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Electric charge is one of the most fundamental quantities in physics. All electrical phenomena arise from the existence and movement of electric charge. In this lesson we explore the nature of charge, how current is defined, and the microscopic model of charge carriers moving through a conductor.
Spec mapping: This lesson maps to AQA 7408 specification section 3.5.1.1 — Basics of electricity. It establishes the formal definitions of electric charge and current, the quantisation condition Q = ne, and the microscopic transport relationship I = nAvq linking macroscopic current to charge-carrier density, drift velocity and conductor geometry. (Refer to the official AQA specification document for exact wording.)
Synoptic links:
- 3.2.1 — Particles and quantum phenomena: the same elementary charge e that quantises macroscopic current appears in atomic ionisation calculations and in the photoelectric effect (eV_s = hf − φ). Treating charge as a quantised property of fundamental particles knits this electricity work into the particle-physics paper.
- 3.5.2 — Resistivity: the number density n introduced here is the lever that explains why metals, semiconductors and insulators differ by 30 orders of magnitude in resistivity — the same I = nAvq derivation underpins the Lesson 5 microscopic interpretation of ρ.
- 3.7.3 — Magnetic fields on moving charges: the F = BIL force on a current-carrying conductor reduces, microscopically, to the F = Bqv force on each drifting carrier; problems that combine I = nAvq with the Hall effect (a common A* synoptic move) sit on exactly this bridge.
Electric charge is a property of matter that causes it to experience a force when placed in an electromagnetic field. There are two types of charge: positive and negative.
| Property | Detail |
|---|---|
| SI unit of charge | Coulomb (C) |
| Elementary charge (e) | 1.60 × 10⁻¹⁹ C |
| Charge of an electron | −1.60 × 10⁻¹⁹ C |
| Charge of a proton | +1.60 × 10⁻¹⁹ C |
| Charge is quantised | Q = ne, where n is an integer |
Charge is conserved — it cannot be created or destroyed. In any process the total charge before equals the total charge after.
Charge is quantised — it always comes in whole-number multiples of the elementary charge e. You cannot have a charge of 0.5e.
Exam Tip: When a question says "charge flows", always think about which charge carriers are actually moving. In metals, it is free electrons. In electrolytes, it is positive and negative ions. In semiconductors, it is electrons and holes.
Current is the rate of flow of electric charge past a point in a circuit.
I=ΔtΔQ
where:
Rearranging: Q = It
The SI unit of current is the ampere (A). One ampere means one coulomb of charge flows past a point per second.
In circuit analysis, we always use conventional current direction. This is a historical convention established before the electron was discovered, but it is perfectly consistent and used universally in physics and engineering.
Common Misconception: Students sometimes think conventional current is "wrong" and electron flow is "right". Both are valid descriptions. Conventional current is the standard used in all circuit equations and diagrams at A-Level.
A current of 3.5 A flows through a lamp for 2 minutes. Calculate the total charge that flows.
Solution:
t = 2 minutes = 2 × 60 = 120 s
Q = It = 3.5 × 120 = 420 C
How many electrons pass through the lamp in the worked example above?
Solution:
Each electron carries charge e = 1.60 × 10⁻¹⁹ C.
Number of electrons: n = Q/e = 420 / (1.60 × 10⁻¹⁹) = 2.63 × 10²¹ electrons
Exam Tip: Always convert time to seconds before using Q = It. A very common error is to leave time in minutes.
In a metallic conductor, the charge carriers are free electrons (also called conduction electrons or delocalised electrons). These electrons are not bound to any particular atom and can move through the metal lattice.
When no potential difference is applied:
When a potential difference is applied across the conductor:
Common Misconception: Students often think electrons travel at the speed of light through a wire. They do not. The drift velocity is extremely slow. What travels at close to the speed of light is the electric field (the signal), which causes all free electrons throughout the wire to start drifting almost simultaneously.
