Electric Current and Charge

Spec mapping: OCR H556 Module 4.1 — Charge and current (electric current as the rate of flow of charge, $I = \Delta Q / \Delta t$I=ΔQ/Δt; the coulomb as the SI unit of charge; the elementary charge $e = 1.60 \times 10^{-19}$e=1.60×10−19 C; charge quantisation $Q = ne$Q=ne; conventional current vs electron flow; the ampere as an SI base unit). Refer to the official OCR H556 specification document for exact wording.

Electric charge and electric current sit at the foundation of the entire Module 4 syllabus. Every later topic — drift velocity (lesson 2), Kirchhoff's laws (lessons 3 and 11), Ohm's law (lesson 5), I-V characteristics (lesson 6), resistivity (lesson 7), internal resistance (lesson 9), potential dividers (lesson 12), capacitor charging (Module 6.1) and even the photoelectric effect (Module 4.5) — rests on the two definitions in this single lesson.

This lesson tightens those definitions to A* depth. We treat charge as a quantised property of matter rather than a fluid; we treat current as the time-derivative of charge transported across a surface, not "the amount of electricity"; we untangle the confusing-but-conventional sign convention that Franklin saddled us with in 1747; and we end with worked OCR-style problems and a specimen 6-mark question.

---

What is Electric Charge?

Electric charge (symbol $Q$Q) is one of the small handful of fundamental properties of matter — alongside mass, spin, and lepton number — that is conserved in every known interaction. It is measured in coulombs (C) in SI. Charge comes in two varieties, positive and negative: like charges repel via the Coulomb interaction, unlike charges attract, and the magnitude of the force is governed by Coulomb's law (Module 6.2).

The charge carriers you will encounter at A-Level are summarised below.

Particle	Sign	Magnitude of charge
Electron	Negative	$e = 1.60 \times 10^{-19}$e=1.60×10−19 C
Proton	Positive	$+e = +1.60 \times 10^{-19}$+e=+1.60×10−19 C
Singly-charged ion (e.g. Na$^+$+, Cl$^-$−)	Either	$e$e
Doubly-charged ion (e.g. Cu$^{2+}$2+)	Either	$2e$2e
Up quark	$+2e/3$+2e/3 (never observed in isolation)	—
Down quark	$-e/3$−e/3 (never observed in isolation)	—

The magnitude $e = 1.60 \times 10^{-19}$e=1.60×10−19 C is called the elementary charge. Since 2019 the SI base units have been redefined so that the numerical value of $e$e is fixed exactly at $1.602176634 \times 10^{-19}$1.602176634×10−19 C, and the coulomb is then derived from it. You don't need to memorise the extra digits, but you should know that $e$e is a defining constant rather than something you measure.

Quark charges are fractions of $e$e, but quarks are permanently confined inside hadrons by the strong force — they have never been observed in isolation in any experiment. The smallest free charge anywhere in the universe (as far as we know) is therefore $e$e itself.

---

Charge Quantisation

One of the deepest empirical facts of physics is that electric charge is quantised. Every observed isolated charge is an integer multiple of $e$e:

$Q = \pm n e \qquad n = 0, 1, 2, 3, \dots$Q=±nen=0,1,2,3,…

You cannot have half an electron's charge on a normal object. Even though quarks carry $\pm e/3$±e/3 and $\pm 2e/3$±2e/3, they never appear in isolation, so the smallest free charge we ever measure is $e$e.

This was established experimentally by Millikan's oil drop experiment (1909). Millikan balanced tiny charged oil droplets in an electric field between two horizontal plates, varied the field until the droplet hung motionless (so that the upward electrical force balanced gravity), and measured the charge from the equilibrium condition. Repeating with hundreds of droplets, every measured charge came out as an integer multiple of $1.6 \times 10^{-19}$1.6×10−19 C.

Modern precision experiments confirm Millikan's result to about one part in $10^{20}$1020: charge is genuinely quantised, not "nearly quantised". This is in contrast to mass, which (so far as we know) takes a continuous range of values.

Worked Example 1 — Counting Electrons

A small isolated metal sphere carries a net charge of $-4.8 \times 10^{-18}$−4.8×10−18 C. How many excess electrons does it have?

$n = \frac{|Q|}{e} = \frac{4.8 \times 10^{-18}}{1.60 \times 10^{-19}} = 30 \text{ electrons}.$n=e∣Q∣​=1.60×10−194.8×10−18​=30 electrons.

