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charge past a point.","correct":false,"explanation":"This describes the amount of charge passing a point, not how quickly it passes. Electric current is the rate of flow of charge past a point."},{"text":"The energy transferred per unit charge past a point.","correct":false,"explanation":"This confuses current with potential difference, which provides energy per unit charge. Current instead measures the rate at which charge flows past a point."},{"text":"The opposition to the flow of charge past a point.","correct":false,"explanation":"This describes the resistance of a component, which opposes charge flow. Electric current is the rate of flow of charge past a point."}]},{"id":"2d4d13c3-1216-4bff-a273-492050ae18b6","cardType":"cloze","points":null,"front":"The magnitude of total charge $Q$ for a collection of $N$ electrons is given by the equation $\\text{[...]}$.","back":"The magnitude of total charge $Q$ for a collection of $N$ electrons is given by the equation $\\boxed{$Q = Ne$}$.","mcqOptions":[{"text":"$$Q = Ne$","correct":true,"explanation":"Each electron has charge magnitude $e$, so $N$ electrons have a total magnitude equal to the number of electrons multiplied by $e$: $Q = Ne$."},{"text":"$$Q = \\frac{N}{e}$","correct":false,"explanation":"Dividing by $e$ gives the number of electrons if the total charge is known, rather than the total charge produced by $N$ electrons. The correct relation is $Q = Ne$."},{"text":"$$Q = N + e$","correct":false,"explanation":"Charge contributions from individual electrons add, so multiplication is required. Adding $N$ and $e$ mixes a count with a charge magnitude; the correct relation is $Q = Ne$."},{"text":"$$Q = \\frac{e}{N}$","correct":false,"explanation":"Multiplying the number of electrons by the reciprocal of their charge gives the wrong dependence on $e$. The total charge magnitude is $Q = Ne$."}]},{"id":"e3ece4bf-09bc-472a-b4fa-51a6fbf4d99c","cardType":"basic","points":1,"front":"What does the unit ampère (A) represent in terms of base SI units?","back":"Coulombs per second ($\\text{C s}^{-1}$).","mcqOptions":[{"text":"Coulombs per second ($\\text{C s}^{-1}$).","correct":true,"explanation":"One ampère is one coulomb of charge passing each second, so its equivalent SI unit is coulombs per second, $\\text{C s}^{-1}$."},{"text":"Coulombs per metre ($\\text{C m}^{-1}$).","correct":false,"explanation":"Coulombs per metre describes charge per length, not charge flow per time. An ampère represents coulombs per second, $\\text{C s}^{-1}$."},{"text":"Coulombs per joule ($\\text{C J}^{-1}$).","correct":false,"explanation":"Coulombs per joule describes the reciprocal of potential difference, not current. An ampère represents coulombs per second, $\\text{C s}^{-1}$."},{"text":"Joules per coulomb ($\\text{J C}^{-1}$).","correct":false,"explanation":"Joules per coulomb is the unit of potential difference, not current. An ampère represents coulombs per second, $\\text{C s}^{-1}$."}]},{"id":"f9103a82-4855-4314-a1f1-091239bfea42","cardType":"basic","points":2,"front":"Why are metals good conductors of electricity?","back":"They contain free (delocalised) electrons that are mobile charge carriers.","mcqOptions":[{"text":"They contain free (delocalised) electrons that are mobile charge carriers.","correct":true,"explanation":"Metals have free, or delocalised, electrons that can move through the lattice. These mobile charge carriers allow electric charge to flow readily."},{"text":"They contain mobile positive ions that are charge carriers.","correct":false,"explanation":"Metal atoms are held in a lattice and do not normally move through the solid to carry current. Conductivity comes from free, delocalised electrons."},{"text":"They contain closely packed nuclei that are mobile charge carriers.","correct":false,"explanation":"A high density of fixed nuclei does not provide moving charge. Metals conduct well because their free, delocalised electrons can drift through the lattice."},{"text":"They contain strongly bonded electrons that are mobile charge carriers.","correct":false,"explanation":"Metallic bonding does not make all electrons move freely. Only the delocalised electrons act as mobile charge carriers and produce electrical conduction."}]},{"id":"61aab6d4-9213-4513-a813-216a362d5044","cardType":"basic","points":2,"front":"Contrast conventional current direction with electron flow in a metal.","back":"