The relationship between current and the microscopic properties of charge carriers is given by:
I=nAvq
where:
Consider a section of wire with cross-sectional area A. In time Δt, charge carriers with drift velocity v travel a distance vΔt along the wire.
The volume of wire through which carriers pass in time Δt is:
Volume = A × vΔt
The number of charge carriers in this volume is:
N = n × A × vΔt
The total charge passing through the cross-section in time Δt is:
ΔQ = N × q = nAvΔt × q
Therefore the current is:
I = ΔQ/Δt = nAvq
| Material | n (m⁻³) | Classification |
|---|---|---|
| Copper | 8.5 × 10²⁸ | Metal (good conductor) |
| Silicon (pure) | 1.0 × 10¹⁶ | Semiconductor |
| Glass | ≈ 0 | Insulator |
The enormous difference in n explains why metals are good conductors and insulators are not. Metals have approximately 10¹² times more free charge carriers per unit volume than pure semiconductors.
A copper wire has a cross-sectional area of 1.5 × 10⁻⁶ m² and carries a current of 2.0 A. The number density of free electrons in copper is 8.5 × 10²⁸ m⁻³. Calculate the mean drift velocity of the electrons.
Solution:
I = nAvq
Rearranging: v = I / (nAq)
v = 2.0 / (8.5 × 10²⁸ × 1.5 × 10⁻⁶ × 1.60 × 10⁻¹⁹)
v = 2.0 / (8.5 × 1.5 × 1.60 × 10²⁸⁻⁶⁻¹⁹)
v = 2.0 / (20.4 × 10³)
v = 2.0 / 20400
v = 9.8 × 10⁻⁵ m s⁻¹ (approximately 0.1 mm s⁻¹)
This confirms that drift velocity is extremely small — less than a tenth of a millimetre per second.
The same current flows through a thick wire and a thin wire made of the same material, connected in series. Compare the drift velocities.
Solution:
Since both wires carry the same current I (series circuit) and are made of the same material (same n and q):
From I = nAvq: v = I / (nAq)
The thinner wire has a smaller A, so v must be larger in the thin wire.
This is analogous to water flowing faster through a narrow section of pipe.
Exam Tip: Questions on I = nAvq are common. Remember that for a given current, drift velocity is inversely proportional to cross-sectional area and inversely proportional to number density. Semiconductors have much smaller n than metals, so for the same current in the same size wire, the drift velocity in a semiconductor would be much larger.
At any junction in a circuit, the total current flowing in equals the total current flowing out. This is a consequence of the conservation of charge (and is formalised as Kirchhoff's first law, covered in Lesson 9).
If a current of 5 A flows into a junction and splits into two branches carrying 3 A and 2 A, charge is conserved: 5 = 3 + 2.
A pure silicon strip has cross-sectional area 2.0 × 10⁻⁷ m² and carries a current of 50 µA. The number density of free charge carriers in the silicon is 1.0 × 10¹⁶ m⁻³. Calculate the mean drift velocity of the charge carriers and compare it with the drift velocity in a copper wire of the same dimensions carrying the same current.
Solution:
For the silicon: v = I / (nAq) = (50 × 10⁻⁶) / (1.0 × 10¹⁶ × 2.0 × 10⁻⁷ × 1.60 × 10⁻¹⁹)
Denominator = 1.0 × 2.0 × 1.60 × 10¹⁶⁻⁷⁻¹⁹ = 3.20 × 10⁻¹⁰
v = (5.0 × 10⁻⁵) / (3.20 × 10⁻¹⁰) = 1.56 × 10⁵ m s⁻¹
For copper carrying the same current in the same area: v = I / (nAq) = (50 × 10⁻⁶) / (8.5 × 10²⁸ × 2.0 × 10⁻⁷ × 1.60 × 10⁻¹⁹) ≈ 1.84 × 10⁻⁹ m s⁻¹.
The silicon drift velocity is roughly 10¹⁴ times greater than in copper. This is the central insight: semiconductors carry the same current with vastly fewer charge carriers, so each carrier must drift dramatically faster. It also explains why semiconductor devices are more vulnerable to carrier scattering at high fields — the carriers are already moving close to the limits where the linear drift model breaks down.
