The big idea: Electric charge is quantised — it always comes in whole-number lumps, never a fraction of a lump.
The smallest lump is the elementary charge, e = 1.60 × 10⁻¹⁹ C. This is the charge on one proton (+e) or one electron (−e).
So any charge Q is just a whole number of these lumps: Q = N e, where N is a whole number.
| Idea | Continuous (wrong) | Quantised (real) |
|---|---|---|
| Allowed charges | any value at all | only whole multiples of e |
| Smallest amount | no smallest amount | one elementary charge, e |
| Example charges | 1.5e, 2.3e, π·e … | 1e, 2e, 3e, … (whole numbers) |
Spot it: Quantised means "comes in fixed steps" — like money in whole cents, not any amount you like.
An object is charged because it has extra electrons (negative) or is missing electrons (positive). You can only ever add or remove whole electrons — so the charge can only change in steps of e.
Because charge comes in whole lumps of e, the charge on any object is the number of lumps N multiplied by the size of one lump:
- the total charge on the object (C, coulombs)
- a whole number — how many elementary charges (extra or missing electrons)
- the elementary charge, e = 1.60 × 10⁻¹⁹ C (given in the data booklet)
[Diagram: phys-formula-triangle] - Available in full study mode
Going the other way: To find how many extra (or missing) electrons an object has, rearrange to make N the subject:
N = Q ÷ e — divide the measured charge by e.
The answer must come out as a whole number. If it doesn't, you've made an error — charge is always a whole multiple of e.
Worked example — how many extra electrons?
A tiny plastic sphere carries a charge of −8.0 × 10⁻¹⁹ C. How many extra electrons does it hold? (e = 1.60 × 10⁻¹⁹ C.)
Solution
- The charge is N lots of e. Start with the relationship:
- Rearrange to make N the subject:
- Put in the numbers (use the size of the charge):
- Work it out — a whole number, as expected:
Final answer
N = 5 — the sphere holds 5 extra electrons. (The charge is negative, so it gained electrons.)
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How this is tested: Quantisation links to Millikan's oil-drop experiment — the classic evidence for it.
- Paper 1A (MCQ): identify which charge values are allowed (whole multiples of e), or count electrons with N = Q ÷ e. - Paper 2: part of a longer oil-drop question — deduce a charge using the fact that it must be a whole-number multiple of e (e.g. when a drop splits).
Classic trap: giving N as a decimal. Charge is quantised, so N is always a whole number — round only because of measurement error, never to a fraction.
Millikan's oil-drop experiment — the evidence: Millikan balanced tiny charged oil drops in an electric field and measured the charge on each one.
Every charge he found was a whole-number multiple of one smallest step — and that step was e = 1.60 × 10⁻¹⁹ C.
No drop ever had a fraction of e. That is the experimental proof that charge is quantised.
IB-style question — (a) charge on one oil drop
In an oil-drop experiment a single drop is found to carry a charge of 6.4 × 10⁻¹⁹ C. Deduce how many elementary charges this is. (e = 1.60 × 10⁻¹⁹ C.)
Solution
- The charge must be a whole-number multiple of e. Start with:
- Rearrange for N:
- Put in the numbers:
- Work it out — a whole number, as it must be:
Final answer
N = 4 — the drop carries 4 elementary charges (a charge of 4e).
IB-style question — (b) the drop splits in two
The same drop (charge 6.4 × 10⁻¹⁹ C) splits into two equal halves, and the charge shares equally between them. Deduce the charge on each half, and check it is allowed by quantisation.
Solution
- The charge shares equally, so each half gets half the total:
- Work it out:
- Check it is a whole number of elementary charges (N = Q ÷ e):
- N = 2 is a whole number, so this charge is allowed by quantisation.
Final answer
Each half carries 3.2 × 10⁻¹⁹ C = 2e. This is a whole number of elementary charges (N = 2), so it is an allowed charge.