Visual walkthrough — FEEP, MEMS thrusters — micro-propulsion

We will build up in this order: a charged liquid on a tip → the cone it forms → one ion falling through a voltage → its final speed → a whole stream of ions → the force that stream delivers → the edge cases. Nothing is used before it is drawn.

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Step 1 — A liquid metal sitting on a sharp needle

WHAT. Picture a tiny drop of liquid metal (say indium) resting on the very tip of a needle, inside a spacecraft. Nothing is happening yet — no field switched on. The drop is rounded because of surface tension.

WHY start here. Every symbol later (the field $E$, the voltage $V$, the ion) is born on this tip. If we don't picture the surface first, the cone in Step 2 appears from nowhere.

PICTURE. Look at figure s01. The blue blob is the liquid; the grey wedge is the needle. Two words to define right now:

At this stage there is no electricity. The drop is a calm rounded cap. Remember its shape — the next step deforms it.

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Step 2 — Switch on the field: the Taylor cone

WHAT. Now we place a second metal plate a short distance away and put a big voltage across the gap. This creates an electric field $E$ that pulls the positive surface charges outward.

WHY this step. Two opposite pulls on the same surface will settle at a shape where they balance. That balance shape is not a round cap — it is a cone. We need to see it because the sharp cone tip is where ions escape.

PICTURE. In figure s02 the inward mint arrows are surface tension; the outward lavender arrows are the electric pull. Where they balance the surface pinches into the Taylor cone — a cone with a half-angle of about $49.3^{\circ }$ (see Taylor Cone and Electrospray & Colloid Thrusters).

The outward electric pull acts like a pressure on the surface: $P_{\text{elec}}=\underset{\text{geometry}}{\underbrace{\frac{1}{2}}}\underset{\begin{array}{c}\text{vacuum}\\ \text{constant}\end{array}}{\underbrace{{\epsilon }_0}}\underset{\begin{array}{c}\text{field}\\ \text{squared}\end{array}}{\underbrace{E^2}}$

Each piece: the $\frac{1}{2}$ is a fixed geometric factor for field pressure; ${\epsilon }_0$ is the constant of free space ($8.85\times 10^{-12}$); $E^2$ says the pull grows with the square of the field, so doubling the voltage quadruples the outward push. When this pressure matches the inward pull of $\gamma$, the cone is stable.

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Step 3 — One ion tears off the tip

WHAT. At the needle-sharp apex the field is so strong it pulls one ion off the liquid surface. Call its charge $q$ and mass $m$. The moment it leaves, it is essentially at rest.

WHY this step. We can't compute the speed of "the beam" all at once — that's too much. So we zoom in to one ion and ask: how fast does this single ion end up going? Get one right, and a stream is just many copies.

PICTURE. Figure s03 shows the lone coral dot leaving the apex, with the field lines (lavender) pointing the way it will accelerate.

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Step 4 — The ion falls through a voltage $V$ and picks up speed

WHAT. The ion, once free, is dragged across the voltage gap $V$. Crossing that gap converts stored electric potential energy into motion energy (kinetic energy).

WHY energy conservation, and not force × time? We could use $F=ma$ and track the ion step by step through a messy, non-uniform field — painful. Energy conservation skips all of that: it cares only about the start (at rest, energy $qV$ available) and the end (moving, energy $\frac{1}{2}m{v}_e^2$). No path details needed. That's the tool for the job.

PICTURE. Figure s04 is the "energy ramp": the ion starts high (all potential energy $qV$) and arrives at the bottom (all kinetic energy $\frac{1}{2}m{v}_e^2$), like a marble rolling down a slope.

$\underset{\begin{array}{c}\text{energy given}\\ \text{by the voltage}\end{array}}{\underbrace{qV}}=\underset{\text{½}}{\underbrace{\frac{1}{2}}}\underset{\text{ion mass}}{\underbrace{m}}\underset{\begin{array}{c}\text{exit speed}\\ \text{squared}\end{array}}{\underbrace{v_e^2}}$

Solve for the exit speed $v_e$ (the exhaust velocity):

$\boxed{v_e=\sqrt{\frac{2qV}{m}}}$

Term by term: the $2$ comes from cancelling the $\frac{1}{2}$; $q$ and $V$ on top say more charge or more voltage → faster; $m$ underneath says heavier ions → slower (they're harder to speed up). The square root is there because the energy held $v^2$, not $v$.

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Step 5 — From one ion to a whole stream: current $I$

WHAT. A thruster fires not one ion but a steady stream. Two ways to count the stream:

• by mass per second, the mass flow $\dot{m}$ ($\text{kg/s}$);
• by charge per second, the beam current $I$ ($\text{amps}=\text{C/s}$).

WHY link them. We can measure and dial the current $I$ electronically (that's what makes FEEP precise), but thrust is about momentum, which needs mass. So we must translate current into mass flow.

