Charge, Current and Potential Difference

Electric circuits are at the heart of modern technology. Every phone, computer, car, and medical device relies on the controlled flow of electric charge through circuits. Before you can analyse any circuit, you need to understand three foundational quantities: charge, current, and potential difference. These definitions underpin every equation and every circuit calculation at A-Level.

Electric Charge

All matter is made of atoms, and atoms contain charged particles. Protons carry a positive charge and electrons carry a negative charge. The fundamental unit of charge is the coulomb (C).

The charge on a single electron is extremely small:

e = 1.60 × 10⁻¹⁹ C

This means that one coulomb of charge is equivalent to roughly 6.25 × 10¹⁸ electrons. In everyday circuits, we routinely deal with charges of several coulombs flowing every second.

Charge is quantised — it always comes in whole-number multiples of the elementary charge e. You cannot have half an electron's worth of charge. This quantisation was demonstrated experimentally by Robert Millikan in his famous oil drop experiment.

Conservation of Charge

Charge is a conserved quantity. It cannot be created or destroyed — it can only be transferred from one place to another. This principle underpins Kirchhoff's first law, which you will meet in a later lesson. At every junction in every circuit, the total charge entering per second must equal the total charge leaving per second.

Electric Current

Current is the rate of flow of electric charge past a point in a circuit. It is defined mathematically as:

I = Q / t

Or equivalently:

Q = It

where:

I = current in amperes (A)
Q = charge in coulombs (C)
t = time in seconds (s)

One ampere means one coulomb of charge flows past a point every second.

Worked Example 1

A current of 3.0 A flows through a lamp for 2.0 minutes. Calculate the total charge that flows.

Solution:

t = 2.0 × 60 = 120 s
Q = It = 3.0 × 120 = 360 C

Worked Example 2: Finding the Number of Electrons

A current of 0.15 A flows through a thin wire for 30 s. Calculate (a) the total charge that passes and (b) the number of electrons that flow past a point.

Solution:

(a) Q = It = 0.15 × 30 = 4.5 C
(b) Number of electrons = Q / e = 4.5 / (1.60 × 10⁻¹⁹) = 2.81 × 10¹⁹ electrons

Worked Example 3: Current from Electron Count

In a cathode ray tube, 5.0 × 10¹⁵ electrons strike the screen every second. What is the current?

Solution:

Charge per second = number of electrons × e = 5.0 × 10¹⁵ × 1.60 × 10⁻¹⁹ = 8.0 × 10⁻⁴ C s⁻¹
I = Q/t = 8.0 × 10⁻⁴ A (or 0.80 mA)

Conventional Current vs Electron Flow

Historically, current was defined as the flow of positive charge from the positive terminal of a battery to the negative terminal. This convention was established before anyone knew that in metallic conductors it is actually the negatively charged electrons that move — and they move in the opposite direction, from negative to positive.

We still use the conventional current direction in circuit analysis. This is important to remember: conventional current flows from positive to negative, but electrons flow from negative to positive.

In electrolytes (solutions containing ions), both positive and negative charge carriers can move. Positive ions move in the direction of conventional current, and negative ions move in the opposite direction.

Charge Carriers in Different Materials

Material	Charge Carriers	Typical Current Mechanism
Metals	Free (delocalised) electrons	Drift through lattice of positive ions
Electrolytes	Positive and negative ions	Migration under electric field
Semiconductors	Electrons and holes	Movement through crystal lattice
Gases (ionised)	Electrons and positive ions	Discharge through gas

Potential Difference

Potential difference (p.d.) — often called voltage — is defined as the energy transferred per unit charge as charge moves between two points:

V = W / Q

where:

V = potential difference in volts (V)
W = energy (work done) in joules (J)
Q = charge in coulombs (C)

One volt means one joule of energy is transferred for every coulomb of charge that passes. The volt is therefore equivalent to J C⁻¹.

Worked Example 4

A 12 V battery drives 5.0 C of charge around a circuit. How much energy is transferred by the battery?

Solution:

W = QV = 5.0 × 12 = 60 J

Worked Example 5: Multi-Step Energy Calculation

A battery of p.d. 9.0 V drives a current of 0.40 A through a resistor for 5.0 minutes. Calculate (a) the total charge that flows, and (b) the total energy transferred to the resistor.

