Thomson's plum-pudding model: Worked examples — Chemistry deep dive

Before anything, let us re-anchor the three symbols we will lean on, in plain words, so nothing is used before it is earned.

The single most important geometric fact — the one every case below re-uses — is the shape of $E (r)$ : it climbs straight up as a line inside the ball, peaks at the skin, then falls off as $1/ r^{2}$ outside. Look at the figure above: the red line inside and the orange curve outside meet at the same height at $r = R$ . Keep this picture in your head; every example is just "which part of this graph am I standing on?"

The scenario matrix

Every question this topic can ask lives in exactly one of these cells. The worked examples that follow are tagged with the cell they land in.

#	Case class	What is special	Example
A	Interior, generic $0 < r < R$	linear field, SHM lives here	Ex 1
B	Centre $r = 0$	zero field, zero force — equilibrium	Ex 2
C	At the skin $r = R$	field is maximum; interior formula = exterior formula	Ex 3
D	Outside $r > R$	acts like a point charge, $1/ r^{2}$	Ex 4
E	Doubled input (scaling)	how $F, ω, ν$ respond to $\times 2$	Ex 5
F	Limiting behaviour ( $R \to 0$ , $R \to \infty$ )	where the model breaks / stiffens	Ex 6
G	Real-world word problem	frequency ⇒ colour of emitted light	Ex 7
H	Exam-style twist	many electrons, neutrality bookkeeping, sign trap	Ex 8

We inherit from the parent, valid only inside the ball ( $0 \leq r \leq R$ ):

For cell D we also need the exterior field, which the parent never wrote but which follows from the same Gauss argument — once you are outside, the entire charge $e$ is enclosed:

Worked examples

Worked example Example 1 — Interior, generic point ·

Cell A An electron sits at $r = 0.5 \times 1 0^{- 10} m$ inside a pudding of charge $+ e$ , radius $R = 1.0 \times 1 0^{- 10} m$ . Find the field magnitude and the force magnitude on it.

Forecast: the electron is halfway out. Field grows linearly, so guess the field is about half of the skin value. Will the force point in or out?

Step 1 — Field. Why this step? We are strictly between centre and skin, so cell A → use $E_{in}$ . $E_{in} = 9.0 \times 1 0^{9} \cdot \frac{( 1.6 \times 1 0 ^{- 19} ) ( 0.5 \times 1 0 ^{- 10} )}{( 1.0 \times 1 0 ^{- 10} ) ^{3}} .$ Numerator $= 9.0 \times 1 0^{9} \times 1.6 \times 1 0^{- 19} \times 0.5 \times 1 0^{- 10} = 7.2 \times 1 0^{- 20}$ . Divide by $1 0^{- 30}$ : $E_{in} = 7.2 \times 1 0^{10} N/C .$ Step 2 — Force. Why this step? Force on the electron is $F = (- e) E$ ; we report magnitude $F = e E$ . $F = 1.6 \times 1 0^{- 19} \times 7.2 \times 1 0^{10} = 1.15 \times 1 0^{- 8} N, pointing toward the centre .$ Verify: the skin field (Ex 3) is $1.44 \times 1 0^{11}$ N/C; ours is exactly half — matches the "linear, halfway ⇒ half" forecast. Direction inward because the enclosed positive charge lies between the electron and the centre and pulls the negative electron back. Units: $\frac{1}{4 π ε _{0}}$ carries N·m²/C², times C times m over m³ ⇒ N/C. ✓

Worked example Example 3 — Exactly at the skin ·

Cell C Find the field at $r = R = 1.0 \times 1 0^{- 10} m$ , using both the interior and exterior formulas, and confirm they agree.

Forecast: this is the boundary. Should the two formulas give the same number here?

Step 1 — Interior formula at $r = R$ . Why? Set $r = R$ in $E_{in}$ ; the $R$ 's partly cancel. $E_{in} (R) = \frac{1}{4 π ε _{0}} \frac{e R}{R ^{3}} = \frac{1}{4 π ε _{0}} \frac{e}{R ^{2}} = 9.0 \times 1 0^{9} \cdot \frac{1.6 \times 1 0 ^{- 19}}{1 0 ^{- 20}} = 1.44 \times 1 0^{11} N/C .$ Step 2 — Exterior formula at $r = R$ . Why? Set $r = R$ in $E_{out} = \frac{1}{4 π ε _{0}} \frac{e}{r ^{2}}$ : $E_{out} (R) = 9.0 \times 1 0^{9} \cdot \frac{1.6 \times 1 0 ^{- 19}}{1 0 ^{- 20}} = 1.44 \times 1 0^{11} N/C .$ Verify: identical. Why they must match: the field is continuous — no charge sits on the infinitely-thin skin, so there is no jump. This is why in figure s01 the red line and orange curve touch at $r = R$ . This peak is the largest field an electron ever feels here. ✓

Worked example Example 6 — Limiting behaviour ·

Cell F What happens to $k$ (and hence $ω$ ) as (a) $R \to 0$ and (b) $R \to \infty$ ?

Forecast: which limit makes the spring infinitely stiff?

