AQA A-level Physics (7408) · 3.7 Fields and their consequences
Mini-Lesson
Fields and their consequences
This mini-lesson covers AQA 3.7 — Fields and their consequences: gravitational and electric fields (and the deep analogy between them), orbits, capacitance and RC charge/discharge, magnetic fields (F = BIL, F = BQv), electromagnetic induction (Faraday and Lenz), transformers and alternating current.
Section 3.7 is compulsory and appears on Paper 2.
Where this sits: AQA sections 3.1–3.8 are compulsory for every A-level Physics student. (Sections 3.9–3.13 are the five options — you study just one of those.)
Work through each screen, answer the questions as you go (many are full A-level calculations) and collect ⭐ stars. Press Start when you're ready.
3.7.2 · gravitational fields
Gravitational fields, potential and orbits
A field is a region where a mass (or charge) feels a force. Gravity is always attractive.
F = Gm1m2 ÷ r² · g = GM ÷ r² · V = −GM ÷ rg = force per unit mass (N kg−1) · V = potential energy per unit mass (J kg−1), always negative
Inverse square for force and field strength; inverse (1/r) for potential.
Gravitational potential is zero at infinity and negative everywhere else, because you must do work to escape a mass.
Escape velocity: ½mv² = GMm/r → v = √(2GM/r).
Orbits: gravity provides the centripetal force. GMm/r² = mv²/r gives v = √(GM/r), and combining with v = 2πr/T gives T² ∝ r³ (Kepler’s third law).
A geostationary satellite has a period of exactly 24 hours, orbits above the equator, and travels west→east (with the Earth’s rotation) — ideal for TV broadcasting.
Worked example
Earth: M = 6.0×1024 kg, r = 6.4×106 m, G = 6.67×10−11.
g = GM/r² = (6.67×10−11 × 6.0×1024) ÷ (6.4×106)² = 4.00×1014 ÷ 4.10×1013 = 9.8 N kg−1. ✓
Calculate
Your turn — calculation 1
1Calculate the gravitational field strength at the Earth’s surface. (M = 6.0×1024 kg, r = 6.4×106 m, G = 6.67×10−11 N m² kg−2)
?Which statement about a geostationary satellite is correct?
3.7.3 · electric fields
Electric fields, Coulomb’s law and potential
F = (1 ÷ 4πε0) × Q1Q2 ÷ r² · E = F ÷ Q · V = (1 ÷ 4πε0) × Q ÷ r1 ÷ 4πε0 = 8.99×109 N m² C−2 · E in N C−1 (= V m−1)
Uniform field between parallel plates: E = V ÷ d, directed from + to −. A charged particle in it accelerates like a projectile in gravity — a parabolic path.
The analogy with gravity: both obey an inverse-square law and both have potential ∝ 1/r. The key difference: charge can be positive or negative, so electric forces can repel; gravity only ever attracts.
Work done moving a charge between two points: W = QΔV. Along an equipotential, no work is done.
2Two point charges of +2.0 nC and +3.0 nC are 5.0 cm apart in a vacuum. Calculate the magnitude of the force between them, in µN. (1/4πε0 = 8.99×109 N m² C−2)
µN
Hint: F = 8.99×109 × (2.0×10−9 × 3.0×10−9) ÷ (0.050)² = 2.16×10−5 N. Now convert to µN.
Sort it
Gravitational, electric, or both?
Tap a statement, then tap the box it belongs in.
🌍 Gravitational only
⚡ Electric only
🔁 True of both
3.7.4 · capacitance
Capacitance, energy stored and RC circuits
C = Q ÷ V · E = ½QV = ½CV² = ½Q²/CC in farads (F) · energy stored = area under a Q–V graph (hence the factor of ½)
discharge: Q = Q0e−t/RCV and I decay in exactly the same way · time constant τ = RC (in seconds)
After one time constant (t = RC), the charge has fallen to e−1 = 37% of its initial value.
The half-life of the discharge is T½ = 0.69RC.
Charging: Q = Q0(1 − e−t/RC) — the charge rises towards its final value.
A dielectric between the plates increases the capacitance (its molecules become polarised, which reduces the field for the same charge).
