HookWeighing the Earth in a garden shed
In 1798 Henry Cavendish hung a six-foot wooden rod from a thin wire in an outbuilding at his house on Clapham Common, fixed a small lead ball to each end, and brought two 158 kg lead spheres close to them. The gravitational pull between the balls twisted the rod through about four millimetres — a deflection he read with a telescope from outside the room, because the warmth of his own body would have stirred the air enough to swamp the signal. From that twist he calculated the density of the Earth to within roughly one per cent of today's value, which is equivalent to measuring \(G\), the constant in Newton's law of gravitation.
Notice what did all the work: nothing touched anything. A force crossed empty space, and Cavendish measured it. That is the entire subject of 3.7 — fields, the machinery of action at a distance — and the section is one long exploitation of a single template: an inverse-square force law, a field strength defined as force per unit something, a potential, and an energy bookkeeping rule. Gravity runs the template on mass; electrostatics runs it on charge; capacitors store energy in an electric field; magnetic fields push on moving charge; and changing fields induce EMFs, which is why the National Grid exists. Learn the gravitational version properly and the electric version comes almost free — AQA asks you to make exactly that comparison.
ModelOne template, two forces — the field idea
A force field is a region in which an object experiences a non-contact force: gravitational fields act on mass, electric fields on charge, magnetic fields on moving charge and current. We draw them with field lines — the arrow gives the direction of force on a test mass (or a positive test charge), and the line spacing encodes strength: converging lines mean an intensifying field.
The gravitational and electric cases are structurally parallel, and the examiners' favourite synoptic question is to make you say precisely how. Similarities: both obey inverse-square force laws between point objects; both define field strength as force per unit property (\(g = F/m\), \(E = F/Q\)); both have a potential defined via work done from infinity, with equipotential surfaces at right angles to field lines. Differences: gravity is always attractive, while electric forces attract and repel; you can shield a region from electric fields but not from gravity; and the strengths are absurdly mismatched — between two protons, electric repulsion beats gravitational attraction by a factor of about \(1.2 \times 10^{36}\). Gravity only wins at planetary scales because matter in bulk is electrically neutral.
ModelGravitational fields — Newton's inverse square law
Newton's law of gravitation: two point masses attract with \(F = \frac{Gm_1 m_2}{r^2}\), where \(G = 6.67 \times 10^{-11}\ \text{N m}^2\text{kg}^{-2}\) — the number Cavendish's experiment pins down. A uniform sphere behaves as if all its mass sat at its centre, which is why the law runs the Solar System, not just laboratory lead balls.
Gravitational field strength is force per unit mass, \(g = F/m\), in \(\text{N kg}^{-1}\). Around a point or spherical mass the field is radial and \(g = \frac{GM}{r^2}\); close to a planet's surface, over distances small compared with its radius, the field lines are effectively parallel and the field is uniform — the flat-Earth approximation your mechanics module quietly used all year. The habitual error is the meaning of \(r\): it is measured from the centre of the mass, so a satellite's altitude must always have the planet's radius added before anything is squared.
Astronauts on the ISS float, so a depressing number of students write that \(g \approx 0\) there. Check it. The ISS orbits about 410 km up, so \(r = 6.37 \times 10^6 + 4.1 \times 10^5 = 6.78 \times 10^6\ \text{m}\).
\[g = \frac{GM}{r^2} = \frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{(6.78 \times 10^6)^2} = \frac{3.98 \times 10^{14}}{4.60 \times 10^{13}} = 8.7\ \text{N kg}^{-1}\]
That is 88% of the surface value. Astronauts are not beyond gravity; they are in free fall, accelerating at 8.7 m s⁻² towards the Earth's centre at all times, along with their spacecraft — so nothing presses on anything. The two-mark explanation is the phrase 'both the astronaut and the station are in free fall with the same acceleration', not any claim about g being zero.
ModelGravitational potential — the well you have to climb out of
Gravitational potential \(V\) at a point is the work done per unit mass to bring a small mass from infinity to that point. Since infinity is defined as the zero and gravity attracts, every real potential is negative: \(V = -\frac{GM}{r}\), in \(\text{J kg}^{-1}\). At the Earth's surface \(V = -62.5\ \text{MJ kg}^{-1}\) — meaning every kilogram you want to remove from Earth's influence entirely costs 62.5 MJ, drag not included. The negative sign is not decoration; it is the physics of being in a well.