Step 1 — Shrink the ball, $R \to 0$ . Why? $k = \frac{e ^{2}}{4 π ε _{0} R ^{3}}$ ; as $R \to 0$ , $R^{3} \to 0$ , so $k \to \infty$ . $k \to \infty, ω \to \infty.$ Physically: cram the same charge into a vanishing sphere → an infinitely stiff spring → infinitely fast oscillation. This is the seed of why a point-like positive charge behaves utterly differently — a hint toward Rutherford's nuclear model. Step 2 — Spread the ball, $R \to \infty$ . Why? $R^{3} \to \infty$ , so $k \to 0$ . $k \to 0, ω \to 0.$ Infinitely diffuse charge → no restoring pull → the electron barely feels anything and stops oscillating. Verify: both limits agree with Ex 5's trend ( $ν \propto R^{- 3/2}$ ): smaller $R$ ⇒ larger $ν$ , larger $R$ ⇒ smaller $ν$ . Consistent. ✓

Worked example Example 7 — Real-world word problem ·

Cell G Using the parent's numbers ( $R = 1.0 \times 1 0^{- 10}$ m), the electron oscillates at $ν \approx 2.6 \times 1 0^{15} Hz$ . An oscillating charge radiates light of this frequency. Q: what wavelength is this, and is it visible? ( $c = 3.0 \times 1 0^{8} m/s$ .)

Forecast: visible light spans roughly $400$ – $700$ nm. Guess: will this land inside, or just outside, that window?

Step 1 — Convert frequency to wavelength. Why this step? Light obeys $c = ν λ$ , so $λ = c / ν$ turns a wiggle-rate into a colour. $λ = \frac{c}{ν} = \frac{3.0 \times 1 0 ^{8}}{2.6 \times 1 0 ^{15}} = 1.15 \times 1 0^{- 7} m = 115 nm .$ Step 2 — Classify. Why? Compare to the visible band $400$ – $700$ nm. $115 nm < 400 nm \Rightarrow ultraviolet .$ Verify: $115$ nm is shorter than violet ( $\sim 400$ nm), so it is UV — the "right order of magnitude for light" claim of the parent holds (same power of ten as visible), which is exactly why Thomson's model felt convincing even though it was ultimately wrong. ✓

Worked example Example 8 — Exam-style twist (many electrons + sign trap) ·

Cell H A student writes: "A neutral helium atom in the plum-pudding model has $2$ electrons, so the pudding carries charge $+ 2 e$ , and each electron feels force $F = + \frac{( 2 e ) e}{4 π ε _{0} R ^{3}} r$ pushing it outward." Find the two errors and give the correct restoring-force constant $k$ felt by one electron sitting at displacement $r$ (ignore electron–electron repulsion).

Forecast: spot the sign error and the charge error before reading.

Step 1 — Neutrality bookkeeping is fine. Why? Neutral atom, $2$ electrons ( $- 2 e$ total) ⇒ pudding is $+ 2 e$ . That part is correct. Step 2 — Fix the enclosed-charge factor. Why? Only the fraction $r^{3} / R^{3}$ of the pudding lies inside radius $r$ , so the enclosed positive charge is $2 e \cdot r^{3} / R^{3}$ , giving field $E_{in} = \frac{1}{4 π ε _{0}} \frac{2 e r}{R ^{3}}$ . The student used the total $2 e$ with a bare $r$ , mixing the interior geometry up. Step 3 — Fix the sign. Why? The electron has charge $- e$ , so $F = (- e) E_{in}$ points toward the centre (restoring), not outward. $F = - \frac{2 e ^{2}}{4 π ε _{0} R ^{3}} r \Rightarrow k = \frac{2 e ^{2}}{4 π ε _{0} R ^{3}} .$ Verify: setting the pudding charge back to $+ e$ (single-electron case) recovers the parent's $k = \frac{e ^{2}}{4 π ε _{0} R ^{3}}$ . Doubling the pudding charge doubles $k$ — consistent scaling. Sign is negative ⇒ restoring ⇒ genuine SHM. ✓ The two errors: outward instead of inward sign, and using total charge instead of the $r^{3} / R^{3}$ -enclosed charge.

Recall Quick self-test across all cells

At $r = 0$ , what is the field? ::: Zero — the equilibrium point (cell B). Interior field law? ::: $E \propto r$ (linear), cell A. Exterior field law? ::: $E \propto 1/ r^{2}$ (point-charge), cell D. Do interior and exterior formulas agree at $r = R$ ? ::: Yes, both give $\frac{1}{4 π ε _{0}} \frac{e}{R ^{2}}$ — the field is continuous (cell C). Double the displacement — force does what? ::: Doubles (cell E). Double the radius — frequency does what? ::: Falls by $2^{- 3/2} \approx 0.354$ (cell E/F). As $R \to 0$ , the spring becomes? ::: Infinitely stiff, $ω \to \infty$ (cell F). The emitted light for $R \sim 1 0^{- 10}$ m is what type? ::: Ultraviolet, $λ \approx 115$ nm (cell G).

Connections

Parent: Plum-pudding model — the derivation these examples exercise.
Gauss's law — the reason only enclosed charge counts (cells A–D).
Simple Harmonic Motion — the $F = - k r$ behaviour of cells A, B, F.
Rutherford's nuclear model — where the $R \to 0$ limit (cell F) secretly points.