark between the flame and the $+ve$ plate and surely it will look $amazing$, it would look like if flame is being transmitted through the plates, WOW!, Hope that answers your question. :)


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