Always check: the answer must be a whole number. If you get $30.04$30.04, you have a rounding error or a wrong exponent — go back to your working.

Worked Example 2 — Ionic Charge

A copper ion Cu$^{2+}$2+ in an electrolyte cell loses two outer electrons. What is its charge in coulombs?

$Q = +2e = +2 \times 1.60 \times 10^{-19} = +3.20 \times 10^{-19} \text{ C}.$Q=+2e=+2×1.60×10−19=+3.20×10−19 C.

Doubly-ionised metals are common in electrolysis (Module 4 extends into electrochemistry implicitly through the Faraday constant $F = N_A e \approx 9.65 \times 10^4$F=NA​e≈9.65×104 C mol$^{-1}$−1).

---

Defining Electric Current

Electric current (symbol $I$I) is the rate of flow of electric charge across a chosen surface. Its SI unit is the ampere (A), one of the seven SI base units.

The precise definition is a time derivative:

$I = \frac{dQ}{dt} \qquad \text{or, for steady current,} \qquad I = \frac{\Delta Q}{\Delta t}$I=dtdQ​or, for steady current,I=ΔtΔQ​

For a steady current $I$I flowing for time $t$t, the total charge that has crossed the surface is:

$Q = I t.$Q=It.

One coulomb is the charge transported by a steady current of one ampere in one second:

$1 \text{ C} = 1 \text{ A s}.$1 C=1 A s.

The ampere itself is a base unit (one of seven), and the coulomb is a derived unit built from it.

Conceptual point: Current is a flow rate, not "an amount of electricity". A river analogy is helpful: charge is the volume of water that has passed a bridge; current is the litres-per-second rate. You cannot store current — you store charge. You drive current by maintaining a potential difference.

Worked Example 3 — Charge from Current

A current of $0.25$0.25 A flows through a resistor for $2.0$2.0 minutes. Calculate the total charge transferred and the corresponding number of electrons.

• Convert time to seconds: $t = 2.0 \times 60 = 120$t=2.0×60=120 s.
• $Q = It = 0.25 \times 120 = 30$Q=It=0.25×120=30 C.
• $n = Q/e = 30 / (1.60 \times 10^{-19}) = 1.88 \times 10^{20}$n=Q/e=30/(1.60×10−19)=1.88×1020 electrons.

About $2 \times 10^{20}$2×1020 electrons cross every cross-section of the wire in just two minutes — twenty thousand billion billion charge carriers. Electrical circuits handle staggering numbers of carriers per second.

Worked Example 4 — Non-Steady Current and the Integral Form

A capacitor is charging through a resistor. The current is $I(t) = 0.10 \, e^{-t/2.0}$I(t)=0.10e−t/2.0 A, with $t$t in seconds. How much charge has flowed in the first $4.0$4.0 s?

$Q = \int_0^{4.0} I(t)\,dt = \int_0^{4.0} 0.10 \, e^{-t/2.0}\,dt = 0.10 \times [-2.0 e^{-t/2.0}]_0^{4.0}$Q=∫04.0​I(t)dt=∫04.0​0.10e−t/2.0dt=0.10×[−2.0e−t/2.0]04.0​

$= 0.10 \times 2.0 \times (1 - e^{-2.0}) = 0.20 \times (1 - 0.135) = 0.173 \text{ C}.$=0.10×2.0×(1−e−2.0)=0.20×(1−0.135)=0.173 C.

You won't be asked to evaluate exponential integrals at AS — but the graphical interpretation matters: the area under an $I$I-$t$t graph equals the charge transferred. This idea generalises Q = It to currents that vary with time, and it returns in lesson 6 (Module 6.1, capacitor discharge).

---

Conventional Current vs Electron Flow

When Benjamin Franklin defined the direction of current flow in the 1740s, the electron had not yet been discovered (J. J. Thomson identified it in 1897). Franklin made an arbitrary choice — that current flows from positive to negative — and it stuck. It turned out to be the opposite of the direction in which the negatively-charged electron carriers actually drift in a metal.

We have two compatible-but-inverted ways to describe the same physical reality:

• Conventional current: flows from + to $-$− (from the positive terminal of the battery, through the external circuit, back to the negative terminal). This is what every circuit symbol, equation and right-hand rule is calibrated to.
• Electron flow (physical reality in metals): from $-$− to + (electrons are repelled by the negative terminal and attracted to the positive). For positive carriers (e.g. positive ions in an electrolyte), conventional current and physical motion coincide.