Solution:

(a) Q = It = 0.40 × (5.0 × 60) = 0.40 × 300 = 120 C
(b) W = QV = 120 × 9.0 = 1080 J (or 1.08 kJ)

Energy Transfer in Circuits

The battery (or power supply) is an energy source. It does work on the charge carriers, giving them electrical potential energy. As charge carriers move through components in the circuit, they transfer this energy to other forms:

In a resistor: electrical energy → thermal energy (heating)
In a lamp: electrical energy → thermal energy + light
In a motor: electrical energy → kinetic energy
In a loudspeaker: electrical energy → sound energy
In a charging battery: electrical energy → chemical energy

The key principle is that energy is conserved around any closed loop. The total energy given to the charges by the battery equals the total energy transferred by the charges to the components. This is the basis of Kirchhoff's second law.

flowchart LR
    A[Battery] -->|Provides energy to charges| B[Charge carriers gain electrical PE]
    B -->|Flow through circuit| C{Components}
    C -->|Resistor| D[Thermal energy]
    C -->|Lamp| E[Light + heat]
    C -->|Motor| F[Kinetic energy]
    C -->|Speaker| G[Sound energy]
    D --> H[Energy conserved around loop]
    E --> H
    F --> H
    G --> H

Measuring Current and Potential Difference

Ammeters measure current and must be connected in series — the current you want to measure must flow through the ammeter. An ideal ammeter has zero resistance so it does not affect the circuit.

Voltmeters measure potential difference and must be connected in parallel across the component — they measure the difference in energy per unit charge between two points. An ideal voltmeter has infinite resistance so it draws no current from the circuit.

Instrument	Connection	Ideal Resistance	Why?
Ammeter	Series	Zero	Must not impede current flow
Voltmeter	Parallel	Infinite	Must not draw current from circuit

Common exam mistake: Confusing the connection requirements. Remember — an ammeter in parallel would short-circuit the component (because it has near-zero resistance). A voltmeter in series would block the current (because it has very high resistance).

The Relationship Between Charge, Current and p.d.

These three quantities are deeply interconnected. Current tells you how fast charge is flowing. Potential difference tells you how much energy each unit of charge transfers. Together, they determine the power delivered to or by a component (which you will explore in a later lesson on electrical energy and power).

Key Relationships at a Glance

Relationship	Equation	Use When You Know
Charge and current	Q = It	Current and time
P.d. and energy	V = W/Q	Energy and charge
Combined	W = VIt	Voltage, current, and time

Worked Example 6: Comprehensive Calculation

An LED draws a current of 20 mA from a 3.0 V supply. In one hour of operation, calculate (a) the charge that flows and (b) the total energy transferred.

Solution:

I = 20 mA = 0.020 A; t = 1 hour = 3600 s
(a) Q = It = 0.020 × 3600 = 72 C
(b) W = QV = 72 × 3.0 = 216 J

Understanding these definitions precisely is essential because every circuit calculation you perform at A-Level builds on them. When you see V = IR or P = IV, remember that these are consequences of the fundamental definitions of charge, current, and potential difference.

A-Level Deep Dive: Charge, Current and Potential Difference

Spec mapping

Edexcel 9PH0 specification, Topic 3 — Electric circuits, sub-topic 3.1 (charge, current and potential difference) establishes the definitions of charge $Q$ , current $I = \Delta Q / \Delta t$ and potential difference $V = W/Q$ , together with the microscopic conduction model $I = nAvq$ relating macroscopic current to charge-carrier number density $n$ , cross-sectional area $A$ , drift velocity $v$ and carrier charge $q$ (refer to the official Pearson Edexcel specification document for exact wording). Although this is the opening sub-topic of Topic 3, it is examined throughout 9PH0-01 (Paper 1, Advanced Physics I) and 9PH0-03 (Paper 3, General and Practical Principles in Physics), and underpins every subsequent circuits sub-topic — Ohm's law (3.2), resistivity (3.3), series/parallel networks (3.4) and emf with internal resistance (3.5). The Edexcel formula booklet does list $I = \Delta Q / \Delta t$ and $I = nqvA$ , but the interpretation — what each symbol means physically and why drift velocity is so small — must be understood, not just substituted.

