Eduqas A-level Physics (A420QS) · Component 2: Electricity and the Universe
Mini-Lesson
Component 2: Electricity and the Universe
This mini-lesson covers all eight areas of Eduqas Component 2: conduction, resistance, d.c. circuits, capacitance, solids under stress, electrostatic and gravitational fields, using radiation to investigate stars, and orbits and the wider universe.
Where it sits: Component 2 is a 2 h written paper worth 31.25% of the A level, and contains synoptic questions. The unifying thread is the inverse square law — it governs Coulomb's law, Newton's law of gravitation and the intensity of starlight.
Press Start when you are ready.
1. Conduction of electricity
Current, charge carriers and drift velocity
Current is the rate of flow of charge:
I = ΔQ / Δt1 A = 1 C s⁻¹. The electron's charge e = 1.60 × 10⁻¹⁹ C — a tiny fraction of a coulomb.
In a metal, conduction is the drift of free electrons. They already move at enormous random speeds; the field superimposes a slow drift along the wire:
I = nAven = number of free electrons per m³ (the number density), A = cross-sectional area, v = drift velocity, e = electronic charge.
The counter-intuitive bit: the drift velocity in a typical wire is under a millimetre per second — yet the lamp comes on instantly. The electric field is established through the circuit at close to the speed of light, so all the electrons everywhere start drifting at once.
Calculate
Your turn — drift velocity
1A copper wire of cross-sectional area 1.0 × 10⁻⁶ m² carries a current of 2.0 A. The free electron density is 8.5 × 10²⁸ m⁻³ and e = 1.60 × 10⁻¹⁹ C. Calculate the drift velocity in units of 10⁻⁴ m s⁻¹.
× 10⁻⁴ m s⁻¹
Hint: v = I / (nAe) = 2.0 ÷ (8.5 × 10²⁸ × 1.0 × 10⁻⁶ × 1.60 × 10⁻¹⁹) = 2.0 ÷ 1.36 × 10⁴ = 1.47 × 10⁻⁴ m s⁻¹. Enter just the number in front.
2. Resistance · 3. D.C. circuits
Resistance, resistivity and circuits
V = IR · P = IV = I²R = V²/R · R = ρL / APotential difference is energy transferred per coulomb: 1 V = 1 J C⁻¹. Resistance is in ohms: 1 Ω = 1 V A⁻¹.
Why resistance rises with temperature (in a metal): free electrons collide with the lattice ions; hotter ions vibrate with greater amplitude, so collisions are more frequent and the electrons are impeded more. Those collisions also transfer energy to the ions — which is exactly why a current heats a wire. Metals vary almost linearly with temperature over a wide range.
Superconductivity: below its transition temperature a superconductor has zero resistance. Most metals only do this a few degrees above absolute zero; high-temperature superconductors have transition temperatures above the boiling point of nitrogen (−196 °C), which makes them practical. Uses: MRI scanner magnets and particle accelerators.
Circuit rules (both are conservation laws):
Current into a junction = current out (conservation of charge). Series: R = R₁ + R₂. Parallel: 1/R = 1/R₁ + 1/R₂.
The p.d.s around a loop add to the supply p.d. (conservation of energy). Components in parallel have the same p.d.
Vout = Vin × R₂ / (R₁ + R₂) · V = E − IrPotential divider · e.m.f. and internal resistance. Specified practicals: I–V characteristics; determination of resistivity; variation of resistance with temperature; determination of the internal resistance of a cell.
Quick check
Cold enough to be perfect
?A sample is cooled below its superconducting transition temperature. What happens to its resistance?
Quick check
Internal resistance in action
?A cell of e.m.f. 1.50 V and internal resistance 0.30 Ω drives a current of 0.50 A. What is the terminal p.d.?
4. Capacitance
Capacitors, energy and RC circuits
A parallel-plate capacitor is two plates separated by a vacuum, air or a dielectric. Charging transfers charge from one plate to the other, so the plates carry equal and opposite charge and the net charge is zero.
C = Q / V · C = ε₀A / d · E = V / dε₀ = 8.85 × 10⁻¹² F m⁻¹. Adding a dielectric increases the capacitance. The field between the plates is uniform.