Moving a mass between two points costs \(\Delta W = m\Delta V\), and only the difference matters. Field strength and potential are locked together: \(g\) equals the negative of the potential gradient, \(g = -\frac{\Delta V}{\Delta r}\), so where potential changes steeply the field is strong — on a graph of \(V\) against \(r\), the gradient hands you \(g\). Equipotential surfaces connect points of equal potential and always cut field lines at right angles; moving along one costs zero work. That is why a satellite in a circular orbit, gliding around an equipotential, needs no engine to maintain altitude.
MechanismOrbits — gravity exactly paying the centripetal bill
A circular orbit is what happens when gravity is the only force and it exactly supplies the required centripetal force: \(\frac{GMm}{r^2} = \frac{mv^2}{r}\), giving \(v^2 = \frac{GM}{r}\). Substitute \(v = 2\pi r/T\) and you get Kepler's third law, \(T^2 = \frac{4\pi^2 r^3}{GM}\): period squared scales with radius cubed, and the satellite's own mass cancels out entirely.
Two orbits carry the application marks. A geostationary satellite has \(T = 24\) hours, orbits above the equator moving west-to-east, and therefore hangs over one longitude — which is why a satellite dish can be bolted rigidly to a wall. Low polar orbits (\(T \approx 90\) min, a few hundred kilometres up) pass over the whole rotating Earth beneath them, ideal for imaging and weather. Energy bookkeeping: kinetic energy is positive, potential negative and larger in magnitude, so a bound satellite's total energy is negative. Air drag on a low satellite reduces the total energy, dropping it to a smaller \(r\) — where, counter-intuitively, it moves faster, because \(v^2 = GM/r\) grows as \(r\) shrinks.
Find the radius of a geostationary orbit. Rearranging Kepler's third law: \(r^3 = \frac{GMT^2}{4\pi^2}\), with \(T = 24\ \text{h} = 86\,400\ \text{s}\) and \(GM = 3.98 \times 10^{14}\ \text{N m}^2\text{kg}^{-1}\).
\[r^3 = \frac{3.98 \times 10^{14} \times (86\,400)^2}{4\pi^2} = \frac{3.98 \times 10^{14} \times 7.46 \times 10^{9}}{39.5} = 7.53 \times 10^{22}\ \text{m}^3\]
\[r = (7.53 \times 10^{22})^{1/3} = 4.22 \times 10^{7}\ \text{m}\]
That is 42,200 km from the Earth's centre — subtract the radius and every geostationary satellite sits about 35,900 km above the equator, in one crowded, precious ring. The exam habits that pay: cube-root at the end (not square root — the single most common slip), and state that the satellite mass cancelled, which is why the answer needed no information about the satellite.
ModelElectric fields — the same machinery running on charge
Swap mass for charge and the template repeats. Coulomb's law: \(F = \frac{1}{4\pi\varepsilon_0}\frac{Q_1 Q_2}{r^2}\), where \(\frac{1}{4\pi\varepsilon_0} = 8.99 \times 10^{9}\ \text{N m}^2\text{C}^{-2}\) and air counts as a vacuum to a very good approximation. Electric field strength is force per unit positive charge, \(E = F/Q\), in \(\text{N C}^{-1}\) — identically \(\text{V m}^{-1}\). Around a point charge the field is radial with \(E = \frac{Q}{4\pi\varepsilon_0 r^2}\); between parallel plates it is uniform with \(E = \frac{V}{d}\), the geometry used to deflect charged particles and, next block, to store energy.
Electric potential completes the parallel: \(V = \frac{Q}{4\pi\varepsilon_0 r}\), zero at infinity, taking the sign of \(Q\) — positive around a positive charge, because the field would do work pushing your positive test charge away, so you must do work to bring it in. Work done moving charge through a potential difference is \(\Delta W = Q\Delta V\), and field strength is again the negative potential gradient. Where the analogy snaps is sign and shielding: gravitational potential is always negative because gravity only attracts, while electric potential comes in both signs — and a conducting box screens its interior from external electric fields, a trick gravity permits nowhere in the universe.