Worked example with full mark scheme

Question (8 marks):

A copper wire of cross-sectional area $1.5 \times 10^{-6}\ \text{m}^2$ carries a steady current of $2.4\ \text{A}$ . The number density of free electrons in copper is $n = 8.5 \times 10^{28}\ \text{m}^{-3}$ and the elementary charge is $e = 1.60 \times 10^{-19}\ \text{C}$ .

(a) State what is meant by electric current. (1)

(b) Calculate the charge that flows past a point in the wire in $3.0\ \text{minutes}$ . (2)

(d) Comment on the magnitude of this drift velocity, given that a domestic light comes on apparently instantaneously when the switch is closed. (2)

Solution with mark scheme:

(a) B1 — current is the rate of flow of (electric) charge, or equivalently $I = \Delta Q / \Delta t$ with symbols defined. A bare $I = Q/t$ without naming the quantities is borderline; examiners reward "rate of flow of charge" as the unambiguous statement.

(b) Step 1 — convert time. $t = 3.0 \times 60 = 180\ \text{s}$ .

Step 2 — apply $Q = It$ .

$Q = It = 2.4 \times 180 = 432\ \text{C}$

M1 — substituting correct values into $Q = It$ with consistent SI units. A1 — $Q = 432\ \text{C}$ (or $4.3 \times 10^2\ \text{C}$ to 2 s.f.). Common error: leaving $t = 3.0\ \text{min}$ unconverted and quoting $Q = 7.2\ \text{C}$ — this loses both marks because the numerical answer is wrong, even though the formula is right.

$v = \dfrac{I}{nAe}$

M1 — correct rearrangement of $I = nAve$ for drift velocity.

Step 2 — substitute.

$v = \dfrac{2.4}{(8.5 \times 10^{28}) \times (1.5 \times 10^{-6}) \times (1.60 \times 10^{-19})}$

The denominator: $8.5 \times 1.5 \times 1.60 = 20.4$ , with powers $10^{28} \times 10^{-6} \times 10^{-19} = 10^{3}$ . So the denominator is $2.04 \times 10^{4}$ .

$v = \dfrac{2.4}{2.04 \times 10^{4}} \approx 1.18 \times 10^{-4}\ \text{m s}^{-1}$

M1 — correct substitution of all four quantities in SI units.

A1 — $v \approx 1.2 \times 10^{-4}\ \text{m s}^{-1}$ to 2 s.f., matching the printed answer. Because this is a "show that" the candidate must produce an answer with at least one extra significant figure than the printed value (so $1.18 \times 10^{-4}$ , not $1.2 \times 10^{-4}$ , secures the A1).

(d) B1 — the drift velocity is very slow (about 0.1 mm per second, so an electron takes minutes to traverse a domestic wire).

B1 — yet the lamp lights up almost instantly because the electric field propagates through the wire at close to the speed of light, so all electrons throughout the wire begin to drift simultaneously when the switch closes. The energy transfer is mediated by the field, not by individual electrons travelling from battery to bulb.

Total: 8 marks (B1 + M1 A1 + M1 M1 A1 + B1 B1).

Specimen question modelled on the Edexcel 9PH0 Paper 1 format

Question (6 marks): A metal strip used in a sensor has rectangular cross-section $2.0\ \text{mm} \times 0.50\ \text{mm}$ . When a potential difference is applied across its length, a current of $80\ \text{mA}$ flows. The number density of free electrons in this metal is $n = 6.0 \times 10^{28}\ \text{m}^{-3}$ .

(a) Calculate the cross-sectional area of the strip in $\text{m}^2$ . (1)

(b) Calculate the drift velocity of the free electrons in the strip. (3)

(c) Explain, using the equation $I = nAve$ , why a wire of smaller cross-section carrying the same current must have a larger drift velocity. (2)

Mark scheme decomposition by AO:

(a)

B1 (AO2.2) — $A = (2.0 \times 10^{-3}) \times (0.50 \times 10^{-3}) = 1.0 \times 10^{-6}\ \text{m}^2$ . The conversion mm to m is the AO2 step; many candidates leave the answer in mm² and lose the mark.