A standard domestic lighting circuit uses copper wire of cross-sectional area 1.5 mm² (i.e. 1.5 × 10⁻⁶ m²) and carries a current of 5.0 A when the lights are on. Take the free-electron number density in copper as n = 8.5 × 10²⁸ m⁻³ and the electronic charge as q = 1.60 × 10⁻¹⁹ C. Calculate the drift velocity of the electrons, comment on how small it is, and reconcile this with the everyday observation that a light bulb appears to come on instantly when the switch is flipped, even though the switch may be several metres of wire away from the bulb.
Solution — Step 1: Convert the cross-sectional area to SI units.
A = 1.5 mm² = 1.5 × 10⁻⁶ m². It is essential to convert mm² to m² by multiplying by 10⁻⁶, not 10⁻³ — squaring the length conversion is a perennial source of errors.
Step 2: Apply I = nAqv and rearrange.
v = I / (nAq)
Step 3: Substitute values.
v = 5.0 / (8.5 × 10²⁸ × 1.5 × 10⁻⁶ × 1.60 × 10⁻¹⁹)
Denominator = 8.5 × 1.5 × 1.60 × 10²⁸⁻⁶⁻¹⁹ = 20.4 × 10³ = 2.04 × 10⁴
v = 5.0 / (2.04 × 10⁴) ≈ 2.45 × 10⁻⁴ m s⁻¹
Step 4: Sanity check and interpretation.
The drift velocity is roughly 0.25 mm per second, or about 1 metre every 70 minutes. An electron starting at the switch would take well over an hour to physically reach the bulb. Yet the lamp appears to light "instantly" — typically within milliseconds of closing the switch.
Step 5: Reconcile the paradox.
The resolution lies in distinguishing three different velocities that operate simultaneously inside the wire. First, the random thermal velocity of the conduction electrons is enormous — of the order of 10⁵ to 10⁶ m s⁻¹ — but this motion is isotropic and carries no net charge in any direction, so it contributes nothing to the current. Second, the drift velocity we have just calculated (~2.5 × 10⁻⁴ m s⁻¹) is the tiny net forward bias superimposed on that random thermal motion when an electric field is applied. Third, and crucially, the electromagnetic signal velocity — the speed at which the electric field itself propagates along the wire when the switch closes — is close to the speed of light in the medium, typically around 2 × 10⁸ m s⁻¹.
When the switch closes, the field travels at near-light speed throughout the entire circuit, and every electron in the wire — including those already adjacent to the bulb filament — begins drifting almost simultaneously. The current at the bulb does not have to wait for any specific electron to travel from the switch; instead, the existing free electrons everywhere in the conductor start moving at once. This is closely analogous to opening a tap on a hose that is already full of water: water emerges from the far end essentially instantly, even though no single water molecule has travelled the length of the hose in that time.
Note: This worked example is a classic A-Level synoptic question because it forces candidates to separate three different speeds (thermal, drift, signal) that are often conflated in casual discussion. AQA examiners regularly reward explicit mention of the field propagation speed when candidates explain why the bulb lights instantly. The figure of 10⁻⁴ m s⁻¹ for drift velocity in a typical domestic copper wire is one worth committing to memory — it is the order of magnitude you should always sanity-check against when working through a current-and-drift-velocity problem.
Extension — sensitivity to current: if the same wire instead carried 10 A (e.g. a high-power kitchen appliance circuit), the drift velocity would double to roughly 5 × 10⁻⁴ m s⁻¹. The relationship is linear in I: v ∝ I for fixed n, A, q. This is why thicker wires (larger A) are mandated for high-current circuits — they reduce the drift velocity for a given current, which in turn reduces I²R heating per unit volume of conductor and keeps the wire safely below its insulation's temperature rating.
Specimen question modelled on the AQA paper format (6 marks).
