[ballenarosada]

If you read the exposition you linked, equation 14 gives an expression for the field which is linear in the area. Again, it's pretty important that the capacitor be infinite in extent, otherwise it behaves differently.

What's the dimension that you're proposing to increase of the capacitor? The total work done across the capacitor will be fixed regardless of distance across.

[T-A]

Eq. (14) is just the expression for a dipole. Keep reading to see the full solution in the simplest case (circular plates), Eq. (19).

Since you insist: your derivation goes wrong right at the start by, as you say, "ignoring curvature effects", i.e. by considering radial distance only. By doing that, you are effectively imposing spherical symmetry; you are not doing parallel plates, you are doing concentric spherical shells. That makes the whole exercise pointless, since it's obvious from symmetry alone that such a device could never produce a net thrust: there is no preferred direction for the thrust to act along.

To answer your final question, just look at Eq. (19): make the plate radius (R) larger.

That answer should also be perfectly obvious from the limit case of infinite plates. A correct derivation for the general case must reproduce that result in that limit. Yours does the opposite; it gets worse the larger you make the capacitor. In the infinite limit, it is infinitely wrong.

[ballenarosada]

You're right, my derivation is mistaken for failing to take the z component of the force. EQ. 19 is more like it although you'll note, also goes as 1/z^3.

You're also right that all that is moot from sheathing. But an ion still begins it's journey out at the bottom of a large potential well; one which is particularly steep because of the debye length, but still just as deep.