U = ½QV = ½CV²= the area under the Q–V graph. The ½ appears because the p.d. rises from zero as the charge accumulates.
parallel: C = C₁ + C₂ · series: 1/C = 1/C₁ + 1/C₂Note this is the opposite way round to resistors.
discharging: Q = Q₀e−t/RC · charging: Q = Q₀(1 − e−t/RC)RC is the time constant (in seconds) — the time to fall to 1/e ≈ 37% of the initial charge (or to rise to 63% of the final charge).
Specified practicals: charging and discharging a capacitor to find the time constant (plot ln V against t — gradient = −1/RC), and investigating the energy stored in a capacitor.
Calculate
Your turn — parallel-plate capacitor
2A parallel-plate capacitor has plates of area 0.020 m² separated by 0.50 mm of air. Calculate its capacitance in pF. (ε₀ = 8.85 × 10⁻¹² F m⁻¹)
Stress, strain, the Young modulus and material behaviour
F = kx · σ = F/A · ε = Δl/l · E = σ/εThe work done deforming a solid is the area under the force–extension graph, which is ½Fx if Hooke's law is obeyed.
Eduqas classifies solids as crystalline (metals), amorphous (glass, ceramics) and polymeric (rubber, polythene) — and expects you to know the shape of the graph for each:
Ductile metal (copper): elastic (Hooke's law) → plastic. Plastic flow happens because dislocations in the lattice move. Metals are strengthened by putting barriers in the way of dislocations: foreign atoms (alloying), other dislocations (work hardening) and more grain boundaries. Eventually the wire necks and suffers ductile fracture.
Brittle (glass): obeys Hooke's law right up to fracture — no plastic region. Fails by crack propagation from surface imperfections. Removing those flaws (thin fibres) or putting the surface into compression (toughened glass, pre-stressed concrete) raises the breaking stress dramatically.
Rubber: only approximately Hookean, very low Young modulus, huge extensions as tangled chain molecules straighten out. Loading and unloading follow different curves — hysteresis — and the area enclosed is the energy converted to internal energy each cycle (which is why tyres get warm).
Specified practicals: determination of the Young modulus of a metal wire; investigation of the force–extension relationship for rubber.
Calculate
Your turn — Young modulus
3A wire of diameter 0.80 mm and length 1.60 m extends by 0.90 mm under a force of 60 N. Calculate the Young modulus, in GPa.
GPa
Hint: A = π(0.40 × 10⁻³)² = 5.03 × 10⁻⁷ m². σ = 60 ÷ 5.03 × 10⁻⁷ = 1.19 × 10⁸ Pa. ε = 0.90 × 10⁻³ ÷ 1.60 = 5.63 × 10⁻⁴. E = σ/ε, then ÷ 10⁹.
Quick check
Why does rubber get warm?
?A rubber band is stretched and then released, and the loading and unloading curves enclose a loop (hysteresis). What does the area of the loop represent?
6. Fields of force
Electrostatic and gravitational fields
F = Q₁Q₂ / 4πε₀r² · F = GM₁M₂ / r²Coulomb's law and Newton's law of gravitation — both inverse square laws. Handy: 1/4πε₀ ≈ 9 × 10⁹ F⁻¹ m.
E = Q / 4πε₀r² · g = GM / r²Field strength = force per unit charge / per unit mass. Outside a spherical body, the field is the same as if all the mass were at the centre.
VE = Q / 4πε₀r · Vg = −GM / rPotential = work done per unit charge (or mass) bringing it from infinity. Note the minus sign for gravity: gravitational potential is always negative because gravity is always attractive. Field strength = −slope of the potential–r graph.
Similarities and differences: both are inverse square; both have field strength defined as force per unit "something"; both have zero potential at infinity. But the electric force can attract or repel, while gravity is only attractive — and gravity is unimaginably weaker (the electric repulsion between two protons is about 10³⁶ times their gravitational attraction).