Conventional current: positive to negative
Electron flow: negative to positive

","mcqOptions":[{"text":"

Conventional current: positive to negative
Electron flow: negative to positive

","correct":true,"explanation":"Conventional current is defined as the direction positive charge would move, from positive to negative. Electrons are negatively charged, so they drift in the opposite direction."},{"text":"

Conventional current: negative to positive
Electron flow: positive to negative

","correct":false,"explanation":"The electron direction has been reversed. Conventional current runs from positive to negative, while electrons in a metal flow from negative to positive."},{"text":"

Conventional current: positive to negative
Electron flow: positive to negative

","correct":false,"explanation":"Electrons do not flow in the same direction as conventional current because their charge is negative. Conventional current is positive to negative; electron flow is negative to positive."},{"text":"

Conventional current: negative to positive
Electron flow: negative to positive

","correct":false,"explanation":"Conventional current is not directed according to electron motion. It goes from positive to negative, whereas electron flow goes from negative to positive."}]},{"id":"76e75e34-5769-4b0c-aea1-f839524d26d3","cardType":"basic","points":1,"front":"What is the definition of number density ($n$)?","back":"The number of charge carriers per unit volume.","mcqOptions":[{"text":"The number of charge carriers per unit volume.","correct":true,"explanation":"Number density describes how many charge carriers are present in a given volume. It is therefore the number of charge carriers per unit volume."},{"text":"The volume occupied by charge carriers per unit number.","correct":false,"explanation":"This gives the volume occupied by each carrier rather than the number of carriers in a volume. Number density is the number of charge carriers per unit volume."},{"text":"The total charge per unit volume.","correct":false,"explanation":"This describes charge per unit volume, which is charge density, not number density. Number density counts charge carriers per unit volume."},{"text":"The distance between charge carriers per unit volume.","correct":false,"explanation":"This describes the spacing between carriers rather than their concentration. Number density is the number of charge carriers per unit volume."}]},{"id":"3f29eac4-fc93-4d99-b6b1-8f22de47e7a4","cardType":"cloze","points":null,"front":"The SI unit for number density ($n$) is $\\text{[...]}$.","back":"The SI unit for number density ($n$) is $\\boxed{$\\text{m}^{-3}$}$.","mcqOptions":[{"text":"$$\\text{m}^{-3}$","correct":true,"explanation":"Number density counts charge carriers in a volume. Since volume has units of $\\text{m}^3$, the corresponding unit is $\\text{m}^{-3}$."},{"text":"$$\\text{m}^{-2}$","correct":false,"explanation":"$$\\text{m}^{-2}$ represents a quantity per unit area. Number density is per unit volume, so its SI unit is $\\text{m}^{-3}$."},{"text":"$$\\text{m}^{-1}$","correct":false,"explanation":"$$\\text{m}^{-1}$ represents a quantity per unit length. Number density is measured per unit volume, giving the unit $\\text{m}^{-3}$."},{"text":"$$\\text{m}^{3}$","correct":false,"explanation":"$$\\text{m}^{3}$ is a unit of volume, not a unit per volume. Number density is the number of carriers divided by volume, so its unit is $\\text{m}^{-3}$."}]},{"id":"0de7936e-3c17-4d76-8c4e-9dbb6f108e9e","cardType":"basic","points":1,"front":"How is drift velocity ($v$) defined?","back":"The average velocity of charge carriers along a conductor due to a potential difference.","mcqOptions":[{"text":"The average velocity of charge carriers along a conductor due to a potential difference.","correct":true,"explanation":"Drift velocity is the average velocity acquired by charge carriers because of the electric field produced by a potential difference across the conductor."},{"text":"The maximum velocity of charge carriers along a conductor due to a potential difference.","correct":false,"explanation":"The random thermal motion of carriers occurs even without a potential difference. Drift velocity is the average velocity caused by the applied potential difference."