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If I have let’s say an ionized gas, composed purely of singly charged positive ions; what would be the permitivity of the material? Is it infinite, as is the case with a perfect conductor, or am I missing out on something here?

Let’s say I place this ionized gas in a STATIC electric field, then the ions would rearrange in such a way to cancel the field, and hence the permitivity constant is essentially infinite - is my reasoning correct?

Now consider the case with an OSCILLATING electric field - how would the dielectric permitivity vary then for the ionized gas?

Just to be clear, I am NOT referring to a quasi-neutral plasma: I am referring to let’s say a cloud of O+ ions, with the ejected electrons completely removed from the cloud.

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  • $\begingroup$ Interesting question, and I don't know the answer, but my thoughts go to kinetic dispersion relations of waves in plasmas. Perhaps one could try to linearize the Vlasov-Poisson equations (assuming electrostatic approximation) and derive an expression for the permittivity? Works for electron plasma waves. Perhaps also in the absence of the electrons? $\endgroup$ – sigvaldm Jul 15 '19 at 11:58
  • $\begingroup$ Interesting suggestion, will research. Thank you! $\endgroup$ – Joeseph123 Jul 15 '19 at 13:57
  • $\begingroup$ @Joseph123: You're welcome. $\endgroup$ – sigvaldm Jul 15 '19 at 15:03
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I am referring to let’s say a cloud of O+ ions, with the ejected electrons completely removed from the cloud.

You're describing a bomb. This system will undergo what's known as a Coulomb explosion, and if the quantity of ions in this cloud is macroscopically significant, then the amount of energy that this explosion will detonate is on magnitudes that you only reach with nuclear weapons.

Worrying about the behaviour of this system under external fields is pretty much completely pointless.

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  • $\begingroup$ Coulomb explosion occurs when in the presence of a source of electron ablation, either a laser or electron beam ~50-70eV. That requires a spacing between two molecules in the order of ~0.2-3nm (i.e. a solid lattice.) $\endgroup$ – Joeseph123 Jul 15 '19 at 10:26
  • $\begingroup$ The ionized gas in my case is bound by a container of equal and opposing charge, so ignore the details for now. Assume an ideal case. $\endgroup$ – Joeseph123 Jul 15 '19 at 10:28
  • $\begingroup$ This does not answer the OP's question. $\endgroup$ – sigvaldm Jul 15 '19 at 12:01
  • $\begingroup$ @sigvaldm The OP's question pertains to an unphysical scenario and is essentially meaningless. There is no other answer. $\endgroup$ – Emilio Pisanty Jul 15 '19 at 12:02
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    $\begingroup$ Strictly speaking, this scenario does occur in magnetrons, cyclotrons, ion beam collimators etc...where the coulombic repulsion (not explosion.) does have a significant effect which is cancelled out by oscillating electric fields $\endgroup$ – Joeseph123 Jul 15 '19 at 13:58
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There are two cases for plasma (which would be your gas if it had electrons):

  1. low frequency EM waves, the permettivity is infinite meaning that these EM waves would reflect from the edge of this media

  2. high frequency EM waves, the permettivity is lower then infinite (like metals), so these EM waves could propagate through this gas

Note, from Eq. (1151), that the plasma frequency is proportional to the square root of the number density of free electrons.

So basically if this gas (which has no electrons) could exist for some time, and not explode, then EM waves could propagate in it (like in air), permettivity would be around 1.

The dialectric constant means the ability of a medium to separate its charges inside it. Since your gas has no separable charges, it has no ability to separate them, and does not need to separate them at all. EM waves could propagate in it like in air (close to vacuum).

http://farside.ph.utexas.edu/teaching/em/lectures/node100.html

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  • $\begingroup$ The equation above is valid for a quasi-neutral plasma as far as I understand, so the number density of electrons is the same as that of the positive ions. Substituing n=0 wouldn’t work in this case, as then the plasma would be non-existent, which would give you back the permitivity of free space. $\endgroup$ – Joeseph123 Jul 15 '19 at 14:47
  • $\begingroup$ @Joeseph123 correct. very close to free space. $\endgroup$ – Árpád Szendrei Jul 15 '19 at 15:02
  • $\begingroup$ The plasma wouldn’t exist as in neither the free electrons nor the positive ions would be present so you still have a vacuum. $\endgroup$ – Joeseph123 Jul 15 '19 at 15:05
  • $\begingroup$ @Joeseph123 it is only about the separation of charges. In your ion gas there is nothing to separate, and there is no need to either, so EM waves would just travel through it, so permettivity is close to 1. $\endgroup$ – Árpád Szendrei Jul 15 '19 at 18:11
  • $\begingroup$ To be clear I’m referring to the real component of permitivity - the electrostatic case - that is. If you place the ionized gas in a field, the charges would instantly be pulled in the direction of the electric flux, thus cancelling the field, resulting in a theoretically infinite dielectric constant I assume. Is this the case, or am I missing out on something? $\endgroup$ – Joeseph123 Jul 15 '19 at 21:35

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