Visual walkthrough — Discovery of electron (Thomson, cathode rays), proton (Goldstein), neutron (Chadwick)

Before any symbol appears, meet the cast in plain words:

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Step 1 — The stage: a beam that we can bend

WHAT. A sealed tube with almost no air. A high voltage makes a thin ray shoot from the negative plate (cathode) to the positive plate (anode), hitting a screen and leaving a bright dot.

WHY. We need a steerable beam. If we can nudge the dot up or down with known pushes and measure how far it moves, the amount of movement secretly carries the numbers $e$ and $m$ inside it.

PICTURE. Follow the cyan beam. It enters from the left at speed $v$, passes between two plates (length $L$), and lands on the screen. Right now, with no fields, it flies dead straight to the centre dot (amber).

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Step 2 — The electric push, drawn as an arrow

WHAT. Charge up the two plates. Now there is an electric field $E$ between them. Any charge inside feels a force.

WHY. This is the first of our two tools. We choose the electric force because it acts on charge whether the charge moves or not — it is the simplest handle on the particle.

$\underset{\text{electric force}}{\underbrace{F_E}}=\underset{\begin{array}{c}\text{how much}\\ \text{charge}\end{array}}{\underbrace{e}}\times \underset{\begin{array}{c}\text{how steep}\\ \text{the field}\end{array}}{\underbrace{E}}$

Read it like a sentence: the push equals the charge times the steepness. Double the charge → double the push. Double the field → double the push.

PICTURE. The amber arrow $F_E$ points from the negative particle toward the positive plate. Because the electron is negative, it is pulled up toward the top (+) plate here. The beam starts to bow.

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Step 3 — The magnetic push, and why it is different

WHAT. Now switch on the magnetic field $B$, crossed with $E$ (perpendicular to both the beam and the electric field).

WHY. We need a second, independent force so we can play the two against each other. Magnetism is perfect because it acts only on moving charge — its strength contains the speed $v$ we are hunting for.

$\underset{\text{magnetic force}}{\underbrace{F_B}}=\underset{\text{charge}}{\underbrace{e}}\times \underset{\begin{array}{c}\text{how fast}\\ \text{it moves}\end{array}}{\underbrace{v}}\times \underset{\begin{array}{c}\text{field}\\ \text{strength}\end{array}}{\underbrace{B}}$

The new ingredient is $v$: a stationary charge ($v=0$) feels zero magnetic force. This is the clue that lets us extract speed.

PICTURE. We arrange $B$ so its force (cyan arrow) points opposite to the electric force (amber arrow) — one pushes the beam up, the other down.

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Step 4 — Balance the two, and speed pops out

WHAT. Tune $E$ and $B$ until the dot returns to the exact centre — the beam is undeflected. The two arrows are now equal in length and cancel.

WHY. When up-push = down-push, we can write $F_E=F_B$. Watch what cancels:

$eE=evB$

The $e$ appears on both sides — divide it out. It vanishes. The stubbornness $m$ never even entered a straight-line beam. What is left is pure geometry of the fields:

$\boxed{v=\frac{E}{B}}$

Read it: speed equals the electric steepness divided by the magnetic strength — both of which we set on our own dials and can read off directly.

PICTURE. The two equal opposing arrows and the perfectly straight cyan beam through the centre. The caption shows $e$ crossing itself out — the "magic cancellation."

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Step 5 — Switch off $B$: the beam falls like a thrown ball

WHAT. Turn the magnet off. Only the electric push remains. Inside the plates the particle drifts along at unchanged speed $v$, while being pushed sideways the whole time.

WHY. This is exactly a ball thrown horizontally under gravity — constant forward speed, steady sideways acceleration. We already know that motion, so we can borrow its formula.

First, the time spent between the plates: $\underset{\text{time inside}}{\underbrace{t}}=\underset{\begin{array}{c}\text{length}÷\text{speed}\end{array}}{\underbrace{\frac{L}{v}}}$

The sideways acceleration (from Step 2's force $÷$ stubbornness): $\underset{\text{acceleration}}{\underbrace{a}}=\frac{F_E}{m}=\frac{eE}{m}$

The sideways drop $y$ (the "fall" distance): $\underset{\text{deflection}}{\underbrace{y}}=\frac{1}{2}a{t}^2=\frac{1}{2}\cdot \underset{a}{\underbrace{\frac{eE}{m}}}\cdot \underset{t^2}{\underbrace{\frac{L^2}{v^2}}}$

PICTURE. The parabola. Horizontal cyan arrow = steady $v$; amber arrows growing longer down the path = the accelerating sideways push; the final gap $y$ marked on the screen.