The amount of charge transferred per second is identical in both descriptions; only the labelling differs. Since 1900 the convention has been to always draw arrows in the conventional-current direction in circuit diagrams. This is the convention OCR mark schemes enforce.

flowchart LR
  BP["Battery (+)"] --> W1[Wire] --> R[Resistor] --> W2[Wire] --> BN["Battery (-)"]
  BN -.->|"actual electron drift"| BP

The dashed line above shows the physical electron drift direction; the solid arrows show the conventional current direction. Both are correct descriptions of the same circuit.

Why the Convention Survives

You might ask: now that we know electrons move the "wrong way", why don't we just redefine the convention? Several reasons:

1. Half the carriers move the "right" way. In ionic conduction (electrolytes, plasmas, biological cells) the positive ions drift in the conventional-current direction. In semiconductor physics, "hole" current (the conduction-band absence of an electron) also moves with conventional current.
2. Existing equations are calibrated. $\vec{F} = q\vec{v} \times \vec{B}$F=qv×B, the right-hand rule for magnetic fields, $\vec{E} = -\nabla V$E=−∇V, and every mark scheme are all set up for conventional current. Changing the sign of one would cascade through every textbook ever printed.
3. Equivalent descriptions. Negative charges drifting in the $-x$−x direction are electrodynamically indistinguishable from positive charges drifting in the $+x$+x direction. The physics doesn't care; the labels do.

---

The Ampere as an SI Base Unit

Since the 2019 SI redefinition, the ampere is defined by fixing the numerical value of the elementary charge:

$e = 1.602176634 \times 10^{-19} \text{ C exactly}.$e=1.602176634×10−19 C exactly.

The ampere is then "the current corresponding to a flow of $1/e \approx 6.241 \times 10^{18}$1/e≈6.241×1018 elementary charges per second". The coulomb is derived: $1 \text{ C} = 1 \text{ A s}$1 C=1 A s.

Before 2019 the ampere was defined by the magnetic force between two parallel current-carrying wires, but that definition has now been retired. You are not expected to recite the new SI definition at A-Level, but you should know:

• The ampere (A) is a base unit.
• The coulomb (C) is a derived unit: $1 \text{ C} = 1 \text{ A s}$1 C=1 A s.
• "Express the coulomb in SI base units" → $\text{C} = \text{A s}$C=A s.

---

Measuring Current — The Ammeter

To measure current you use an ammeter, connected in series with the component whose current you want to measure. An ideal ammeter has zero resistance, so that inserting it does not disturb the circuit.

Real ammeters have a small but non-zero resistance: a cheap multimeter on the 10 A range may have $\sim 0.01$∼0.01 $\Omega$Ω, while a laboratory digital meter may be effectively zero at the precision needed for A-Level work.

flowchart LR
  B[Battery] --> A((Ammeter)) --> R[Resistor] --> B

Never connect an ammeter in parallel across a cell or a resistor. Because the ammeter has very low resistance, doing so creates a near short-circuit, drawing a huge current that can destroy the meter, burn its fuse, melt insulation, or damage the cell.

The contrast with voltmeter use is sharp: ammeters go in series, voltmeters go in parallel (lesson 4). Mixing them up is one of the classic OCR experimental-design errors.

---

Direct and Alternating Current

For Module 4 we deal almost entirely with direct current (DC): the current flows in one direction with a constant (or slowly varying) magnitude. Batteries, dry cells, solar cells and lab DC power supplies all provide DC.

Alternating current (AC) changes direction periodically — in the UK mains supply at 50 Hz, so 100 reversals every second. The current is sinusoidal: $I(t) = I_0 \sin(2\pi f t)$I(t)=I0​sin(2πft) with $f = 50$f=50 Hz. AC is studied in detail in Module 6.3 (alternating currents). For Module 4 all our calculations are DC unless stated otherwise.

A useful unifying point: at any instant, the AC current $I(t)$I(t) has a well-defined value, and Kirchhoff's first law (lesson 3) and the definition $I = dQ/dt$I=dQ/dt both still apply. AC is just DC with $I$I varying sinusoidally with time.

---

Worked Example 5 — Charging a Capacitor

A capacitor is charged by a steady current of $15$15 mA for $4.0$4.0 s. Calculate:

(a) The charge delivered.

(b) The number of electrons that passed through the wire.

(c) If the capacitor has capacitance $C = 100$C=100 $\mu$μF, what is the final pd across its plates?

