A basic ion engine uses a positively charged plate and passes positively charged ions over it to accelerate them. How does the charged plate draw current to accelerate the ions? It seems that a positively charged ion passing over the plate would just cause an oscillation in the charges in the plate, and not a current.

  • $\begingroup$ Although there are some errors in this question's description (typical ion engines have two grids, a screen, charged positive, and an accelerator, charged negative, though some have more), the basic question is a good one that delves into the fundamental physics of acceleration of charged particles. When a static electric field (which requires no current to maintain) accelerates a charged particle, the particle gains kinetic energy. Where does that energy come from? $\endgroup$ – Tom Spilker May 4 '18 at 3:52

Rephrasing to "what's the source of the kinetic energy when a bit of charge is accelerated by the electric field between two plates?"

The energy comes from the decrease in voltage across the capacitor formed by the two plates or, if there's a voltage source (i.e. battery or power supply attached) it comes from that source.

To see that, consider the two plates as a parallel-plate capacitor of capacitance $C$, initially with voltage $V$. The charge on each plate (plus on one, minus on the other) necessary for that voltage is $Q = CV$.

Now move a small charge $\Delta Q$ from one plate to the other. The small charge gains energy $V \Delta Q$. But now the plates have only charge $Q-\Delta Q$. The energy stored is in general $E = Q^2/(2C)$, so it becomes $(Q-\Delta Q)^2/(2C)$ which (for small $\Delta Q$) becomes

$(Q-\Delta Q)^2/(2C) = Q^2 (1-\Delta Q/Q)^2/(2C) \simeq Q^2/(2C) - Q\Delta Q/C = Q^2/(2C) - V\Delta Q$

which means the energy lost by the plates is $V\Delta Q$, exactly equal to what was gained by the moving charge.

If there's a battery or power supply to keep the voltage $V$ constant, it'll have to add charge to replace the $\Delta Q$ that moved, and it will have to provide $V \Delta Q$ energy to do that. In that case, the energy comes out of the battery, and goes back into the capacitor to make it even.

There's also some interesting of bits physics arising from the charge moving past the plate and continuing off into space, but that's probably another question.

| improve this answer | |
  • $\begingroup$ Bob's answer is exactly right (good job!) if the first plate, the one the particle is leaving, charges the particle, i.e. part of the charge from that plate goes to the particle, and that charge is then deposited on the second plate. But for ion engines, the ionization (charging) doesn't occur at the first plate (screen grid), it occurs farther upstream, so it doesn't take any charge from that grid. And designers actively try to avoid having the ions inpinge on the accelerator grid (largely successfully), so no intentional charge transfer there. I'll add an answer. $\endgroup$ – Tom Spilker May 4 '18 at 22:23

The source of the accelerated ion's energy involves the electric potential field in the engine. Comparison to a more intuitive potential field, the gravitational field, can help with the understanding.

Imagine a mass suspended beneath a balloon, at equilibrium so its altitude is constant. This system is within the Earth's gravitational potential field, where the gravitational potential is low (very negative) at the surface, grading to higher (less negative) as you go up. The kinetic energy of both parts, the balloon and the mass, is zero. The upward force exerted by the balloon is equal in magnitude to the downward force on the mass. Now if you cut the cable connecting them, the mass screams downward and the balloon zooms upward. The mass responds to the gradient in the surrounding potential field, moving from higher potential to lower potential, and accelerates downward. The balloon, with a "negative mass", accelerates upward.

[I know, I know, somewhat contrived: you have to treat the surrounding atmosphere as zero mass density to get the balloon's lower mass density to appear as a "negative mass", but I have to do something to connect the analogy with the +/- polarity dichotomy in electrodynamics that gravity doesn't have!]

Where did the kinetic energy of the falling mass come from? Answer: gravitational potential energy. Merely as a result of its position within the potential field it had that potential energy. Cutting the cable allowed the mass to move from higher potential to lower potential, converting the potential energy into kinetic energy.

Now look at the ion engine and its electrical potential field. For ease of making this analogy, point the engine's exhaust plume downward. In the discharge chamber and at the screen grid the electrical potential is high and positive. At the accelerator grid the potential is negative, so there's a strong decrease in potential between the chamber and screen and the accelerator. Having an atom in the chamber is like having the connected balloon and mass in the gravitational potential field: the positive charges want to "fall" in one direction (downward, toward the grids and space) and the negative (buoyant?? ;-)) charges want to float upward, toward the anode. As long as the charge on the atom is neutral you have equal numbers of positive and negative charges, so the net force from the potential gradient is zero, and the atom doesn't accelerate.

But when you knock an electron from the atom the equilibrium is ended. You've "cut their cable". The positive ion accelerates downward, in the direction of decreasing potential, just like the mass from the balloon analogy. And the electron heads upward. Unlike the gravity example, where the change in the gravitational potential is a smooth function of altitude, the electrical potential field is more uneven: the gradient is much, much stronger between the two grids, so the bulk of the acceleration happens there.

Note: no charges were transferred from or to either of the grids in this interaction. All the ion's kinetic energy came from its fall through the potential field.