(b)

M1 (AO1.1b) — rearranging $I = nAve$ to $v = I/(nAe)$ .
M1 (AO1.1b) — substituting $v = 0.080 / [(6.0 \times 10^{28}) \times (1.0 \times 10^{-6}) \times (1.60 \times 10^{-19})]$ .
A1 (AO1.1b) — $v \approx 8.3 \times 10^{-6}\ \text{m s}^{-1}$ (denominator $= 9.6 \times 10^{3}$ ).

(c)

M1 (AO2.1) — recognising that $I$ , $n$ and $e$ are constant, so $v \propto 1/A$ .
A1 (AO2.5) — concluding that smaller $A$ implies larger $v$ to maintain the same charge-flow rate, with a clear physics argument (e.g. "the same number of charges per second must squeeze through a smaller area, so they must move faster").

Total: 6 marks split AO1 = 3, AO2 = 3. This question is more synoptic than the worked example: AO2 marks reward physical reasoning about the $nAve$ equation, not just algebraic rearrangement. Edexcel routinely tests "explain why" questions on this sub-topic precisely because the equation has four variables, three of which are material properties — a perfect platform for proportional reasoning.

Synoptic links

Connects to:

Topic 3 sub-topic 3.3 — Resistivity: the equation $R = \rho L / A$ for a uniform conductor follows directly from combining $I = nAve$ with the definition of resistance and the assumption that drift velocity is proportional to applied field. The two equations share $A$ , and the microscopic origin of $\rho$ is the scattering rate of free electrons off the lattice — a story that begins with $nAve$ .
Topic 7 — Fields (electric and magnetic): charged particles moving through a wire experience magnetic forces if the wire sits in a magnetic field. The force on a current-carrying wire $F = BIL$ is derived by summing the Lorentz force $F = qvB$ over all the drifting charges in length $L$ — and the substitution $I = nAve$ is exactly what closes the algebra. This is why the Hall effect (Topic 7) is most clearly explained at A-Level by viewing it as a transverse force on drifting electrons.
Topic 5 — Waves (electromagnetic radiation): the contrast between drift velocity ( $\sim 10^{-4}\ \text{m s}^{-1}$ ) and field-propagation speed ( $\sim 3 \times 10^{8}\ \text{m s}^{-1}$ ) only makes sense once you know that the energy in an electrical signal travels as an electromagnetic disturbance in the field around the wire, not in the kinetic energy of the electrons. This is a beautiful synoptic link to Topic 5's wave equation $c = f\lambda$ and the recognition that EM waves do not need a material medium.
Topic 9 — Thermodynamics (resistive heating): the power dissipated in a resistor, $P = I^2 R$ , has a microscopic interpretation in terms of drift electrons losing kinetic energy in collisions with the lattice and depositing it as thermal energy. The rate of heating is set by the rate of charge flow — and so by $nAve$ — multiplied by the energy lost per charge per unit length. This bridges the macroscopic Topic 3 picture with the atomic-scale Topic 9 picture.
Topic 8 — Nuclear and particle physics: the elementary charge $e$ that appears in $I = nAve$ is the same $e$ that quantises charge in beta decay, in pair production, and in the quark model. Recognising that $1.60 \times 10^{-19}\ \text{C}$ is a universal constant — not a property of metals — links Topic 3 to Topic 8 and prepares candidates for the deep idea that charge, like energy and momentum, is conserved in every interaction in nature.

Mark-scheme literacy

Charge/current/p.d. questions on 9PH0 Paper 1 split AO marks across all three assessment objectives but in characteristic proportions:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	50–60%	Recalling definitions; substituting into $Q = It$ , $V = W/Q$ , $I = nAve$ with consistent SI units; quoting standard values such as $e = 1.60 \times 10^{-19}\ \text{C}$
AO2 (application / interpretation)	25–35%	Choosing the right equation when the question is phrased in everyday language; converting units (mm to m, minutes to seconds); commenting on the physical meaning of an answer (e.g. "drift velocity is small but the field propagates quickly")
AO3 (analysis / evaluation)	10–20%	Comparing materials; explaining limitations of the free-electron model; designing experiments to test $I = nAve$

Examiner-rewarded phrasing: "current is the rate of flow of charge" (preferred to "current equals charge over time"); "potential difference is the energy transferred per unit charge between two points"; "drift velocity is the average velocity of charge carriers along the wire, in addition to their random thermal motion". Phrases that lose marks: "current is how much charge flows" (no mention of rate); "voltage is the push" (no energy concept); writing $I = nAve$ as $I = nAv$ (the carrier charge $q$ or $e$ is essential and is part of the formula booklet entry).