},{"text":"The random velocity of charge carriers along a conductor due to a potential difference.","correct":false,"explanation":"Charge carriers already have random motion, but drift velocity refers specifically to their average directed velocity caused by a potential difference."},{"text":"The acceleration of charge carriers along a conductor due to a potential difference.","correct":false,"explanation":"The potential difference produces a directed drift of carriers, not their acceleration as the definition. Drift velocity is their average velocity along the conductor."}]},{"id":"491451e4-cb59-40f9-8cd7-9680bfa5f2fc","cardType":"basic","points":2,"front":"Describe the motion of free electrons in a metal before a potential difference is applied.","back":"They move randomly at high speeds with no overall net flow.","mcqOptions":[{"text":"They move randomly at high speeds with no overall net flow.","correct":true,"explanation":"Without an applied potential difference, conduction electrons have rapid, random thermal motion. Their directions cancel statistically, so there is no overall drift or net flow of charge."},{"text":"They move slowly in one direction with no overall net flow.","correct":false,"explanation":"An applied field produces a small average drift, but before the potential difference there is no preferred direction. The electrons still move rapidly and randomly because of thermal energy."},{"text":"They remain stationary at all times with no overall net flow.","correct":false,"explanation":"Free electrons in a metal are not stationary before connection; thermal energy keeps them moving rapidly. Their random directions cancel, giving no net current rather than no motion."},{"text":"They move randomly at high speeds with a steady net flow.","correct":false,"explanation":"The high-speed random motion is correct, but without a potential difference the random directions produce no net flow. A net flow requires an average drift caused by an electric field."}]},{"id":"3f9a7872-a791-442b-94e5-cd914362f7b9","cardType":"cloze","points":null,"front":"When a potential difference is applied to a metal, an $\\text{[...]}$ is established, exerting a force on free electrons and causing them to $drift$.","back":"When a potential difference is applied to a metal, an $\\boxed{electric field}$ is established, exerting a force on free electrons and causing them to $drift$.","mcqOptions":[{"text":"electric field","correct":true,"explanation":"A potential difference establishes an electric field throughout the conductor. The field exerts a force on free electrons, giving their random motion a small average drift."},{"text":"potential difference","correct":false,"explanation":"A potential difference establishes an electric field, not another potential difference. The field exerts the force that causes free electrons to drift through the metal."},{"text":"current","correct":false,"explanation":"Current is the rate of charge flow and results from drifting electrons; it is not the field established directly by applying a potential difference."},{"text":"magnetic field","correct":false,"explanation":"A magnetic field is not the general cause of electron drift in this situation. The applied potential difference establishes an electric field inside the metal."}]},{"id":"626b24f4-3518-4b4d-aa22-f87b39797d32","cardType":"cloze","points":null,"front":"When a potential difference is applied to a metal, an $electric field$ is established, exerting a force on free electrons and causing them to $\\text{[...]}$.","back":"When a potential difference is applied to a metal, an $electric field$ is established, exerting a force on free electrons and causing them to $\\boxed{drift}$.","mcqOptions":[{"text":"drift","correct":true,"explanation":"The electric field causes free electrons to drift: their rapid random motion remains, but they gain a small average velocity in the direction opposite the field."},{"text":"accelerate","correct":false,"explanation":"An electric field causes electrons to acquire a small average drift velocity, rather than continually accelerating without collisions. Frequent collisions prevent sustained acceleration through the conductor."},{"text":"vibrate","correct":false,"explanation":"Thermal motion is already random before the field is applied, and the field produces a directional drift rather than merely making electrons vibrate."