Edexcel A-level Physics (9PH0) · Topic 7: Electric and Magnetic Fields
Mini-Lesson
Electric & Magnetic Fields
Topic 7 packs three big ideas into one: electric fields, capacitors and magnetic fields with electromagnetic induction.
F = kQ₁Q₂/r² · E = V/d · C = Q/V · τ = RC · F = BIL · F = BQv · ε = −N dΦ/dtseven equations that between them run every electric motor, generator and transformer on the planet
The structural parallel worth spotting: Coulomb's law and Newton's law of gravitation are both inverse-square laws with identical mathematical shape. The differences: gravity is only attractive and unimaginably weaker; the electric force can attract or repel.
Work through each screen, answer the questions as you go (several are full A-level calculations) and collect ⭐ stars. Press Start when you're ready.
Topic 7 · electric fields
Coulomb's law and electric field strength
F = (1/4πε₀) · Q₁Q₂/r² · E = F/Qk = 1/4πε₀ = 8.99 × 10⁹ N m² C⁻² · field strength E is force per unit POSITIVE charge, in N C⁻¹ (= V m⁻¹)
Around a point charge the field is radial: E = kQ/r², so it obeys an inverse-square law. Field lines point away from a positive charge and towards a negative one.
Between parallel plates the field is uniform: the lines are straight, parallel and equally spaced.
uniform field: E = V/dV = pd between the plates, d = their separation — hence the unit V m⁻¹
Electric potential V = kQ/r is the work done per unit charge in bringing a charge from infinity. Unlike the gravitational case, it can be positive or negative. A charge released in a field always moves so as to reduce its potential energy.
Calculate
Your turn — Coulomb's law
1Two point charges of 2.0 nC and 3.0 nC are 5.0 cm apart in a vacuum. Calculate the force between them. Give your answer in µN to 3 significant figures. (k = 8.99 × 10⁹ N m² C⁻²)
µN
Hint: F = kQ₁Q₂/r² = (8.99 × 10⁹ × 2.0 × 10⁻⁹ × 3.0 × 10⁻⁹) ÷ (0.050)². The numerator is 5.394 × 10⁻⁸ and r² = 2.5 × 10⁻³. Convert the answer to µN by multiplying by 10⁶.
Calculate
Your turn — uniform field
2Two parallel plates are 2.0 cm apart with a potential difference of 250 V between them. Calculate the electric field strength between them. Give your answer in V m⁻¹.
V m⁻¹
Hint: E = V/d = 250 ÷ 0.020.
Topic 7 · capacitors
Capacitance and energy stored
C = Q / Vcapacitance in farads (F) — one farad is one coulomb per volt, which is enormous, hence µF and nF
A capacitor stores charge — but crucially it stores energy in the electric field between its plates. Because the pd rises as the charge builds, you cannot simply write W = QV:
W = ½QV = ½CV² = ½Q²/Cthe ½ comes from the AREA UNDER the charge–pd graph, which is a triangle
Series: 1/C = 1/C₁ + 1/C₂ — the reciprocals add. (Exactly the opposite of resistors.)
Worked example
C = 220 µF charged to V = 12 V.
Q = CV = 220 × 10⁻⁶ × 12 = 2.64 × 10⁻³ C = 2.64 mC
W = ½CV² = ½ × 220 × 10⁻⁶ × 12² = ½ × 220 × 10⁻⁶ × 144 = 0.0158 J = 15.8 mJ
Calculate
Your turn — energy stored
3A 220 µF capacitor is charged to 12 V. Calculate the energy it stores. Give your answer in mJ to 3 significant figures.
mJ
Hint: W = ½CV² = 0.5 × 220 × 10⁻⁶ × 144 = 0.01584 J. Multiply by 1000 for mJ.
Topic 7 · charge & discharge
Charging, discharging and the time constant
Q = Q₀ e^(−t/RC) · V = V₀ e^(−t/RC) · I = I₀ e^(−t/RC)discharge through a resistor — all three decay exponentially with the same time constant
The time constant is τ = RC, measured in seconds. After one time constant the charge has fallen to 1/e = 37% of its starting value. After 5τ the capacitor is essentially fully discharged (< 1%).
Exponential decay means the fractional drop in each equal time interval is constant — the quantity never quite reaches zero.
Taking natural logs straightens it: ln Q = ln Q₀ − t/RC, so a graph of ln Q against t has gradient −1/RC.
When charging, Q rises towards Q₀ as Q = Q₀(1 − e^(−t/RC)) while the current decays exponentially.
Worked example
R = 47 kΩ, C = 220 µF.
τ = RC = 47 000 × 220 × 10⁻⁶ = 10.3 s
Calculate
Your turn — time constant
4A 220 µF capacitor discharges through a 47 kΩ resistor. Calculate the time constant. Give your answer in s to 3 significant figures.
s
Hint: τ = RC = 47 × 10³ × 220 × 10⁻⁶.
