Thomson's plum-pudding model: Visual walkthrough — Chemistry deep dive

Step 1 — Draw the pudding and place one electron

WHAT. Picture a solid ball of positive charge. It is not a point and not a shell — it is a filled ball, the same "positive-ness" packed at every point inside. We call its radius $R$ (the distance from the dead centre to the surface). Somewhere inside sits one tiny electron, at a distance $r$ from the centre.

WHY. Before we can compute any force we must agree on the setup. The whole model hinges on the positive charge being spread out evenly (that word "evenly" is the pudding's defining property — see the parent). Everything below is a consequence of that one word.

PICTURE. Look at the figure. The blue disc is the pudding, radius $R$ (blue arrow). The orange dot is the electron. The green arrow $r$ is how far out from the centre it currently sits.

Step 2 — Only the charge inside $r$ pulls — the shell trick

WHAT. Split the pudding into two parts: the positive charge that lies closer to the centre than the electron (inside a ball of radius $r$ ), and the rest that lies farther out (a thick shell between $r$ and $R$ ). Claim: the outer shell pulls the electron with exactly zero net force. Only the inner ball matters.

WHY. This is Gauss's law doing its one job: for anything spherically symmetric, the outer shell's pulls from all directions cancel perfectly, and the inner ball acts as if its whole charge sat at the centre. We use Gauss here (rather than adding up Coulomb forces piece by piece) precisely because that sum is horrible and Gauss makes the cancellation obvious. It answers the exact question we have: "how much charge effectively acts on the electron at radius $r$ ?"

PICTURE. In the figure, the inner region (darker blue, radius $r$ ) is the part that pulls. The outer shell (light blue) is shaded out — its tugs cancel. The red arrows on opposite sides of the shell show equal-and-opposite pulls killing each other.

Step 3 — How much charge is enclosed? Count by volume

WHAT. Compute $q_{in}$ , the positive charge inside radius $r$ . Since the charge is spread evenly, it is shared in proportion to volume.

WHY. "Evenly spread" means charge-per-unit-volume is the same everywhere. So the fraction of charge inside is the fraction of volume inside. A ball of radius $r$ has volume $\frac{4}{3} π r^{3}$ ; the whole pudding has $\frac{4}{3} π R^{3}$ . This step turns the word "uniform" into an equation.

PICTURE. Two nested balls; the small one is a fraction $r^{3} / R^{3}$ of the big one's volume, and carries that same fraction of the total charge $+ e$ .

q_{in} = total positive charge e \cdot fraction of the volume inside r \frac{\frac{4}{3} π r ^{3}}{\frac{4}{3} π R ^{3}} = e \frac{r ^{3}}{R ^{3}}

Term by term: $e$ is the pudding's whole positive charge; the ratio is "small ball ÷ big ball"; the $\frac{4}{3} π$ cancels top and bottom, leaving the clean cube ratio $r^{3} / R^{3}$ .

Recall Sanity check the two ends

Put $r = 0$ : $q_{in} = 0$ (no charge enclosed at the exact centre). Put $r = R$ : $q_{in} = e$ (you've enclosed the whole pudding). Both correct — good sign.

Step 4 — The field at the electron — watch $r^{2}$ meet $r^{3}$

WHAT. Find the electric field $E$ at radius $r$ . "Field" = the pull-per-unit-positive-charge that the enclosed ball produces at the electron's location.

WHY. We split into two moves on purpose: first find the field (independent of what charge feels it), then multiply by the electron's charge (Step 5). Gauss again lets us treat $q_{in}$ as a point at the centre, so the field is just Coulomb's $q / r^{2}$ shape — and now the magic happens.

PICTURE. The enclosed charge collapses to a point at the centre; a single field arrow of length $E$ points outward at the electron.

E (r) = Coulomb constant \frac{1}{4 π ε _{0}} \frac{q _{in}}{r ^{2}} = \frac{1}{4 π ε _{0}} \frac{e r ^{3} / R ^{3}}{r ^{2}} = \frac{1}{4 π ε _{0}} \frac{e r}{R ^{3}}

The star of the show: $\frac{r ^{3}}{r ^{2}} = r$ . The cube from "charge grows with volume" beats the square from "field weakens with distance," and one power of $r$ survives. So the field grows linearly from the centre — the exact opposite of the outside-a-charge $1/ r^{2}$ you may expect.

Step 5 — Turn the field into a force, and read off the minus sign

WHAT. The electron carries charge $- e$ . Its force is $F = (charge) \times (field) = (- e) E$ .