},{"text":"flow","correct":false,"explanation":"Electrons drift through the metal, while conventional current is the associated charge flow. The sentence asks for the motion caused directly by the field."}]},{"id":"8417e436-2b7d-4955-acf5-1cd6920b9b5b","cardType":"basic","points":1,"front":"In the derivation of $I = nAve$, what is the volume of electrons that pass a cross-section in time $\\Delta t$?","back":"$$Av\\Delta t$","mcqOptions":[{"text":"$$Av\\Delta t$","correct":true,"explanation":"In time $[AvackslashDelta t]$, electrons travelling at drift speed $v$ move a distance $vackslashDelta t$. Multiplying this length by cross-sectional area $A$ gives the swept volume."},{"text":"$$\\frac{Av}{\\Delta t}$","correct":false,"explanation":"Dividing by time gives the wrong dimensions for a volume. The distance travelled is $v\\Delta t$, which must be multiplied by area $A$."},{"text":"$$\\frac{A\\Delta t}{v}$","correct":false,"explanation":"This uses the reciprocal of the distance travelled. In time $\\Delta t$, the electrons move $v\\Delta t$, so the volume is area times that distance."},{"text":"$$\\pi Av\\Delta t$","correct":false,"explanation":"The extra factor of $\\pi$ has no physical basis because the cross-sectional area $A$ is already included. The swept volume is simply area multiplied by distance."}]},{"id":"6ffbc341-68a0-4080-ade0-78fd4a4092ce","cardType":"basic","points":2,"front":"In the derivation of $I = nAve$, how is the total charge ($\\Delta Q$) in a volume expressed using number density ($n$)?","back":"$$\\Delta Q = n(Av\\Delta t)e$","mcqOptions":[{"text":"$$\\Delta Q = n(Av\\Delta t)e$","correct":true,"explanation":"The number of electrons in the swept volume is $n(Av\\Delta t)$. Multiplying that number by the charge magnitude $e$ gives the total charge $\\Delta Q$."},{"text":"$$\\Delta Q = n(Av\\Delta t)$","correct":false,"explanation":"Number density times volume gives the number of electrons, not their charge. Multiplication by the elementary charge $e$ is required to obtain $\\Delta Q$."},{"text":"$$\\Delta Q = \\frac{n(Av\\Delta t)e}{\\Delta t}$","correct":false,"explanation":"Dividing by $\\Delta t$ calculates a rate rather than the charge passing in the interval. The charge is found by multiplying the electron count by $e$."},{"text":"$$\\Delta Q = n(v\\Delta t)e$","correct":false,"explanation":"The cross-sectional area and drift speed determine the swept volume, so both must be present. Omitting $A$ undercounts the electrons and therefore the charge."}]},{"id":"1a03047a-7ae1-4dee-a051-e9cb54e8ba29","cardType":"basic","points":1,"front":"State typical magnitudes for drift velocity in metal wires.","back":"Very small, often fractions of a millimetre per second.","mcqOptions":[{"text":"Very small, often fractions of a millimetre per second.","correct":true,"explanation":"Drift velocity is an average directional velocity superimposed on rapid random electron motion. In ordinary metal wires it is typically extremely small, often fractions of a millimetre per second."},{"text":"Very small, often fractions of a metre per second.","correct":false,"explanation":"Fractions of a metre per second are generally far too large for ordinary electron drift in a metal wire. Typical drift speeds are fractions of a millimetre per second."},{"text":"Very small, often several millimetres per second.","correct":false,"explanation":"Several millimetres per second overestimates the usual drift speed in a metal wire. The typical scale is smaller, commonly fractions of a millimetre per second."},{"text":"Very small, often tens of centimetres per second.","correct":false,"explanation":"Tens of centimetres per second confuses drift speed with the much faster random thermal motion of electrons. The average directed drift is usually fractions of a millimetre per second."}]},{"id":"5cb50188-17f5-43e8-9bd8-5478a0fb62ac","cardType":"basic","points":1,"front":"How do you calculate the cross-sectional area ($A$) of a circular wire from its diameter ($d$)?","back":"$$A = \\pi (\\frac{d}{2})^2$ or $A = \\frac{\\pi d^2}{4}$","mcqOptions":[{"text":"$$A = \\pi (\\frac{d}{2})^2$ or $A = \\frac{\\pi d^2}{4}$","correct":true,"explanation":"A circular wire has radius $d/2$. Using the circle area formula $A=\\pi r^2$ gives $A=\\pi(d/2)^2$, equivalently $\\pi d^2/4$."