Quick check
One time constant later
?A charged capacitor discharges through a resistor. After a time equal to one time constant, the charge remaining is approximately what fraction of the original?
Topic 7 · magnetic fields
Magnetic flux density: F = BIL and F = BQv
F = BIL sin θ · F = BQv sin θB = magnetic flux density in tesla (T) · θ = angle between the field and the current/velocity
Magnetic flux density B is defined by these equations: one tesla is the flux density that produces a force of 1 N on a wire of length 1 m carrying 1 A perpendicular to the field.
The force is maximum when the current (or velocity) is perpendicular to B (sin 90° = 1), and zero when it is parallel (sin 0° = 0).
Fleming's left-hand rule (motor rule): first finger = Field, second finger = Current, thuMb = Motion (force).
For a negative charge such as an electron, point the second finger along the conventional current — i.e. opposite to the electron's velocity.
Charged particle in a magnetic field: the force is always perpendicular to the velocity, so it does no work. The speed never changes; the particle moves in a circle of radius r = mv/(BQ). That is the basis of the cyclotron and the mass spectrometer.
Calculate
Your turn — force on a wire
5A wire of length 0.12 m carrying a current of 3.0 A lies perpendicular to a magnetic field of flux density 0.25 T. Calculate the force on it. Give your answer in mN.
mN
Hint: F = BIL = 0.25 × 3.0 × 0.12 = 0.090 N. Convert to mN.
Calculate
Your turn — force on a moving charge
6An electron travels at 2.0 × 10⁶ m s⁻¹ perpendicular to a magnetic field of 0.15 T. Calculate the force on it. Give your answer as a multiple of 10⁻¹⁴ N — for example, type 3.5 if the answer is 3.5 × 10⁻¹⁴ N. (e = 1.60 × 10⁻¹⁹ C)
× 10⁻¹⁴ N
Hint: F = BQv = 0.15 × 1.60 × 10⁻¹⁹ × 2.0 × 10⁶ = 4.8 × 10⁻¹⁴ N.
Topic 7 · induction
Flux, Faraday's law and Lenz's law
Φ = BA · flux linkage = NΦ · ε = −N dΦ/dtflux in webers (Wb); flux linkage in Wb turns
Faraday's law: the magnitude of the induced EMF equals the rate of change of flux linkage. Move faster, use more turns, or use a stronger field → bigger EMF.
Lenz's law (the minus sign): the induced current flows in the direction that opposes the change producing it. Push a magnet into a coil and the coil pushes back — you must do work, and that work becomes the electrical energy. Lenz's law is conservation of energy.
A conductor of length L moving at speed v perpendicular to B generates ε = BLv.
Transformers exploit induction directly. An alternating current in the primary produces a changing flux in the iron core, which links the secondary and induces an alternating EMF:
Vs/Vp = Ns/Npand for an ideal (100% efficient) transformer, VpIp = VsIs
Why transformers need a.c.: d.c. gives a constant flux, so dΦ/dt = 0 and no EMF is induced at all. Real losses are cut by using a laminated core (reduces eddy currents), a soft-iron core (easily magnetised and demagnetised) and thick low-resistance windings.
Calculate
Your turn — transformer turns
7A transformer steps 230 V down to 12 V. The primary coil has 1150 turns. Calculate the number of turns on the secondary coil.
turns
Hint: Ns/Np = Vs/Vp, so Ns = 1150 × (12 ÷ 230).
Quick check
Lenz's law
?The north pole of a magnet is pushed into a coil connected to a closed circuit. What happens?
Sort it
Electric, magnetic, or induction?
Tap a statement, then tap the part of Topic 7 it belongs to.
⚡ Electric fields
🧲 Magnetic fields
🔁 Electromagnetic induction
Quick check
Angle matters
?A current-carrying wire is placed parallel to a magnetic field. What force acts on it?
Quick check
Capacitors in parallel
?Two capacitors of 4.0 µF and 6.0 µF are connected in parallel. What is the total capacitance?
Match it
Match the quantity to its equation
Tap an item on the left, then its partner on the right.
Quantity
Equation
Recap
The big ideas to know
Coulomb: F = kQ₁Q₂/r², inverse square, attractive OR repulsive · E = F/Q · radial field E = kQ/r²
Uniform field: E = V/d between parallel plates (V m⁻¹ = N C⁻¹)
Capacitors: C = Q/V · W = ½QV = ½CV² · parallel adds, series uses reciprocals
Discharge: Q = Q₀e^(−t/RC); τ = RC; 37% left after one time constant; ln Q against t gives gradient −1/RC
Magnetic force: F = BIL sin θ on a wire · F = BQv sin θ on a charge · left-hand rule · circular path r = mv/BQ
Induction: Φ = BA · ε = −N dΦ/dt (Faraday) · minus sign = Lenz = conservation of energy