WHY. Field tells us the pull on a positive test charge (outward, away from centre). The electron is negative, so it feels the opposite way — inward, toward the centre. That inward direction, no matter which side the electron is on, is what makes the force restoring.

PICTURE. Electron on the right of centre → force points left (toward centre). Electron on the left → force points right. Always back toward home. This is the "spring" behaviour drawn out for both signs of $r$ .

F = (- e) E = - call this k \frac{e ^{2}}{4 π ε _{0} R ^{3}} r = - k r

The minus = "opposite to the displacement $r$ " = points home.
$e^{2}$ = electron charge times pudding charge (both magnitude $e$ ).
$R^{3}$ downstairs = the pudding's fixed size; a bigger, fluffier atom gives a gentler spring.

Step 6 — Recognise Hooke's law ⇒ it oscillates (SHM)

WHAT. Any force of the form $F = - k r$ produces simple harmonic motion: the object wiggles back and forth with a fixed angular frequency $ω = k / m$ .

WHY. We didn't guess this — it is the definition of SHM. Once we proved $F = - k r$ (Step 5), the oscillation follows automatically. Here the mass is the electron's mass $m_{e}$ .

PICTURE. The electron traces a back-and-forth path through the centre; a small sine wave beside it shows its position vs. time.

ω = \frac{k}{m _{e}} = \frac{e ^{2}}{4 π ε _{0} m _{e} R ^{3}}, ν = \frac{ω}{2 π}

$ω$ is how fast (radians per second) it swings; $ν$ converts that to full wiggles per second (Hz). An oscillating charge radiates light of frequency $ν$ — that is why Thomson cared.

Step 7 — The degenerate cases: centre, surface, and beyond

WHAT. Check the endpoints, because a good derivation must survive them.

WHY. If the formula misbehaved at $r = 0$ or $r = R$ , we'd distrust it. Let's confirm all three regimes.

PICTURE. A graph of force magnitude vs. $r$ : rising straight line inside ( $F \propto r$ ), a peak at the surface $r = R$ , then falling as $1/ r^{2}$ outside.

At the centre, $r = 0$ : $F = - k \cdot 0 = 0$ . No enclosed charge, no pull — the equilibrium point. ✔
At the surface, $r = R$ : $F = - \frac{e ^{2}}{4 π ε _{0} R ^{3}} \cdot R = - \frac{e ^{2}}{4 π ε _{0} R ^{2}}$ — the ordinary Coulomb force between two charges $e$ a distance $R$ apart. The inside-formula hands off smoothly to the outside-formula exactly at the edge. ✔
Outside, $r > R$ : now $q_{in} = e$ (all of it), so $F \propto e / r^{2}$ — plain $1/ r^{2}$ again. The electron would no longer be doing SHM; it's just a distant Coulomb attraction. ✔

So the SHM story is valid only while the electron stays inside the pudding ( $r \leq R$ ) — which, in Thomson's picture, it always does.

The one-picture summary

Everything above in a single frame: enclosed charge $\propto r^{3}$ , field $\propto r$ , force $= - k r$ inward, motion = SHM.

Recall Feynman: the whole walk in plain words

Picture a soft glowing ball that is "positive" all the way through, and drop a tiny "negative" bead inside it. To figure out the pull on the bead, here's the trick: the ball-stuff farther out than the bead pulls it equally in every direction, so it all cancels — you can ignore it. Only the ball-stuff closer in than the bead does anything, and it pulls as if squeezed into the very centre. Now, as the bead moves outward, the amount of stuff pulling it grows fast (it grows with volume, like $r^{3}$ ), while the distance-weakening only grows like $r^{2}$ — so the net pull actually gets stronger the farther out you go, in a dead-straight line: double the distance, double the pull. And because the bead is negative, the pull always points home to the centre. "A pull straight back home, proportional to how far you strayed" is precisely a spring. So the bead wiggles in place, forever, at one fixed rate — and a wiggling charge glows with light. That glow was Thomson's hope for explaining atomic colours. (The gold-foil experiment later showed the ball wasn't soft after all, but that's the next story.)

Connections

Thomson's plum-pudding model — the parent whose result we drew out here.
Gauss's law — Steps 2 & 4: why the shell cancels and $E \propto r$ .
Simple Harmonic Motion — Step 6: what $F = - k r$ means.
Geiger–Marsden gold foil experiment — the experiment that ended this picture.
Rutherford's nuclear model — the successor.
Bohr model — the later fix for orbits and spectra.