MechanismCapacitors — parking energy in an electric field
A capacitor is two conductors separated by an insulator, and capacitance is charge stored per volt: \(C = Q/V\), in farads. One farad is enormous; real components live in microfarads down to picofarads. For the parallel-plate geometry, \(C = \frac{A\varepsilon_0\varepsilon_r}{d}\): big plates, small gap, and a dielectric — an insulator whose polar molecules rotate to align against the applied field. Their alignment partially cancels the field between the plates, lowering the pd for the same stored charge, and since \(C = Q/V\) that raises the capacitance by the factor \(\varepsilon_r\), the relative permittivity.
Charging a capacitor stores energy in its field, and the amount is the area under the charge–voltage graph. The graph is a straight line through the origin, so the area is a triangle: \(E = \tfrac{1}{2}QV = \tfrac{1}{2}CV^2 = \tfrac{1}{2}\frac{Q^2}{C}\). The half is physical, not cosmetic — the first coulomb arrives when the pd is nearly zero and costs almost nothing; only the final coulomb pays the full price. Writing \(E = QV\) is the standard one-mark donation to the examiner.
A defibrillator capacitor of \(C = 100\ \mu\text{F}\) is charged to 2,000 V, then dumped through the chest in about 5 ms.
Charge: \(Q = CV = 1.0 \times 10^{-4} \times 2000 = 0.20\ \text{C}\). Energy: \(E = \tfrac{1}{2}CV^2 = \tfrac{1}{2} \times 1.0 \times 10^{-4} \times (2000)^2 = 200\ \text{J}\). Mean power during the pulse: \(P = \frac{200}{5 \times 10^{-3}} = 40\ \text{kW}\).
Forty kilowatts — from a component you can hold in one hand. This is why the machine whines for a few seconds before the shock (charging at modest power) and why capacitors, not batteries, deliver it: a battery stores far more energy but cannot release it anywhere near that fast. Capacitors trade capacity for delivery rate — the same reason they fire camera flashes.
DataDischarge and RP9 — the exponential you can measure on a bench
Discharge a capacitor through a resistor and the current is driven by a pd that the current itself is destroying — the recipe for exponential decay. Charge, pd and current all obey the same law: \(Q = Q_0 e^{-t/RC}\). The product \(RC\) is the time constant \(\tau\), in seconds: the time to fall to \(1/e \approx 37\%\) of the starting value. The half-time is \(T_{1/2} = \ln 2 \times RC \approx 0.69RC\), and charging follows the mirror image, \(V = V_0(1 - e^{-t/RC})\). To make the law testable, take logs: \(\ln V = \ln V_0 - \frac{t}{RC}\), so a graph of \(\ln V\) against \(t\) is a straight line of gradient \(-1/RC\) — straightness is the evidence of exponential behaviour.
Required practical 9 does exactly this. Method: charge a capacitor to a known pd, discharge it through a known resistance, and log pd against time with a data logger (or stopwatch and high-resistance voltmeter for slow decays). Independent variable: time; dependent: pd; control: \(R\), the capacitor, and the starting pd. The error budget is where marks live: manual timing adds reaction-time lag, so a data logger beats a stopwatch; a low-resistance voltmeter steals current and speeds up the discharge, so use a high-impedance digital meter; and electrolytic capacitors carry tolerances of ±20%, which is precisely why you extract \(C\) from the measured gradient instead of trusting the printed value.
A student discharges a capacitor labelled \(470\ \mu\text{F}\) through \(R = 100\ \text{k}\Omega\) and plots \(\ln V\) against \(t\), getting a straight line of gradient \(-0.0223\ \text{s}^{-1}\).
