> Regarding multipole expansions: I love multipole analysis, it's a beautiful way to study charge distributions. That said, I think it's massive overkill to actually use it in this case, since at least for circular plates along the axis that process is far more complicated than just using the exact fields themselves.

Yes, yes, in this case the multiple expansion is a conceptual tool which can be used to understand why things work as they do (and it originally entered the discussion with ballenarosada because he or she was trying to use just the dipole term to model a plate capacitor).

The exact solution can be looked up e.g. in the pedagogical presentation which I linked to in that thread. I see now that the original link no longer seems to work, but this one does:

https://www.researchgate.net/publication/228807623_Electric_...

Cutting to the chase: you are still not accounting for plasma screening. As the ion leaves the plate, it's surrounded by a cloud of oppositely charged ions which reduce its effective charge, i.e. its coupling to the plate's electric field. The further away it moves, the smaller its charge seems to be.

To account for that, include a Debye factor: the coupling between electric field E and ion with naked charge q goes from E * q to

E * q * exp(-r/l_D)

where r is distance from the plate and l_D is the Debye length [1].

As you say, integrating the equations of motion for the correct E is a lot of work, but we don't need to: it's quite enough to agree that the field is falling by some power of distance from the plate. The exponential then wins in very short order. At a separation of just ten Debye lengths, the ion's effective charge has been reduced by a factor 0.000045; at a separation of one hundred Debye lengths, the factor is 3.72e-44, for all practical purposes zero.

When the Debye length is short compared to the field gradient (i.e. to the size of your capacitor) you can just let E be constant. It's then trivial to integrate the work done by the electric field as the ion moves out; even if you let the ion run all the way to infinity, the work done is just

E_0 * q * l_D

where E_0 is the electric field at the plate.

Compare with the work done inside the capacitor (where E is actually constant) and you see that even using this maximally pessimistic approximation of constant external field, you've only shaved off one Debye length's worth of the capacitor's plate separation from the total work done on the ion.

In low Earth orbit that's one to a few millimeters [1].

Regarding this,

> rather than falling off to zero very slowly over a long distance as in the unscreened case (giving rise to a small but wide electric field as the gradient), the potential must fall to zero very rapidly over the short screening distance (giving rise to a large but short-ranged electric field)

I think the easy way to think about it is to view the ion as being screened. The plates will of course also attract oppositely charged ions, but they are connected to a generator (solar panels, nuclear reactor...) which is constantly pumping electrons accruing around the anode to the cathode, where they recombine with protons accruing around the cathode. As long as that's running, you have a steady state situation where the net charge of each plate is somewhat reduced relative to the vacuum case, but the field can still be approximated by the vacuum solution (using the reduced net charge). For the ion, this is clearly not true; it is not connected to a generator and will just behave like any other ion in the plasma.

[1] https://en.wikipedia.org/wiki/Debye_length