A specific Edexcel pattern to watch: questions asking "calculate the number of electrons" require dividing total charge by $e$ ; questions asking "calculate the number density" require a different setup involving $nAve$ or atomic-mass arguments. Misreading "number" for "number density" (or vice versa) loses the entire question.

Grade-band model answers

3-mark question

Question: A current of $0.50\ \text{A}$ flows through a wire for $2.0\ \text{minutes}$ . Calculate the number of electrons that pass a point in the wire during this time. ( $e = 1.60 \times 10^{-19}\ \text{C}$ .)

Grade C response (~190 words):

$Q = It = 0.50 \times 120 = 60\ \text{C}$ .

Number of electrons $= Q / e = 60 / (1.60 \times 10^{-19}) = 3.75 \times 10^{20}$ electrons.

Examiner commentary: Full marks (3/3). The candidate converts minutes to seconds correctly (M1), applies $Q = It$ (A1) and divides by the elementary charge (A1). The working is brief but every step is verifiable. This is the standard Grade C answer for a procedural calculation: efficient, correct, with the units shown only at the final answer. A Grade A* candidate would typically include units on every intermediate line and round the final answer explicitly to 2 or 3 significant figures.

Grade A response (~230 words):*

The total charge is given by $Q = It$ where $t$ must be in seconds:

$t = 2.0\ \text{minutes} \times 60\ \text{s minute}^{-1} = 120\ \text{s}$ $Q = 0.50\ \text{A} \times 120\ \text{s} = 60\ \text{C}$

Each electron carries charge $e = 1.60 \times 10^{-19}\ \text{C}$ , so the number of electrons $N$ that pass a point is:

$N = \dfrac{Q}{e} = \dfrac{60}{1.60 \times 10^{-19}} = 3.75 \times 10^{20}$

To 2 s.f., $N \approx 3.8 \times 10^{20}\ \text{electrons}$ .

Examiner commentary: Full marks (3/3). Beyond procedural correctness, the candidate carries SI units through every line, names the elementary charge explicitly rather than just substituting a number, and rounds the final answer to a sensible number of significant figures (matching the data, which are given to 2 s.f.). The closing sentence demonstrates "examination craft" — the kind of self-checking that signals examiner-aware sophistication and would matter on longer multi-step questions where unit slips compound.

6-mark question

Question: A copper wire of diameter $0.80\ \text{mm}$ carries a current of $1.5\ \text{A}$ . The number density of free electrons in copper is $8.5 \times 10^{28}\ \text{m}^{-3}$ . (a) Calculate the cross-sectional area of the wire. (b) Calculate the drift velocity of the free electrons. (c) State and briefly justify one assumption made in deriving $I = nAve$ .

Grade A response (~310 words):*

(a) The radius is $r = 0.40\ \text{mm} = 4.0 \times 10^{-4}\ \text{m}$ .

$A = \pi r^2 = \pi \times (4.0 \times 10^{-4})^2 = 5.03 \times 10^{-7}\ \text{m}^2$

(b) Rearranging $I = nAve$ :

$v = \dfrac{I}{nAe} = \dfrac{1.5}{(8.5 \times 10^{28}) \times (5.03 \times 10^{-7}) \times (1.60 \times 10^{-19})}$

Denominator: $8.5 \times 5.03 \times 1.60 = 68.4\$ , with powers $10^{28-7-19} = 10^{2}$ . So denominator $= 6.84 \times 10^{3}$ .