Gradient \(= -\frac{1}{RC}\), so \(RC = \frac{1}{0.0223} = 44.8\ \text{s}\), and \[C = \frac{44.8}{1.00 \times 10^{5}} = 4.48 \times 10^{-4}\ \text{F} = 448\ \mu\text{F}\]
That is 4.7% below the printed 470 µF — comfortably inside the ±20% manufacturing tolerance, so the component is fine and the measurement is better than the label. Predicted time constant from the label would have been \(\tau = 47\ \text{s}\); the straight-line method both confirmed the exponential model and measured the true value. Quote gradients with units (\(\text{s}^{-1}\)) — omitting them is a routinely dropped mark.
MechanismMagnetic force — F = BIl, F = BQv and RP10 on a top-pan balance
A current-carrying conductor lying across a magnetic field feels a force \(F = BIl\) (current perpendicular to field), which defines magnetic flux density \(B\): one tesla is one newton of force per amp per metre. Directions come from Fleming's left-hand rule — thumb for force, first finger for field, second finger for conventional current — and 'state the rule you used' is frequently a creditable line in itself.
Required practical 10 turns \(F = BIl\) into a measurement. A stiff wire is clamped horizontally through the gap of a magnadur-magnet yoke that sits on a top-pan balance. Pass a current: the field pushes the wire one way, and by Newton's third law the wire pushes the magnets the other, changing the balance reading. Convert the change from grams to kilograms (the classic lost mark) and \(F = \Delta m \times g\). Vary \(I\) as the independent variable, plot \(F\) against \(I\): a straight line through the origin with gradient \(Bl\), so dividing by the measured field length gives \(B\). Control the length of wire in the field and the magnet arrangement; keep the wire perpendicular to the field and centred, because the field fringes weakly at the ends of the yoke; zero the balance before each run and switch off between readings so resistive heating cannot drift the current.
A free charge moving through the field feels \(F = BQv\) (velocity perpendicular to field). Because the force stays perpendicular to the velocity it does no work: speed is constant and the path is a circle, with \(\frac{mv^2}{r} = BQv\) giving \(r = \frac{mv}{BQ}\). That geometry is the cyclotron: two D-shaped electrodes, an alternating pd that kicks the particle at each gap crossing, and the happy accident that the orbital frequency \(f = \frac{BQ}{2\pi m}\) is independent of speed, so a fixed-frequency supply stays in step as the spiral grows — the machine behind medical isotope production and proton-beam therapy.
MechanismInduction — flux, Faraday, Lenz and RP11
Magnetic flux is field through area: \(\Phi = BA\), in webers, and a coil of \(N\) turns at angle \(\theta\) between the field and the normal to its plane links flux \(N\Phi = BAN\cos\theta\). The cos-of-the-wrong-angle error is endemic: \(\theta\) is measured from the normal, so flux linkage is a maximum when the coil's plane is perpendicular to the field.
Faraday's law: the induced EMF equals the rate of change of flux linkage, \(\varepsilon = -N\frac{\Delta\Phi}{\Delta t}\). The minus sign is Lenz's law — induced currents flow to oppose the change creating them — and it is conservation of energy in disguise: drop a magnet down a copper pipe and it crawls, because the eddy currents it induces manufacture fields that resist its fall; the lost gravitational energy reappears as heat in the copper. A straight conductor of length \(l\) moving at speed \(v\) across a field sweeps flux at rate \(Blv\), so \(\varepsilon = Blv\). A coil rotating at angular speed \(\omega\) in a uniform field has \(N\Phi = BAN\cos\omega t\), giving \(\varepsilon = BAN\omega\sin\omega t\) — the generator equation, with peak EMF when the coil plane lies parallel to the field and its sides are slicing flux fastest.
Required practical 11 checks the \(\cos\theta\) dependence directly. A large slotted coil driven by an AC signal generator produces an alternating field; a small search coil connected to an oscilloscope sits at its centre, where the field is most uniform. Rotate the search coil in measured steps and read the peak induced EMF from the trace at each angle. Since the induced EMF is proportional to the flux linkage being alternated, peak EMF plotted against \(\cos\theta\) should give a straight line through the origin. Precision habits: measure peak-to-peak and halve it (doubling the measured deflection halves the fractional reading error), keep the search coil in the same central position throughout, and mount the protractor scale to read the angle without parallax.