(a) $Q = It = 15 \times 10^{-3} \times 4.0 = 0.060$Q=It=15×10−3×4.0=0.060 C $= 60$=60 mC.

(b) $n = Q/e = 0.060 / (1.60 \times 10^{-19}) = 3.75 \times 10^{17}$n=Q/e=0.060/(1.60×10−19)=3.75×1017 electrons.

(c) From $Q = CV$Q=CV (Module 6.1): $V = Q/C = 0.060 / (100 \times 10^{-6}) = 600$V=Q/C=0.060/(100×10−6)=600 V.

Notice the consistent use of base units throughout: mA was converted to A before substitution; $\mu$μF was converted to F before division. OCR mark schemes are strict about base-unit working; forgetting to convert remains one of the single most common errors in calculations at this level.

---

Currents on Very Different Scales

To build intuition, here is a rough ladder of electrical currents you might encounter.

Scale	Typical current
Quantum-coherence experiment (single-electron transistor)	$\sim 10$∼10 fA ($10^{-14}$10−14 A)
Nerve impulse	$\sim 10$∼10 pA ($10^{-11}$10−11 A)
Wristwatch	$\sim 1$∼1 $\mu$μA ($10^{-6}$10−6 A)
LED indicator	$\sim 20$∼20 mA ($0.02$0.02 A)
Torch bulb	$\sim 0.2$∼0.2 A
Domestic kettle (UK mains, 3 kW at 230 V)	$\sim 13$∼13 A
Domestic ring main rating	30 A
Car starter motor	$\sim 200$∼200 A
Arc-welding electrode	$\sim 500$∼500 A
Lightning strike (peak)	$\sim 30$∼30 kA ($3 \times 10^4$3×104 A)
Particle accelerator beam current	$\sim 1$∼1 mA but at $10^9$109 V

The relation $Q = It$Q=It means that even tiny currents over long times can transfer enormous charge (a nanoamp over a year transfers $\sim 0.03$∼0.03 C), and very large currents over short times can do huge amounts of damage — hence the importance of fuses and circuit breakers in real installations. A 30 A fuse will blow within milliseconds if a short pulls a current beyond its rating, isolating the fault before the wiring overheats.

---

Worked Example 6 — Total Charge Through a Domestic Appliance

A 2.0 kW electric kettle is plugged into the UK 230 V mains supply and boils a litre of water in 3.0 minutes. Calculate the total charge that flows through the heating element.

• Current drawn: from $P = VI$P=VI, $I = P/V = 2000/230 = 8.7$I=P/V=2000/230=8.7 A.
• Total charge: $Q = It = 8.7 \times (3.0 \times 60) = 8.7 \times 180 = 1570$Q=It=8.7×(3.0×60)=8.7×180=1570 C $\approx 1.6 \times 10^3$≈1.6×103 C.
• Number of electrons: $n = Q/e = 1570 / (1.60 \times 10^{-19}) = 9.8 \times 10^{21}$n=Q/e=1570/(1.60×10−19)=9.8×1021.

Nearly $10^{22}$1022 electrons cross the heating element to boil one kettle. This is roughly the number of stars in the observable universe — to boil tea.

---

Synoptic Links

• Drift velocity (lesson 2) — current $I = nAvq$I=nAvq is the microscopic picture of $I = dQ/dt$I=dQ/dt. The free-electron number density $n$n and drift velocity $v$v multiplied by carrier charge $q$q give the macroscopic current.
• Kirchhoff's first law (lesson 3) — $\Sigma I_{\text{in}} = \Sigma I_{\text{out}}$ΣIin​=ΣIout​ at any junction is just conservation of charge ($dQ/dt = 0$dQ/dt=0 for steady-state nodes) re-expressed in current language.
• Capacitor charging (Module 6.1) — the exponential current $I(t) = (V_0/R)e^{-t/RC}$I(t)=(V0​/R)e−t/RC in a charging RC circuit is the time-domain extension of the steady-state $I = dQ/dt$I=dQ/dt definition established here. The area under the $I$I-$t$t curve is the total charge stored on the capacitor.
• Photoelectric effect (Module 4.5) — the photoelectric current is $I = n_p e$I=np​e where $n_p$np​ is the rate of photoelectron emission. Both rest on $I = dQ/dt$I=dQ/dt and the quantisation $Q = ne$Q=ne.
• Magnetic forces (Module 6.2) — the force on a current-carrying conductor $F = BIL$F=BIL requires $I$I in the conventional direction; getting the sign wrong reverses the predicted force direction.