$v = \dfrac{1.5}{6.84 \times 10^{3}} \approx 2.2 \times 10^{-4}\ \text{m s}^{-1}$

(c) One assumption is that all free electrons in the metal contribute to conduction (i.e. that $n$ counts the conduction electrons, typically one per atom for copper). This is justified because copper is a monovalent metal: each atom contributes exactly one delocalised electron to the conduction band. An alternative assumption is that drift velocity is uniform across the cross-section — strictly true only for steady DC current in a uniform conductor.

Examiner commentary: Full marks (6/6). Part (a) is a straightforward AO1 mark for area of a circle in SI units. Part (b) earns three marks: rearrangement (M1), substitution (M1), correct numerical answer with units (A1). Part (c) is the AO3 evaluation step — the candidate names a specific assumption and justifies it with reference to atomic structure. Many candidates state "we assume all electrons drift" without explaining why this is reasonable for copper specifically; the A* candidate supplies the monovalent-metal justification. The alternative answer about uniform drift velocity would also score full marks and shows the candidate is thinking about the full set of approximations.

9-mark question

Question: A semiconductor strip and a metal wire of identical cross-sectional area carry the same current. The number density of charge carriers in the semiconductor is $10^{20}\ \text{m}^{-3}$ and in the metal is $10^{29}\ \text{m}^{-3}$ . (a) Compare the drift velocities in the two materials, justifying your answer with reference to $I = nAve$ . (b) Explain why this comparison illustrates the difference between metallic and semiconductor conduction. (c) Suggest one practical consequence of this difference for the design of electronic components.

Grade A response (~390 words):*

(a) From $I = nAve$ , with $I$ , $A$ and $e$ identical for both materials, drift velocity is inversely proportional to number density:

$\dfrac{v_\text{semi}}{v_\text{metal}} = \dfrac{n_\text{metal}}{n_\text{semi}} = \dfrac{10^{29}}{10^{20}} = 10^{9}$

So the drift velocity in the semiconductor is one billion times larger than in the metal. If $v_\text{metal} \sim 10^{-4}\ \text{m s}^{-1}$ , then $v_\text{semi} \sim 10^{5}\ \text{m s}^{-1}$ — comparable to thermal speeds.

(b) In a metal, conduction electrons are abundant (about one per atom) but each drifts slowly. In a semiconductor, charge carriers are scarce — they must be thermally or optically excited from the valence to the conduction band — but those few carriers must move much faster to deliver the same current. This is why semiconductor conductivity is highly temperature-dependent: heating creates more carriers ( $n$ rises) so resistance falls, opposite to the metallic behaviour where heating increases lattice scattering and resistance rises.

(c) One practical consequence: semiconductor devices such as diodes and transistors can be switched rapidly because the small number of carriers can respond quickly to changes in applied field, whereas metal interconnects in the same circuit carry the bulk current with low resistive heating thanks to their large $n$ . Modern electronics exploits both materials in concert.

Examiner commentary: Full marks (9/9). Part (a) earns three marks: identifying the proportionality (M1), correct ratio (M1), correct numerical conclusion with sensible units (A1). Part (b) earns three marks for the physics explanation, with the temperature-dependence point being the AO3 evaluative move that distinguishes A* from A. Part (c) earns three marks for an applied AO2 consequence with technological relevance — many candidates stop at "semiconductors are used in electronics" without explaining why the drift-velocity difference matters; the A* candidate explicitly links the small $n$ to fast switching. The closing remark about metals and semiconductors being used together in real devices is the kind of synthesis Edexcel rewards on the highest-mark questions.

A-Level-depth misconceptions

The errors that distinguish A from A* on charge/current/p.d. questions:

Drift velocity vs signal speed. Drift velocity is $\sim 10^{-4}\ \text{m s}^{-1}$ ; the speed at which a lamp lights up is $\sim 10^{8}\ \text{m s}^{-1}$ . These are different physical quantities: the first is the average velocity of carriers along the wire, the second is the propagation speed of the electromagnetic field. Conflating them ("electrons travel at near the speed of light") is the single most penalised conceptual error on this sub-topic.
Thermal speed vs drift velocity. Free electrons in a metal at room temperature have random thermal speeds of order $10^{5}\ \text{m s}^{-1}$ — a billion times larger than typical drift velocities. The drift is the small bias superimposed on this thermal jiggling. Saying "the electrons aren't moving when there is no current" is wrong; saying "their average velocity is zero when there is no current" is right.
Counting carriers. $n$ in $I = nAve$ is the number density of charge carriers, not the number density of atoms. For copper these coincide (one conduction electron per atom), but for divalent metals they differ by a factor of two, and for semiconductors they can differ by many orders of magnitude.
$Q$ vs $q$ . $Q$ in $Q = It$ is the total charge that flows over a time interval; $q$ in $I = nAve$ (sometimes written $e$ for electrons) is the charge per carrier. They have the same SI unit (the coulomb) but represent different physical quantities, and confusing them produces order-of-magnitude errors.
Sign of conventional current. Conventional current is the direction in which positive charge would flow; in a metal, electrons drift in the opposite direction. The sign convention matters when applying Kirchhoff's laws or interpreting Hall-effect measurements. Writing " $I = -nAve$ for electrons" is the precise statement; the formula booklet form $I = nqvA$ uses unsigned magnitudes.
Potential vs potential difference. Potential is defined relative to a chosen zero (usually earth or infinity); potential difference is the difference between two points and is independent of that choice. Saying "the potential at the positive terminal is 12 V" is meaningful only with respect to a reference; saying "the potential difference across the battery is 12 V" is unambiguous.
Energy per charge vs energy per electron. $V = W/Q$ gives energy per coulomb, so a 12 V supply delivers 12 J of energy per coulomb, not 12 J per electron. The energy per electron is $eV = 1.60 \times 10^{-19} \times 12 = 1.92 \times 10^{-18}\ \text{J}$ (also expressible as 12 eV). This subtle distinction matters in atomic-physics contexts where energies are quoted in electronvolts.

Common errors and mark-loss patterns

Three patterns repeatedly cost candidates marks on Paper 1 charge/current/p.d. questions. They are all about quantitative discipline, not conceptual gaps.

Unit-conversion lapses. Time given in minutes or hours, distances in mm, areas in mm², all need conversion to SI before substitution. Candidates routinely forget that $1\ \text{mm}^2 = 10^{-6}\ \text{m}^2$ (not $10^{-3}$ ) and produce drift velocities that are a thousand times too small. The cure: write the conversion explicitly on a separate line (" $A = 1.5\ \text{mm}^2 = 1.5 \times 10^{-6}\ \text{m}^2$ ") rather than doing it mentally during substitution.
Significant-figure mismatches in "show that". A "show that" question demands that the candidate produce an answer with at least one extra significant figure beyond the printed value. If the question prints $1.2 \times 10^{-4}\ \text{m s}^{-1}$ and the candidate writes $1.2 \times 10^{-4}$ , the A1 is in jeopardy — examiners want to see $1.18 \times 10^{-4}$ to confirm the calculation actually works. The cure: always carry one extra s.f. through to the final line of a "show that".
Forgetting carrier charge in $I = nAve$ . The most common slip is writing $I = nAv$ and forgetting the factor $e$ (or $q$ ). This produces a current that is $10^{19}$ times too small. The formula booklet has $I = nqvA$ in the correct form; the candidate must read it from the booklet rather than reconstruct it from memory.

Going further — university and academic signposting

Charge/current/p.d. and the $I = nAve$ relation point directly toward several undergraduate trajectories:

Solid-state physics (Year 2/3): the free-electron model of Drude (1900) treats conduction electrons as a classical gas; the more sophisticated Sommerfeld model uses Fermi-Dirac statistics and predicts the Fermi velocity $v_F \sim 10^{6}\ \text{m s}^{-1}$ — yet the drift velocity remains tiny because electrons can scatter only into unoccupied states above the Fermi level. The deep insight is that the drift you compute from $I = nAve$ is a small perturbation on top of a quantum sea of fast electrons.
Plasma physics (Year 3): in an ionised gas, both ions and electrons drift, with very different number densities and masses. The $I = nAve$ framework generalises to a sum over species, $I = A \sum_i n_i q_i v_i$ , which is the starting point for analysing currents in fusion plasmas and the ionosphere.
Semiconductor device physics (Year 2): $I = nAve$ in a doped semiconductor includes both electron and hole contributions, with $n$ controlled by dopant concentration. The pn-junction diode equation $I = I_0(e^{eV/k_BT} - 1)$ is derived by tracking how the carrier density $n$ depends exponentially on the applied voltage $V$ .
Superconductivity (Year 3/4): in a superconductor, current flows with zero resistance, which means drift velocity is sustained by an inertial-like response with no scattering. The London equation $\partial j / \partial t = (n_s e^2/m) E$ is the superconducting analogue of $I = nAve$ , and the surprising prediction that $E = 0$ inside a superconductor follows from it.