7. Using radiation to investigate stars
Black bodies, Wien's law and Stefan's law
A star's spectrum is a continuous emission spectrum (from the dense surface) crossed by a line absorption spectrum (from the cooler, tenuous atmosphere absorbing specific photon energies). Stars are excellent approximations to black bodies.
λmax T = 2.90 × 10⁻³ m K (Wien)The peak wavelength is inversely proportional to the absolute temperature — hotter stars peak bluer. T(K) = θ(°C) + 273.15.
P = σAT⁴ (Stefan–Boltzmann)σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴. For a star, A = 4πR², so the luminosity L = 4πR²σT⁴ — the total power it radiates.
intensity I = L / 4πd² (inverse square law)Measure the intensity at Earth and know the distance → get the luminosity. Combine with Wien's law (temperature) and Stefan's law gives you the radius.
Multiwavelength astronomy: observing the same region at different wavelengths (i.e. at different photon energies) reveals different physical processes — cold dust in the infrared, hot gas in X-rays. No single waveband tells the whole story.
Calculate
Your turn — Wien's law
4A star's black-body spectrum peaks at a wavelength of 500 nm. Calculate its surface temperature. (Wien constant = 2.90 × 10⁻³ m K)
5That star has a radius of 7.0 × 10⁸ m and a surface temperature of 5800 K. Calculate its luminosity, in units of 10²⁶ W. (σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴)
× 10²⁶ W
Hint: L = 4πR²σT⁴. 4πR² = 6.16 × 10¹⁸ m². T⁴ = 5800⁴ = 1.13 × 10¹⁵ K⁴. σT⁴ = 6.42 × 10⁷ W m⁻². Enter just the number in front of × 10²⁶.
8. Orbits and the wider universe
Kepler, dark matter, Doppler and Hubble
Kepler's laws: (1) planets move in ellipses with the Sun at one focus; (2) the line from Sun to planet sweeps equal areas in equal times; (3) T² ∝ r³. You can derive the third law for a circular orbit: set GMm/r² = mv²/r with v = 2πr/T, giving T² = (4π²/GM)r³. Rearranged, it lets you calculate the mass of the central object from a satellite's orbit.
Dark matter: stars in the outer parts of spiral galaxies orbit much faster than the visible mass can account for. Either our gravity is wrong, or there is a great deal of matter we cannot see. The favoured explanation is dark matter — and the Higgs boson may be related to it.
Δλ / λ = v / cDoppler shift of a spectral line gives the radial velocity. Red shift → receding; blue shift → approaching. Applied to a star wobbling under an orbiting exoplanet, it lets us weigh both bodies.
v = H₀D · age ≈ 1 / H₀Hubble's law. If a galaxy has always receded at v, it took D/v = 1/H₀ to reach its present distance — hence 1/H₀ estimates the age of the universe.
Critical density: ρc = 3H₀² / 8πG is the density that makes the universe exactly "flat" — derived by setting the total energy (kinetic + gravitational potential) of a galaxy at the edge of an expanding sphere to zero.
Calculate
Your turn — Hubble's law
6A galaxy is 8.0 × 10²⁴ m away. Taking H₀ = 2.2 × 10⁻¹⁸ s⁻¹, calculate its recession speed in units of 10⁶ m s⁻¹.
× 10⁶ m s⁻¹
Hint: v = H₀D = 2.2 × 10⁻¹⁸ × 8.0 × 10²⁴ = 1.76 × 10⁷ m s⁻¹. Express that as a multiple of 10⁶.
Quick check
Evidence for dark matter
?Which observation provides the evidence for dark matter in spiral galaxies?
Match it
Match each law to what it tells you
Tap an item on the left, then its partner on the right.
Equation
What it gives you
Recap
Component 2 — the big ideas
Conduction: I = ΔQ/Δt; I = nAve; drift velocity is slow but the field is established almost instantly
Resistance: V = IR; P = I²R = V²/R; R = ρL/A; superconductors → zero R below Tc
Circuits: series/parallel rules; Vout = VinR₂/(R₁+R₂); V = E − Ir
Capacitance: C = Q/V = ε₀A/d; U = ½QV; time constant RC; parallel adds