Worked example
C = 100 µF charged to 12 V: E = ½CV² = 0.5 × 100×10−6 × 144 = 7.2×10−3 J = 7.2 mJ.
Discharged through 100 kΩ: τ = RC = 105 × 100×10−6 = 10 s.
Calculate
Your turn — calculation 3
3A 100 µF capacitor is charged to 12 V. Calculate the energy stored, in mJ.
mJ
Hint: E = ½CV² = 0.5 × 100×10−6 × 12² = 7.2×10−3 J. Convert J → mJ.
Calculate
Your turn — calculation 4
4The same 100 µF capacitor, charged to 12 V, is discharged through a 100 kΩ resistor. Calculate the pd across it after 20 s.
V
Hint: τ = RC = 10 s, so t/τ = 2. V = 12 × e−2 = 12 × 0.135.
Quick check
Think it through
?A capacitor discharges through a resistor. After a time equal to one time constant, the charge remaining is closest to
3.7.5 · magnetic fields
Magnetic fields: F = BIL and F = BQv
F = BIL sinθ · F = BQv sinθB = magnetic flux density, in tesla (T) · use Fleming’s left-hand rule for the direction (thuMb = Motion/force, First = Field, seCond = Current)
The force is a maximum when the current (or velocity) is perpendicular to the field, and zero when it is parallel.
A charged particle moving perpendicular to a uniform field travels in a circle: BQv = mv²/r → r = mv ÷ BQ. The magnetic force does no work (it is always perpendicular to v), so the speed never changes.
This is the basis of the cyclotron and of mass spectrometry.
Worked example
A wire of length 0.40 m carrying 3.0 A sits perpendicular to a 0.25 T field.
F = BIL = 0.25 × 3.0 × 0.40 = 0.30 N.
Calculate
Your turn — calculation 5
5A straight wire of length 0.40 m carries a current of 3.0 A at right angles to a uniform magnetic field of flux density 0.25 T. Calculate the force on the wire.
N
Hint: F = BIL = 0.25 × 3.0 × 0.40 (sin90° = 1).
3.7.5 · induction
Flux, Faraday, Lenz and transformers
Φ = BA · flux linkage = NΦΦ in webers (Wb) · flux linkage in Wb turns
ε = −N × ΔΦ ÷ ΔtFaraday: induced emf ∝ rate of change of flux linkage · Lenz (the minus sign): the induced current opposes the change producing it
Lenz’s law is conservation of energy: if the induced current helped the change, you would get energy for free.
An a.c. generator: a coil rotating in a field gives a sinusoidal emf; peak emf occurs when the coil is in the plane of the field (flux linkage zero, but changing fastest).
Transformer: Vs/Vp = Ns/Np. For an ideal (100% efficient) transformer, VpIp = VsIs.
Transformer losses: eddy currents (reduced by a laminated core), heating in the coils (thick, low-resistance wire), and hysteresis (a soft magnetic core).
National Grid: step up the voltage to cut the current, because the power wasted in the cables is I²R.
Irms = I0 ÷ √2 · Vrms = V0 ÷ √2the rms value is the d.c. equivalent that would deliver the same mean power
Calculate
Your turn — calculation 6
6A transformer has 1000 turns on the primary and 250 turns on the secondary. The primary is connected to a 230 V a.c. supply. Calculate the secondary voltage.
V
Hint: Vs = Vp × Ns ÷ Np = 230 × 250 ÷ 1000.
Match it
Match each statement to the law or quantity
Tap a card on the left, then its partner on the right.
Statement
Answer
Quick check
Think it through
?The peak voltage of a UK mains supply is about 325 V. What is the rms voltage?
Quick check
Think it through
?A bar magnet is pushed north-pole-first into a coil connected to a sensitive ammeter. Which statement is correct?
Recap
The big ideas to know
Gravitational: F = Gm1m2/r² · g = GM/r² · V = −GM/r · T² ∝ r³
Electric: F = Q1Q2/4πε0r² · E = V/d in a uniform field · W = QΔV
The analogy: both inverse-square; only electric fields can repel
Capacitance: C = Q/V · E = ½CV² · Q = Q0e−t/RC · τ = RC (37% left)
Magnetic: F = BIL sinθ · F = BQv sinθ · r = mv/BQ · Fleming’s left hand