Oxbridge interview prompt: "If electrons in a copper wire drift at 0.1 mm/s, how can a lamp light up apparently instantly when you flick the switch in your bedroom? Estimate, with reasoning, the actual time delay between flicking the switch and the lamp emitting light."

Required practical reference

The charge/current/p.d. sub-topic underpins three of the Edexcel Core Practicals in the circuits suite. CP6 (resistance versus length of a wire) is built on $V = IR$ and the prior measurement of $I$ via an ammeter and $V$ via a voltmeter — the very instruments whose ideal behaviours (zero and infinite resistance respectively) you established using the definitions of current and p.d. CP7 (current–voltage characteristics of components) requires plotting $I$ against $V$ for a metallic conductor, a filament lamp, and a semiconductor diode; the $x$ - and $y$ -axes are exactly the quantities defined in this lesson, and the gradient interpretation depends on $V = W/Q$ . CP8 (emf and internal resistance) uses $V = \varepsilon - Ir$ , which combines energy conservation per unit charge (the $V = W/Q$ definition) with current conservation at a junction. In all three, the experimental technique is the same: connect ammeters in series, voltmeters in parallel, vary one quantity, plot a straight-line graph, extract gradient and intercept. Examiners on Paper 3 (General and Practical Principles) reward candidates who can describe these protocols and explain why each step is taken — and the "why" almost always traces back to the definitions established here.

Edexcel A-Level alignment footer

This content is aligned with the Pearson Edexcel GCE A Level Physics (9PH0) specification, Topic 3: Electric circuits, sub-topic 3.1 — charge, current and potential difference. For the most accurate and up-to-date information, please refer to the official Pearson Edexcel specification document.

Visual summary

graph TD
    A["Microscopic picture<br/>n free carriers per m³<br/>each charge q = e<br/>drift velocity v"] --> B["Macroscopic current<br/>I = nAve"]
    B --> C["Charge over time<br/>Q = It"]
    C --> D["Energy per charge<br/>V = W/Q"]
    D --> E{"What does the<br/>question give you?"}
    E -->|"I and t"| F["Find Q via Q = It<br/>then N = Q/e for<br/>number of electrons"]
    E -->|"I, n, A"| G["Find drift velocity<br/>v = I/(nAe)"]
    E -->|"V and Q"| H["Find energy<br/>W = QV"]
    E -->|"V, I, t"| I["Find total energy<br/>W = VIt"]
    F --> J["Check SI units<br/>and significant figures"]
    G --> J
    H --> J
    I --> J

    style B fill:#3498db,color:#fff
    style J fill:#27ae60,color:#fff

Electric Charge

Conservation of Charge

Electric Current

Worked Example 1

Worked Example 2: Finding the Number of Electrons

Worked Example 3: Current from Electron Count

Conventional Current vs Electron Flow

Charge Carriers in Different Materials

Potential Difference

Worked Example 4

Worked Example 5: Multi-Step Energy Calculation

Energy Transfer in Circuits

Measuring Current and Potential Difference

The Relationship Between Charge, Current and p.d.

Key Relationships at a Glance

Worked Example 6: Comprehensive Calculation

A-Level Deep Dive: Charge, Current and Potential Difference

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the Edexcel 9PH0 Paper 1 format

Synoptic links

Mark-scheme literacy

Grade-band model answers

3-mark question

6-mark question

9-mark question

A-Level-depth misconceptions

Common errors and mark-loss patterns

Going further — university and academic signposting

Required practical reference

Edexcel A-Level alignment footer

Visual summary

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