Note 2: "...in this interaction,", i.e. when the ion successfully goes through grid holes and doesn't impact a grid. In reality, sometimes charged particles will impact the grids, and charges will be moved around. This is not always by a simple charge transfer, where an impacting positive ion steals an electron from a grid atom, changing the grid's charge and thus necessitating a bit of power from the power supply. For example, the ion can steal one and also knock another one off into space, for a two-charge interaction that requires double the power to restore.

The preceding in no way implies that no currents are required in an ideal ion engine. The exhaust plume is a directed flow of positive charges, and that's a current. I calculated the beam current for an ideal Xenon engine with a beam velocity of 30 km/s (Isp = 3059) and thrust of 1 Newton: 24.5 Amps. For each singly-ionized Xenon ion, an electron must go into the discharge chamber's anode, so right there is 24.5 Amps into the anode. But if the engine uses the Kaufman method, a hollow cathode shoots electrons into the chamber to knock other electrons from the Xenon, and those shot-in electrons also must find their way to the anode (precious few make their way into space!), increasing the anode current. For each electron taken from a Xenon atom, an electron must be fired from the beam neutralizer outside of the grids, so there's the 24.5 Amps again. And those 24.5 Amps of electrons must be pumped from a very high positive potential at the anode to a very low potential at the neutralizer. Electrons naturally want to flow the other way, like in the chamber where they head for the anode, so pumping that much current that direction takes a lot of power. If you use NSTAR-like specs that potential difference is somewhere around 1200-1300 Volts, requiring a power of ~31 kW. This doesn't include any of the inefficiencies and losses a real engine must contend with, so the power you'd have to supply to a real engine would be larger.

| improve this answer | |

I think the point of the question is that the ions do not touch the positive screen, which is true, by and large. They also don't touch the negative grid which has already been mentioned. However, they have to come from somewhere. So there is an ion source, responsible for ionising the atoms and releasing them close to the positive plate. There also has to be an electron source, since a beam of electrons is sent out somewhere to stop the spacecraft building up too much negative charge. So the current flows between those two sources. The energy is required to make that current flow, to raise the ions to sufficient positive potential or the electrons to sufficient negative potential.

| improve this answer | |

Let's see if we can find the simplest, yet valid, way to look at this.

Invoke Newton's principle of action and reaction. Introduce an electron between two plates, one positively charged and the other negatively charged, which is the configuration of a capacitor. The electron will experience a force directed to the positive plate (action), and the two plates will experience equal and opposite forces (reaction): the negative plate force is in the direction away from the electron, and the positive plate force is in the direction of the electron. Thus, both plates experience the same force in the same direction.

The force on the electron accelerates the electron towards the positive plate, and the force on the plates accelerate both plates in the direction opposite to that of the force on the electron. By "plates," I mean here all the apparatus connected to the actual capacitor plates. Because of the relative masses, of course the electron accelerates much more than the plates. In these moments of electron acceleration, there is no change in the electric field set up by the two plates.

There is only the gain of kinetic energies of both the electron and the two plates, which must be equal, from conservation of energy. When the electron hits the positive plate, the previous forces experienced by the electron and the plates cease. The negative electron charge then reduces the charge on the positive plate, and the reduced charge on the positive plate then causes a reduction of the negative charge on the negative plate, because of the way capacitor charges are established, by induction. In addition, the momentum of the electron causes a momentum change in the negative plate, reducing the motion that it experienced during the acceleration stage of this process. But the positive plate will continue its motion, which achieves an increase in the separation distance between the two plates.

The net effect is to cause the stored charge of the two plates to be reduced and with an increase in separation distance. Both these results drops the stored charge - energy - in the capacitor (two charged, separated plates).

From this line of reasoning, there's no way to explain a simple transfer of solely electrical energy between the plates and the electron, since mechanical, electrical, and heat (internal) energy of the plates are all involved. (The stage where the electron becomes incorporated into the surface of the positive plate can be considered to be a dissipation of energy.) I presume that the law of Conservation of Energy does apply here, and further investigation would be required in order to prove that is indeed the case.

| improve this answer | |
  • 1
    $\begingroup$ Sources, or people might think you're making it up as you go along. $\endgroup$ – JCRM May 11 '18 at 1:32
  • $\begingroup$ @JCRM none of these answers are sourced ._. Except the one Tom Spiker commented on, I suppose he's a source. $\endgroup$ – Magic Octopus Urn May 11 '18 at 14:50
  • $\begingroup$ @MagicOctopusUrn the other answers have the advantage of going along with my understanding of how an ion engine works. $\endgroup$ – JCRM May 11 '18 at 15:53
  • 1
    $\begingroup$ JCRM, "making it up as you go along"? My approach is to use step by step deductive reasoning, invoking the most basic principles of physics. I suppose my references would be any undergraduate fundamental College textbook on classical physics. The Feynman Lecture series would be one example. I think a reference would be necessary if I invoke knowledge beyond basic concepts, but I don't do that. $\endgroup$ – ttonon Mar 21 at 4:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.