A copper wire of cross-sectional area 1.0 × 10⁻⁶ m² carries a steady current of 4.0 A for 5.0 minutes. The number density of free electrons in copper is 8.5 × 10²⁸ m⁻³ and the magnitude of the electronic charge is 1.60 × 10⁻¹⁹ C.
(a) Calculate the total charge that flows in the 5.0 minute interval. (2 marks)
(b) Estimate the number of free electrons that pass through any cross-section of the wire in this interval. (2 marks)
(c) Calculate the mean drift velocity of the electrons in the wire and comment on the magnitude of your answer. (2 marks)
This specimen question is dominated by AO2 (application of physics) with a brief AO3 evaluative element in part (c). On a typical 6-mark calculation of this kind, AO1 (knowledge and understanding) carries roughly 1 mark for stating the relevant defining equations, AO2 (application to a given context) carries the bulk — around 4 marks — for substituting values correctly with consistent units and obtaining numerical answers to appropriate precision, and AO3 (analysis, interpretation and evaluation) takes the final mark for the qualitative comment on the surprising smallness of the drift velocity. Generic AO descriptors only — refer to the AQA published mark scheme for the authoritative wording of any specific paper.
(a) Q = It = 4.0 × 5.0 × 60 = 4.0 × 300 = 1200 C.
(b) Number of electrons N = Q/e = 1200 / (1.60 × 10⁻¹⁹) = 7.5 × 10²¹.
(c) Using I = nAvq, v = I/(nAq) = 4.0 / (8.5 × 10²⁸ × 1.0 × 10⁻⁶ × 1.60 × 10⁻¹⁹) ≈ 3 × 10⁻⁴ m s⁻¹. The drift velocity is very small.
Examiner commentary: This response secures the calculation marks. M1 is earned for converting 5.0 minutes to 300 s and applying Q = It; M2 for the correct numerical answer 1200 C. M3 is earned for the use of N = Q/e and M4 for the correct figure 7.5 × 10²¹. M5 is earned for substituting into I = nAvq correctly and arriving at the order-of-magnitude correct drift velocity. The response loses the final AO3 mark because the qualitative comment ("very small") is unevaluated — it does not put the figure in context or compare with anything the marker can recognise as physical insight. This is the canonical Grade C pattern: arithmetic is sound, evaluative tail is thin.
(a) Total charge Q = It. The interval in seconds is t = 5.0 × 60 = 300 s, so Q = 4.0 × 300 = 1.20 × 10³ C (to 3 s.f.).
(b) Each electron carries elementary charge e = 1.60 × 10⁻¹⁹ C. The number of electrons crossing any cross-section is therefore N = Q/e = 1.20 × 10³ / (1.60 × 10⁻¹⁹) = 7.50 × 10²¹ electrons.
(c) From the transport equation I = nAvq, v = I/(nAq). Substituting: v = 4.0 / (8.5 × 10²⁸ × 1.0 × 10⁻⁶ × 1.60 × 10⁻¹⁹) = 4.0 / (1.36 × 10⁴) = 2.94 × 10⁻⁴ m s⁻¹, i.e. about 0.3 mm s⁻¹. This is remarkably small — slower than a snail. The apparent paradox that switching on a light gives an instantaneous response is resolved by recognising that what propagates at near-light speed is the electric field (the signal that establishes the drift), not the carriers themselves. The conductor is already populated with ≈ 8.5 × 10²² electrons per cubic centimetre, so even a tiny drift superimposed on huge thermal motion delivers a substantial macroscopic current.
Examiner commentary: All five calculation marks (M1–M5) are secured cleanly, with attention to significant figures and unit consistency throughout. The crucial differentiator is the AO3 evaluative move in part (c). The candidate does not merely state that the drift velocity is small — they (i) quantify it in a familiar unit (mm s⁻¹), (ii) resolve the everyday paradox of instant response, and (iii) close the loop by connecting back to the enormous n that makes the small v sufficient. This three-step evaluation pattern (quantify → contrast → reconnect to underlying physics) is the move that separates Grade A* from Grade A in 6-mark calculation-with-comment items.