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Step 6 — Solve for the prize: $e/m$

WHAT. Rearrange the deflection formula to put $e/m$ alone on one side.

WHY. Every other symbol — $y$, $E$, $L$, $v$ — is measurable. So whatever multiplies them is the ratio.

Start from Step 5: $y=\frac{1}{2}\frac{e}{m}\frac{E{L}^2}{v^2}$

Multiply both sides to free $e/m$: $\boxed{\frac{e}{m}=\frac{2y{v}^2}{E{L}^2}}$

Then swap in $v=E/B$ from Step 4 if you prefer field-only quantities: $\frac{e}{m}=\frac{2y(E/B{)}^2}{E{L}^2}=\frac{2yE}{B^2{L}^2}$

Term by term: bigger deflection $y$ → bigger ratio (a light or highly-charged particle bends more); bigger $E$ or longer plates $L$ → the same particle bends more, so we divide those out to isolate the particle's own property.

PICTURE. The three measured quantities ($y$, and the dials $E$, $B$, $L$) feeding into the box that outputs the number.

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Step 7 — The edge cases (never leave a scenario unshown)

WHAT & WHY. A good derivation must survive its extremes. Here is what each degenerate input does.

• $v=0$ (particle at rest): magnetic force $F_B=evB=0$. Nothing to balance — Step 4 breaks. You cannot measure speed by balancing a stationary charge. This is why the beam must be flying.
• $B=0$ with $E$ on: we are back in Step 5 — pure parabola, maximum bend. This is the deflection stage.
• $E=0$ with $B$ on: magnetic force alone curves the beam into a circular arc (magnetic force always turns, never speeds). A valid alternative method, but then $v$ is unknown again.
• Neutral particle ($e=0$): both forces vanish — the beam ignores every field. Precisely how Chadwick later spotted the neutron: no deflection at all.
• Heavier particle (bigger $m$): for the same push, $a=eE/m$ shrinks, so $y$ shrinks — a heavy positive ion barely bends. This is why canal-ray (proton) deflections are tiny and depend on the gas.

PICTURE. Four beams overlaid: neutral (straight through, amber), light electron (big bend), heavy ion (small bend), and pure-$B$ circular arc.

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The one-picture summary

Everything in a single frame: balance to get $v$ → unbalance to get $y$ → combine to get $e/m$.

Recall Feynman retelling — the whole walkthrough in plain words

Picture a glowing dot flying down a tube. I put two charged plates around it — the dot bends toward the plus plate, so it's negative. Then I add a magnet, arranged to shove it back the other way. I fiddle with both knobs until the dot sits dead centre again: now the electric shove and the magnetic shove are exactly equal. Setting them equal, the charge cancels off both sides and I'm left holding the speed, $v=E/B$ — just two dial readings. Next I kill the magnet. Now only the electric plate pushes sideways, and the dot arcs like a thrown ball, landing a little off-centre by an amount $y$. I know that "thrown ball" formula, so from how far it fell I read off charge-over-mass: $e/m=2y{v}^2/(E{L}^2)$. Same number for hydrogen, helium, any gas, any metal — so it's one universal particle, the electron. And the beautiful part: I never needed to know the charge by itself, and a truly neutral particle would sail straight through both fields — which is exactly how the neutron was caught years later.

Recall

Why does balancing $E$ and $B$ give speed and not $e/m$? ::: Setting $eE=evB$ cancels the charge $e$; mass never enters a straight beam, leaving $v=E/B$. Why is the deflected path a parabola, not a slant? ::: Constant sideways force → constant acceleration → $y\propto t^2$, a parabola. What happens to the beam if the particle is neutral? ::: Both $F_E=eE$ and $F_B=evB$ are zero, so it passes undeflected — the neutron signature. In $e/m=2y{v}^2/(E{L}^2)$, why divide by $E$ and $L^2$? ::: A bigger field or longer plates bend any particle more; dividing them out isolates the particle's own $e/m$.

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Connections

• Millikan Oil Drop Experiment — measures $e$ alone, so mass $m=e÷(e/m)$.
• Charge to Mass Ratio — the general technique this page derives.
• Thomson Plum Pudding Model — the model built once the electron was found.
• Rutherford Nuclear Model — where these particles ended up.
• Isotopes and Mass Number — why the neutron's missing mass mattered.