---

Specimen question modelled on the OCR H556 paper format

Question (6 marks): A car battery is rated as "60 A h" (60 ampere-hours), meaning it can deliver $60$60 A for $1$1 h before its terminal voltage drops below a usable level.

(a) Explain what is meant by "current is the rate of flow of charge". [1]

(b) Calculate the total charge stored in a fully-charged 60 A h battery. [2]

(c) Estimate the total number of electrons that pass through the battery during a complete discharge. The elementary charge is $e = 1.60 \times 10^{-19}$e=1.60×10−19 C. [3]

AO breakdown

Mark	AO	Awarded for
1	AO1	Definition: current $I$I equals charge $\Delta Q$ΔQ transferred per unit time $\Delta t$Δt ($I = \Delta Q / \Delta t$I=ΔQ/Δt).
2	AO2	Conversion: $1$1 h $= 3600$=3600 s, so $Q = It = 60 \times 3600$Q=It=60×3600.
3	AO2	$Q = 2.16 \times 10^5$Q=2.16×105 C.
4	AO2	$n = Q/e$n=Q/e identified as the equation needed.
5	AO2	$n = 2.16 \times 10^5 / 1.60 \times 10^{-19}$n=2.16×105/1.60×10−19.
6	AO3	$n \approx 1.35 \times 10^{24}$n≈1.35×1024 electrons, with appropriate sig figs and units.

AO split: AO1 = 1, AO2 = 4, AO3 = 1.

Mid-band response (3/6)

(a) Current is how much charge flows.

(b) $Q = It = 60 \times 1 = 60$Q=It=60×1=60 C.

(c) Number of electrons $= 60 / 1.6 \times 10^{-19} = 3.75 \times 10^{20}$=60/1.6×10−19=3.75×1020.

Examiner commentary: The next-band move is unit consistency — treating "1 hour" as "1" rather than converting to seconds. The candidate's $Q = 60$Q=60 C is exactly the charge in one second of 60 A current, not in one hour. Mark 1 partial (definition gets the qualitative idea but misses "per unit time" / time-rate), Mark 2 lost (no second-conversion), Mark 3 lost (numerical value wrong by factor of 3600), Mark 4 awarded (correct equation $n = Q/e$n=Q/e), Mark 5 lost (uses the wrong $Q$Q), Mark 6 partial (arithmetic is internally consistent but the wrong starting value). A reliable mid-band trap: failing to convert hours to seconds is the single most common SI-units error in OCR electricity papers.

Stronger response (5/6)

(a) Current is the rate of flow of electric charge, $I = \Delta Q / \Delta t$I=ΔQ/Δt.

(b) Convert: $1$1 h $= 3600$=3600 s. $Q = It = 60 \times 3600 = 216\,000$Q=It=60×3600=216000 C.

(c) Number of electrons: $n = Q/e = 216\,000 / 1.60 \times 10^{-19} = 1.35 \times 10^{24}$n=Q/e=216000/1.60×10−19=1.35×1024 electrons.

Examiner commentary: The next-band move is more precise unit and significant-figure handling — writing $Q = 2.16 \times 10^5$Q=2.16×105 C in standard form, and explicitly stating "approximately" in (c) given the rating is itself an approximation. The candidate has all the physics right and would gain Marks 1, 2, 3, 4, 5 and most of 6 — losing the sig-fig presentation mark only. This is a solid response that with a touch more polish would hit full marks.

Top-band response (6/6)

(a) Electric current is defined as the rate of flow of electric charge: $I = \Delta Q / \Delta t$I=ΔQ/Δt, with $I$I in amperes when $\Delta Q$ΔQ is in coulombs and $\Delta t$Δt in seconds.

(b) Converting time to SI base units: $t = 1$t=1 h $= 3600$=3600 s. Total charge delivered:

$Q = It = 60 \times 3600 = 2.16 \times 10^5 \text{ C}.$Q=It=60×3600=2.16×105 C.

(c) The total charge corresponds to $n = Q/e$n=Q/e elementary charges, each carried by one electron:

$n = \frac{Q}{e} = \frac{2.16 \times 10^5}{1.60 \times 10^{-19}} \approx 1.35 \times 10^{24} \text{ electrons}.$n=eQ​=1.60×10−192.16×105​≈1.35×1024 electrons.