},{"text":"$$A = \\pi d^2$ or $A = \\frac{\\pi d^2}{1}$","correct":false,"explanation":"Using the diameter as the radius makes the calculated area four times too large. The radius is half the diameter, so the diameter must be divided by two before squaring."},{"text":"$$A = \\pi (\\frac{d}{2})^2$ or $A = \\frac{\\pi d^2}{2}$","correct":false,"explanation":"This introduces an incorrect factor of two. Squaring the radius produces a factor of four in the denominator, giving $A=\\pi d^2/4$."},{"text":"$$A = \\pi d$ or $A = \\frac{\\pi d}{2}$","correct":false,"explanation":"The circumference formula has been used instead of the area formula. Cross-sectional area depends on radius squared, not linearly on the diameter."}]},{"id":"8e208c7b-de72-4409-8028-60b9add3e6ad","cardType":"basic","points":2,"front":"Why does a circuit respond almost instantly when switched on despite a slow drift velocity?","back":"The electric field is established through the circuit at nearly the speed of light.","mcqOptions":[{"text":"The electric field is established through the circuit at nearly the speed of light.","correct":true,"explanation":"When switched on, the electric field is established throughout the circuit at nearly the speed of light. Electrons then begin their slow drift, so the response is almost immediate."},{"text":"The electrons travel through the circuit at nearly the speed of light.","correct":false,"explanation":"Individual electrons do not travel through the circuit at nearly light speed; their drift is very slow. It is the electric field that propagates rapidly and initiates the response."},{"text":"The current travels through the circuit at nearly the speed of light.","correct":false,"explanation":"The current is a consequence of the established electric field, not the entity that propagates nearly at light speed to switch the circuit on."},{"text":"The electric field takes several seconds to be established through the circuit.","correct":false,"explanation":"The field is established rapidly rather than taking a long time. A delay of several seconds would contradict the near-light-speed propagation of the electrical signal."}]},{"id":"bb31665f-ff7d-4e61-9739-8d55cd19dbad","cardType":"basic","points":1,"front":"How must an ammeter be connected to measure the current through a component?","back":"In series.","mcqOptions":[{"text":"In series.","correct":true,"explanation":"An ammeter must carry the same current as the component, so it is placed in series with that component. Its very low resistance minimises its effect on the current."},{"text":"In parallel.","correct":false,"explanation":"Parallel connection is used for a voltmeter, not an ammeter. An ammeter must be connected in series so that the component’s current passes through it."},{"text":"Across the component.","correct":false,"explanation":"Connecting an ammeter across the component creates a very low-resistance parallel path and may cause a dangerously large current. It must instead be connected in series."},{"text":"Between the power supply terminals.","correct":false,"explanation":"Placing an ammeter between the supply terminals effectively short-circuits the supply because the meter has very low resistance. It must be connected in series with the component."}]},{"id":"52411320-3055-4ae4-8d3a-3c06ad386004","cardType":"cloze","points":null,"front":"To convert a current of $35\\,\\text{mA}$ into the SI base unit for calculations, it must be written as $\\text{[...]}$ (or $35 \\times 10^{-3}\\,\\text{A}$).","back":"To convert a current of $35\\,\\text{mA}$ into the SI base unit for calculations, it must be written as $\\boxed{$0.035\\,\\text{A}$}$ (or $35 \\times 10^{-3}\\,\\text{A}$).","mcqOptions":[{"text":"$$0.035\\,\\text{A}$","correct":true,"explanation":"The prefix milli means $10^{-3}$, so $35\\,\\text{mA}=35\\times10^{-3}\\,\\text{A}=0.035\\,\\text{A}$. Amperes are the required SI base unit for current."},{"text":"$$0.35\\,\\text{A}$","correct":false,"explanation":"This divides by only $100$, whereas milli means one-thousandth. Dividing $35$ by $1000$ gives the correct conversion, $0.035\\,\\text{A}$."},{"text":"$$0.0035\\,\\text{A}$","correct":false,"explanation":"This divides by $10{,}000$, introducing one extra factor of ten. Since milli means $10^{-3}$, the correct value is $0.035\\,\\text{A}$."