PICTURE. Figure s05: a conveyor of ions. Each ion contributes $q$ to the charge count and $m$ to the mass count. In a small time $\Delta t$, $N$ ions leave, giving current $I=Nq/\Delta t$ and mass flow $\dot{m}=Nm/\Delta t$. Divide:

$\frac{\dot{m}}{I}=\frac{Nm/\Delta t}{Nq/\Delta t}=\frac{m}{q}⟹\dot{m}=\frac{m}{q}I$

Reading it: $\frac{m}{q}$ is the "mass carried per unit charge." Multiply the current by it and you convert charge-per-second into mass-per-second.

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Step 6 — The thrust: momentum leaving each second

WHAT. Thrust is Newton's third law in action: the ions shoot out one way, the spacecraft is pushed the other. The push equals the momentum carried away per second.

WHY combine now. We have every piece: $v_e$ (Step 4) and $\dot{m}=\frac{m}{q}I$ (Step 5). Substitute both into $F=\dot{m}{v}_e$ and simplify.

PICTURE. Figure s06: the beam pushing the satellite, with the substitution chain shown as arrows. Watch the algebra:

$F=\dot{m}{v}_e=\underset{\dot{m}}{\underbrace{\frac{m}{q}I}}\cdot \underset{v_e}{\underbrace{\sqrt{\frac{2qV}{m}}}}=I\sqrt{\frac{m^2}{q^2}\cdot \frac{2qV}{m}}=I\sqrt{\frac{2mV}{q}}$

$\boxed{F=I\sqrt{\frac{2mV}{q}}}$

Term by term: $I$ out front — more ions per second, more push (this is the knob we turn). $V$ under the root — higher voltage, faster ions, more push. $m$ under the root — heavier ions carry more momentum per charge, so more thrust per amp (but lower $v_e$, so lower $I_{sp}$: the trade-off the parent flagged). $q$ underneath — more charge per ion means fewer kg per amp, less thrust.

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Step 7 — Edge cases: what if a knob goes to zero (or huge)?

WHAT. A formula is only trustworthy if it behaves sanely at its extremes. Let's push each knob.

WHY. The reader must never meet a case we didn't check. Micro-thrusters really do get dialled to near-zero current for the gentlest nudges — so the limits are physical, not academic.

PICTURE. Figure s07 plots $F=I\sqrt{2mV/q}$ against $I$ (a straight line) and against $V$ (a $\sqrt{V}$ curve), with the special points marked.

• $I\to 0$ (no ions fired): $F\to 0$. No stream, no push. Correct — the thruster is idling. This is exactly how FEEP reaches sub-µN resolution: turn $I$ down to nanoamps.
• $V\to 0$ (no accelerating voltage): $v_e=\sqrt{2qV/m}\to 0$ and $F\to 0$. Ions barely move; no useful push. You need the kilovolts.
• $V$ very large: $F$ grows like $\sqrt{V}$ — slowly. Doubling $V$ only multiplies thrust by $\sqrt{2}\approx 1.41$. So you can't cheat huge thrust from voltage alone; the honest way to more thrust is more current $I$ (a straight line, no diminishing returns).
• Heavier ion, $m$ large: $v_e\downarrow$ (worse $I_{sp}$) but $F$ per amp $\uparrow$. Neither "always good" nor "always bad" — a genuine trade. This is why the parent warns against calling heavy ions useless.

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Worked numbers, re-derived on the picture

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The one-picture summary

Figure s08 compresses the entire chain onto a single strip: surface → Taylor cone → one ion → falls through $V$, gains $v_e$ → stream of current $I$, mass flow $\frac{m}{q}I$ → thrust $F=I\sqrt{2mV/q}$. Follow the arrows left to right and you have re-derived FEEP.

Recall Feynman retelling — say it in plain words

A blob of liquid metal sits on a needle. Switch on a strong electric field: the field pulls the surface outward, surface tension pulls it inward, and where they balance the blob pinches into a sharp cone. From that sharp tip the field yanks off individual charged atoms — ions. Each ion, once free, slides "downhill" through a voltage $V$; energy conservation says the energy it's given, $qV$, becomes motion energy $\frac{1}{2}m{v}_e^2$, so it exits at $v_e=\sqrt{2qV/m}$ — tens of km/s. Now fire a whole stream: the current $I$ counts charges per second, and since each ion carries mass $m$ per charge $q$, the mass leaving per second is $\frac{m}{q}I$. Thrust is just momentum leaving per second, mass-flow times speed, which tidies up to $F=I\sqrt{2mV/q}$. Turn the current down to nanoamps and the push drops to sub-micro-newtons — precise enough to nudge a floating leaf.

Recall Self-test

Why energy conservation instead of $F=ma$ for the ion? ::: We only need start and end energies, not the messy non-uniform field along the path. Which knob sets thrust, which sets $I_{sp}$? ::: Current $I$ sets thrust; voltage $V$ sets $v_e$ and hence $I_{sp}$. Why does doubling $V$ not double thrust? ::: Thrust $\propto \sqrt{V}$, so it only rises by $\sqrt{2}\approx 1.41$. What happens to $F$ as $I\to 0$? ::: $F\to 0$ — this is how FEEP reaches sub-µN resolution.