This is roughly Avogadro's number times 2 — an entire two moles of electrons cycling through the battery in a single full discharge, which is consistent with the lead-acid battery storing energy electrochemically by displacing two moles of charge per mole of PbO$_2$2​ reduced.

Examiner commentary: Full marks. The candidate has converted units carefully, used scientific notation consistently, and made the synoptic connection to electrochemistry (mole-counts of electrons) without being asked. The depth shown in the final remark goes beyond what is needed for marks but indicates secure command of the topic.

---

Common errors and mark-loss patterns

• Confusing $Q$Q and $I$I. Charge and current are different quantities with different units (C vs A). Mark schemes reject answers where these are interchanged.
• Forgetting to convert minutes (or hours) to seconds. The formula $Q = It$Q=It requires $t$t in seconds. Failing to convert is the single most common arithmetic error.
• Drawing the current arrow from $-$− to $+$+ in the external circuit. Always from $+$+ to $-$− in conventional current.
• Treating charge as continuous. For counting problems, $Q = ne$Q=ne with $n$n an integer. A non-integer answer indicates an error.
• Quoting $e$e without units. Always $e = 1.60 \times 10^{-19}$e=1.60×10−19 C, not just "$1.60 \times 10^{-19}$1.60×10−19".
• Stating current as "amount of electricity". Current is a rate, not an amount. The amount is the charge $Q$Q.
• Stating the ampere as derived from the coulomb. It is the other way round: the ampere is base, the coulomb is derived.

---

A-Level misconceptions

• Current is the FLOW rate of charge, not the "amount of electricity". Current is a derivative ($dQ/dt$dQ/dt); charge is the cumulative quantity. The river analogy: charge $=$= litres of water past a bridge; current $=$= litres per second.
• Conventional current flows + to $-$− even though electrons flow $-$− to +. Both descriptions are physically equivalent; OCR mark schemes enforce conventional-current arrows in diagrams.
• Charge is quantised, not continuous. Any free charge is an integer multiple of $e$e. Quark fractions exist but are confined.
• The ampere is a base unit; the coulomb is derived. Not the other way round. $1$1 C $= 1$=1 A s.
• Current is not "used up" by components. In a series circuit, the same current flows through every component. It is energy that is dissipated, not current — see lesson 4 (EMF and pd) and lesson 8 (electrical power).

---

Going further

• The Drude model (undergraduate). Treats the metal as a classical gas of free electrons drifting under an applied field with a characteristic relaxation time $\tau$τ. The drift velocity comes out as $v_d = eE\tau/m$vd​=eEτ/m, which yields Ohm's law and gives $n = Q/(eA L)$n=Q/(eAL) in the wire-volume picture (lesson 2).
• Quantum Hall effect. Under strong magnetic fields and low temperatures, the Hall conductance is quantised in units of $e^2/h$e2/h — a macroscopic manifestation of charge quantisation, with experimental precision now better than one part in $10^{10}$1010. This underpins the SI definition of resistance.
• Single-electron transistors. Devices small enough that adding even one electron measurably changes the conductance, allowing direct measurement of $e$e in a solid-state device.
• Faraday's electrolysis laws. Mass deposited at an electrode is $m = (Q/F) \cdot M/z$m=(Q/F)⋅M/z where $F = N_A e$F=NA​e is the Faraday constant. Counting electrons in a wire is equivalent to counting moles of ions in solution — a beautiful synoptic link between current, charge, and chemistry.
• Oxbridge interview prompt: "Explain why charge is conserved but mass is not — and why this difference matters for nuclear physics."
• Reading: Feynman Lectures on Physics Vol II, Chapters 1-2; Griffiths Introduction to Electrodynamics Chapter 5.

---

Summary

• Charge $Q$Q: a fundamental conserved property of matter, measured in coulombs, quantised in units of $e = 1.60 \times 10^{-19}$e=1.60×10−19 C.
• Current $I$I: the rate of flow of charge, $I = dQ/dt$I=dQ/dt, measured in amperes (SI base unit). $1$1 A $= 1$=1 C s$^{-1}$−1.
• For steady current: $Q = It$Q=It (with $t$t in seconds).
• Conventional current flows + to $-$−; electron flow is $-$− to + in metals. Both describe the same physics.
• Ammeter in series, zero resistance ideally; never in parallel across a component.
• Foundation for all of Module 4: drift velocity, Kirchhoff's laws, Ohm's law, internal resistance, potential dividers.