Many candidates lose marks on this topic by failing to convert time to SI units before applying Q = It; leaving an answer in "ampere-minutes" or "ampere-hours" wipes out both calculation marks even if the arithmetic is otherwise sound. A second common slip is to confuse the elementary charge magnitude e with the signed electron charge −e in problems asking for the number of carriers — the count is always a positive integer, so use |e|. A third recurring error is to treat drift velocity as comparable with random thermal velocity; candidates sometimes assert that electrons "move at 10⁶ m s⁻¹ through the wire" when in fact that figure is the chaotic thermal speed and the drift is six to ten orders of magnitude smaller. Finally, when applying I = nAvq, candidates often substitute diameter where they need radius (or vice versa) in the area calculation, or forget to square the radius. The teacher fix is simple: always write A = π(d/2)² as an explicit intermediate step, never collapse it into a single line. Verify units throughout — a wrong area will not necessarily flag a wrong answer because the orders of magnitude can still look plausible.
For students who want to push beyond the spec, the Drude free-electron model (a classical kinetic-theory treatment of conduction electrons) is the natural next step and is covered in chapter 1 of Ashcroft and Mermin's Solid State Physics. The Drude model derives I = nAvq formally and explains why it must fail for semiconductors at high field strengths. Kittel's Introduction to Solid State Physics (chapter 6) develops the same machinery using the relaxation-time approximation, which is a cleaner setup for later quantum corrections. For applications, the Hall effect — which uses I = nAvq combined with the Lorentz force to measure the sign and density of charge carriers — is a beautiful piece of synoptic physics; it features in many university interview problems and is well treated in Hecht's Physics: Calculus. Reading any of these will deepen the conceptual base without requiring formal quantum mechanics.
The first A-vs-A* error is conflating conventional current direction with electron-flow direction in microscopic arguments. The transport equation I = nAvq uses the magnitude of charge q = e for electrons; the conventional-current direction is opposite to the actual electron drift, and a candidate who tries to "fix" this with a minus sign typically ends up with a sign-confused answer. The convention is sufficient; do not over-correct it.
A second subtle error is treating n as if it depends on the applied voltage. In a metal at fixed temperature, n is a material property — it does not change with V or I. What changes when V is applied is the drift velocity v. In a semiconductor at varying temperature, n does change, but it is the thermal generation that drives this, not the field itself.
A third trap is the assumption that "current is the same in series" implies "drift velocity is the same in series". It is not: from v = I/(nAq), a thinner segment of the same material carrying the same series current must have a larger drift velocity. Candidates lose marks on synoptic questions about non-uniform wires by missing this.
On any question that supplies n, A and I, write I = nAvq before anything else — even if you are going to rearrange immediately. The marker can credit method marks from the rearrangement even if a substitution slips. Always convert all values to SI base units (metres, seconds, square metres, coulombs) before substitution; mixing mm² with m s⁻¹ is one of the most common ways to lose an order of magnitude.
If a question asks for "number of electrons", check whether it wants the number that flow in a given time (use N = Q/e = It/e) or the number present in a given length of wire (use N = nAL). These are different quantities and the wording is the only clue. Read carefully.
Where a question splits into multiple parts feeding the same numerical answer through, carry an unrounded intermediate value into the next step and round only at the end. Premature rounding is a quiet but persistent way to drop the final significant-figure mark in long calculations.
Finally, when a comment or evaluation mark is on offer (the "comment on the magnitude of your answer" type), aim to do three things: quantify in a familiar unit, contrast with an everyday expectation, and close the loop by connecting back to the underlying physics. That three-move template earns the AO3 mark almost every time.
| Equation | Meaning |
|---|---|
| Q = It | Charge = current × time |
| I = ΔQ/Δt | Current = rate of flow of charge |
| Q = ne | Charge is quantised (n is an integer) |
| I = nAvq | The transport equation linking current to charge carrier properties |