Electricity and Magnetism - fields, circuits and induction
Electricity and Magnetism
This section tests electrostatic force and field, capacitance, direct-current circuits (Ohm's law, resistivity, Kirchhoff's rules, Wheatstone bridge and potentiometer), the magnetic field of currents, electromagnetic induction and alternating current including the transformer.
Core rules
- Coulomb's law F = Q₁Q₂/(4πε₀r²); field of a point charge E = Q/(4πε₀r²); potential V = Q/(4πε₀r); uniform field E = V/d.
- Capacitance C = Q/V; parallel plates C = ε₀A/d; series 1/C = 1/C₁ + 1/C₂, parallel C = C₁ + C₂; energy = ½CV² = ½QV.
- V = IR; R = ρl/A; series and parallel resistor rules; Kirchhoff: ΣI = 0 at a junction, Σe.m.f. = ΣIR round a loop; Wheatstone balance P/Q = R/S.
- Magnetism: F = BIl sin θ; B = μ₀I/(2πr) near a long wire; B = μ₀nI inside a solenoid; parallel currents attract when in the same direction.
- Induction: Φ = BA cos θ, ε = −N dΦ/dt; Lenz's law says the induced current opposes the change producing it.
- AC: I_rms = I_peak/√2, V_rms = V_peak/√2; ideal transformer V_s/V_p = N_s/N_p = I_p/I_s.
Worked example
A 4 µF and a 2 µF capacitor are in series across a 12 V supply. Find the charge and the p.d. across each.
1/C = 1/4 + 1/2 = 3/4, so C = 1.33 µF. Charge Q = CV = 1.33 × 10⁻⁶ × 12 = 1.6 × 10⁻⁵ C, the same on both. V across 4 µF = Q/C = 16 µC/4 µF = 4 V; across 2 µF = 16 µC/2 µF = 8 V. The two p.d.s add to 12 V, as they must.
Traps
Treating series capacitors like series resistors; slipping between µF, mA and V; getting Kirchhoff loop signs wrong; predicting the induced current direction by guesswork instead of Lenz's law; using peak values when an rms value is asked for; forgetting the terminal p.d. drop caused by internal resistance; assuming a transformer can raise power.