DataAlternating current and the transformer — the case for 400,000 volts
Mains electricity is sinusoidal, so 'the' voltage needs defining. The rms value is the DC-equivalent for heating: a sinusoidal supply of rms voltage 230 V delivers the same average power to a resistor as a 230 V battery would. For sine waves \(I_{rms} = \frac{I_0}{\sqrt{2}}\) and \(V_{rms} = \frac{V_0}{\sqrt{2}}\), so UK mains at 230 V rms actually peaks at \(230\sqrt{2} = 325\ \text{V}\), swinging 650 V peak-to-peak. On an oscilloscope, the y-gain converts trace height to volts and the time-base converts trace width to time, so one period read off the screen gives frequency via \(f = 1/T\).
The transformer is induction industrialised: alternating current in a primary coil drives an alternating flux around a laminated iron core, and that changing flux linkage induces an EMF in the secondary, with \(\frac{N_s}{N_p} = \frac{V_s}{V_p}\). An ideal transformer passes power unchanged, \(I_p V_p = I_s V_s\) — step the voltage up and the current steps down in exact proportion. Real ones leak a little: resistance of the windings heats the copper, eddy currents induced in the core are choked by laminating it into insulated sheets, the core wastes energy each magnetisation cycle, and some flux misses the secondary entirely. Grid-scale transformers still exceed 99% efficiency, which is why the National Grid is a chain of them: generation at about 25 kV, stepped up to 400 kV for the supergrid, then down through 132 kV and 33 kV and 11 kV to the 230 V at your socket.
Why bother with 400 kV? Send 100 MW down a transmission line of total resistance \(5.0\ \Omega\) and compare.
At 25 kV: \(I = \frac{P}{V} = \frac{1.0 \times 10^{8}}{2.5 \times 10^{4}} = 4000\ \text{A}\), so cable loss \(P = I^2R = (4000)^2 \times 5.0 = 8.0 \times 10^{7}\ \text{W}\) — 80 MW, four-fifths of everything generated, gone as heat.
At 400 kV: \(I = \frac{1.0 \times 10^{8}}{4.0 \times 10^{5}} = 250\ \text{A}\), so loss \(= (250)^2 \times 5.0 = 3.1 \times 10^{5}\ \text{W}\) — 0.31 MW, or 0.3%.
Sixteen times the voltage means one-sixteenth the current and \(16^2 = 256\) times less \(I^2R\) loss. That one calculation is the entire justification for high-voltage transmission, and it is worth writing exactly this way in a 6-marker: same power, lower current, losses scale with current squared.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
This is Paper 2 territory: 85 marks, of which 60 are short and long answer and 25 are multiple choice, with roughly 40% of the whole qualification testing AO2 application — so almost every field question hands you a context (a satellite, a defibrillator, a rail gun) and expects the template applied, not recited. First habit: classify the field before touching the calculator. Radial or uniform? If radial, r runs from the centre — add the planet's radius to any altitude. If the question says 'show that', quote one more significant figure than the target so the examiner can see the value wasn't reverse-engineered.
Second: respect signs and definitions. Gravitational potential is negative and its definition ('work done per unit mass from infinity') is a routinely available mark; ΔW = mΔV wants a potential difference, not a potential. Inverse-square ratio questions are quickest without the calculator at all: halve the distance, quadruple the field. For capacitor decay, work in ln-space — the gradient of ln V against t is −1/RC with units s⁻¹, and stating that the straight line itself is the evidence for exponential decay is an explicit marking point.
Third: the required practicals earn at least 15% of the marks across the papers, and RP9–RP11 are the most heavily mined in this section. For each, be able to name the independent, dependent and two control variables, plus one genuine accuracy improvement with its reason: data logger over stopwatch because it removes reaction time (RP9); converting the balance reading from grams to kilograms before F = Δmg, and keeping the wire perpendicular and centred in the field (RP10); reading peak-to-peak and halving it to reduce fractional reading error (RP11). Direction questions want the rule named and applied — 'by Fleming's left-hand rule the force is vertically upwards', 'by Lenz's law the induced current opposes the flux increase, so…' — and AC questions want rms and peak values converted before any power calculation, never after.