},{"text":"$$35\\,\\text{A}$","correct":false,"explanation":"This leaves the numerical value unchanged when replacing milliamperes with amperes. Because $1000\\,\\text{mA}=1\\,\\text{A}$, the correct current is $0.035\\,\\text{A}$."}]},{"id":"bcc6a667-c9d6-4e9b-8918-cf84a1237428","cardType":"basic","points":2,"front":"Calculate the charge that passes a point if a current of $0.25\\,\\text{A}$ flows for $4.0$ minutes.","back":"$$60\\,\\text{C}$ ($0.25 \\times 240\\,\\text{s}$)","mcqOptions":[{"text":"$$60\\,\\text{C}$ ($0.25 \\times 240\\,\\text{s}$)","correct":true,"explanation":"Using $Q=It$, first convert $4.0$ minutes to $240\\,\\text{s}$. Therefore $Q=0.25\\times240=60\\,\\text{C}$."},{"text":"$$1.0\\,\\text{C}$ ($0.25 \\times 4.0\\,\\text{s}$)","correct":false,"explanation":"This treats the numerical value in minutes as though it were already in seconds. Converting $4.0$ minutes to $240\\,\\text{s}$ gives $Q=0.25\\times240=60\\,\\text{C}$."},{"text":"$$15\\,\\text{C}$ ($0.25 \\times 60\\,\\text{s}$)","correct":false,"explanation":"This uses only $60\\,\\text{s}$, equivalent to one minute rather than four minutes. The full time is $240\\,\\text{s}$, so the charge is $60\\,\\text{C}$."},{"text":"$$240\\,\\text{C}$ ($4.0 \\times 60\\,\\text{s}$)","correct":false,"explanation":"This gives the elapsed time in seconds but fails to multiply it by the current. Using $Q=It$ gives $0.25\\times240=60\\,\\text{C}$."}]},{"id":"6cabef52-b585-46c2-9f8e-0a4858cba7c4","cardType":"cloze","points":null,"front":"The elementary charge $e$ has a magnitude of approximately $\\text{[...]}$.","back":"The elementary charge $e$ has a magnitude of approximately $\\boxed{$1.60 \\times 10^{-19}\\,\\text{C}$}$.","mcqOptions":[{"text":"$$1.60 \\times 10^{-19}\\,\\text{C}$","correct":true,"explanation":"The elementary charge is the magnitude of the charge carried by one proton or electron. Its accepted approximate value is $1.60\\times10^{-19}\\,\\text{C}$."},{"text":"$$1.60 \\times 10^{19}\\,\\text{C}$","correct":false,"explanation":"The positive exponent reverses the scale and produces an impossibly large charge for one particle. The elementary charge has magnitude $1.60\\times10^{-19}\\,\\text{C}$."},{"text":"$$1.60 \\times 10^{-16}\\,\\text{C}$","correct":false,"explanation":"This is one thousand times larger than the elementary charge, typically from misplacing the exponent by three powers. The correct magnitude is $1.60\\times10^{-19}\\,\\text{C}$."},{"text":"$$9.11 \\times 10^{-31}\\,\\text{C}$","correct":false,"explanation":"The value $9.11\\times10^{-31}$ is associated with the electron’s mass in kilograms, not its charge. The elementary charge is $1.60\\times10^{-19}\\,\\text{C}$."}]},{"id":"95575da1-e186-4597-b09e-c897e762fb6a","cardType":"basic","points":2,"front":"What happens to the drift velocity of electrons if the current is doubled in the same wire?","back":"The drift velocity doubles (since $v \\propto I$).","mcqOptions":[{"text":"The drift velocity doubles (since $v \\propto I$).","correct":true,"explanation":"For the same wire, the charge-carrier density and cross-sectional area remain constant. From $I=nAqv$, drift velocity is directly proportional to current, so it doubles."},{"text":"The drift velocity halves (since $v \\propto 1/I$).","correct":false,"explanation":"Halving the drift velocity would halve the current, not double it. For the same wire, $I=nAqv$, so doubling current requires the drift velocity to double."},{"text":"The drift velocity quadruples (since $v \\propto I^2$).","correct":false,"explanation":"This incorrectly assumes a square relationship between current and drift velocity. The relation $I=nAqv$ is linear, so doubling the current only doubles the drift velocity."},{"text":"The drift velocity stays constant (since $v$ is independent of $I$).","correct":false,"explanation":"An unchanged drift velocity would leave the current unchanged when the wire and charge-carrier density are fixed. Since $I=nAqv$, doubling current doubles the drift velocity."}]}]}],"examCountdown":{"examDate":"2027-05-21","paperLabel":"Component 1"},"heading":"Conduction of Electricity","title":"Conduction of Electricity Flashcards","topbar":{"backHref":"/a-level/physics/eduqas/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"3ccf866b-c911-57e6-85c0-b7e2b61c2ece","walled":"$L39"}],"$L3a","$L3b"]