(a) Find the energy stored in a system of four charges Q₁ = 1 nC, Q₂ = 2 nC, Q₃ = 3 nC and Q₄ = 4 nC placed at the cartesian…

(a) Energy stored in the four-charge system

The electrostatic energy of point charges is U = k Σ(i<j) QᵢQⱼ/rᵢⱼ with k = 1/(4πε₀) = 8.99×10⁹ N m² C⁻². Distances (in m): r₁₂ = 1; r₁₃ = 3; r₁₄ = √2; r₂₃ = √10; r₂₄ = 1; r₃₄ = √5. Each product of charges carries 10⁻¹⁸ C². So U = k×10⁻¹⁸ [ (1×2)/1 + (1×3)/3 + (1×4)/√2 + (2×3)/√10 + (2×4)/1 + (3×4)/√5 ] = k×10⁻¹⁸ [ 2 + 1 + 2.828 + 1.897 + 8 + 5.367 ] = k×10⁻¹⁸ × 21.09 U = 8.99×10⁹ × 10⁻¹⁸ × 21.09 = 1.90×10⁻⁷ J. The energy is positive (about 0.19 μJ) because all charges have the same sign: work is needed to assemble them.

(b) Inductance per unit length of two parallel wires

Let the wires of radius a carry +I and −I, with axes a distance d apart (d > 2a). External flux. At a point between the wires, at distance r from wire 1 and d − r from wire 2, both fields point the same way: B = (μ₀I/2π)[1/r + 1/(d − r)]. Flux per unit length through the plane between the wire surfaces: Φ_ext = ∫ₐ^(d−a) B dr = (μ₀I/2π)[ln((d − a)/a) + ln((d − a)/a)] = (μ₀I/π) ln((d − a)/a). So L_ext = Φ_ext/I = (μ₀/π) ln((d − a)/a). Internal inductance. Inside a wire with uniform current, Ampère's law gives B = μ₀Ir/(2πa²). The energy per unit length is ∫₀^a (B²/2μ₀) 2πr dr = μ₀I²/(16π), and equating this to ½L_iI² gives L_i = μ₀/(8π) per wire, so μ₀/(4π) for the two wires. Total: L = (μ₀/π)[ln((d − a)/a) + 1/4] H/m, which for d ≫ a becomes (μ₀/π)[ln(d/a) + 1/4]. (The internal term holds for uniform, low-frequency current; at high frequencies, skin effect removes it.)

(c) The continuity equation from Maxwell's equations

Take the Ampère-Maxwell law: ∇×B = μ₀J + μ₀ε₀ ∂E/∂t. Take the divergence of both sides. The divergence of a curl is zero, so 0 = μ₀ ∇·J + μ₀ε₀ ∂(∇·E)/∂t. Use Gauss's law ∇·E = ρ/ε₀: 0 = μ₀ ∇·J + μ₀ ∂ρ/∂t. Therefore ∇·J + ∂ρ/∂t = 0, the continuity equation, in integral form ∮J·dA = −dQ/dt. It states the conservation of charge. Without Maxwell's displacement current, the divergence would give ∇·J = 0, which is wrong for time-varying charge density, so the term is needed for consistency.

(d) Zeroth law, temperature and isotherms

Zeroth law: if system A is in thermal equilibrium with C, and B is in thermal equilibrium with C, then A and B are in thermal equilibrium with each other. Thermal equilibrium is therefore transitive, and so systems can be sorted into classes of mutual equilibrium; a number θ attached to each class is the empirical temperature. Two systems have the same temperature if and only if they are in thermal equilibrium. In practice, a thermometer C with a thermometric property X (gas pressure, resistance, emf) is calibrated, for example θ = 100(X − X_ice)/(X_steam − X_ice). An isotherm is the locus of states of one system with the same temperature, given by the equation of state f(P, V, θ) = 0; for an ideal gas PV = nRθ gives hyperbolas on a P-V diagram. To draw isotherms of two systems A and B: bring the thermometer C to a fixed reading θ₁; change the state of A until A is in equilibrium with C at that reading, mark the state, and repeat at many states to join them as the θ₁ isotherm of A. Do the same for B with the same C and θ₁; repeat for θ₂, θ₃, and so on. By the zeroth law, any state on the θ₁ isotherm of A is in equilibrium with any state on the θ₁ isotherm of B, so the isotherms of different systems carry a common label, though their shapes differ.

(e) Fermi-Dirac and Bose-Einstein distributions

Fermi-Dirac (fermions, half-integer spin, Pauli exclusion): f_FD(E) = 1/[exp((E − μ)/k_BT) + 1], with 0 ≤ f ≤ 1. Bose-Einstein (bosons, integer spin): f_BE(E) = 1/[exp((E − μ)/k_BT) − 1], with E > μ; for photons μ = 0. Plots against E: Fermi-Dirac: at T = 0 it is a step, f = 1 for E < μ (the Fermi energy) and 0 above; for T > 0 the step is rounded over about k_BT, passing through f = ½ at E = μ, with a Boltzmann tail e^(−(E − μ)/k_BT) at high energy. Bose-Einstein: f diverges as E → μ⁺ and decreases monotonically with E, lying above the Boltzmann curve at low energy, which shows the tendency of bosons to cluster; it tends to the same tail at high energy. For E − μ ≫ k_BT both reduce to the classical Maxwell-Boltzmann form e^(−(E − μ)/k_BT).