(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) The electrostatic energy of a system of point charges is U = (1/(4πϵ₀)) Σ over pairs i<j of qᵢ qⱼ / r_ij. There is no self-energy. Taking the given coordinates in metres and k = 1/(4πϵ₀) ≈ 8.9875517923×10⁹ N m² C⁻²:

• r₁₂ = |R₂ − R₁| = √((2−1)² + (1−1)²) = 1 m
• r₁₃ = |R₃ − R₁| = √((1−1)² + (4−1)²) = 3 m
• r₁₄ = |R₄ − R₁| = √((2−1)² + (2−1)²) = √2 m
• r₂₃ = |R₃ − R₂| = √((1−2)² + (4−1)²) = √10 m
• r₂₄ = |R₄ − R₂| = √((2−2)² + (2−1)²) = 1 m
• r₃₄ = |R₄ − R₃| = √((2−1)² + (2−4)²) = √5 m

Using Q₁ = 1 nC, Q₂ = 2 nC, Q₃ = 3 nC, Q₄ = 4 nC, each product qᵢ qⱼ carries 10⁻¹⁸ C². Therefore U = k×10⁻¹⁸ [ (1×2)/1 + (1×3)/3 + (1×4)/√2 + (2×3)/√10 + (2×4)/1 + (3×4)/√5 ] J = k×10⁻¹⁸ [ 2 + 1 + 4/√2 + 6/√10 + 8 + 12/√5 ] J. Since 4/√2 = 2√2, U = k×10⁻¹⁸ [ 11 + 2√2 + 6/√10 + 12/√5 ] J. The bracket equals 11 + 2.8284271247 + 1.8973665961 + 5.3665631460 = 21.0923568668. Thus U = 8.9875517923×10⁹ × 10⁻¹⁸ × 21.0923568668 J = 1.8956865×10⁻⁷ J. U ≈ 1.89569×10⁻⁷ J ≈ 0.1896 μJ. The positive sign means work must be done to assemble the system because all charges have the same sign.

(b) Let the two long parallel wires have radii a and axes separated by d. Assume d > 2a, free space permeability μ₀, and uniformly distributed currents +I and −I. Use cylindrical symmetry and Ampere’s law. At a distance r from the axis of wire 1, its field is B₁ = μ₀ I/(2π r). The second wire is at distance d − r from the same point and carries current −I, so its field has the same direction in the region between the wires: B₂ = μ₀ I/(2π(d − r)). Thus the total field between the wires is B = (μ₀ I/(2π)) [ 1/r + 1/(d − r) ]. The external flux per unit length through the rectangle between the wire surfaces r = a and r = d − a is Φ_ext = ∫ from a to d−a of B dr = (μ₀ I/(2π)) [ ∫ from a to d−a dr/r + ∫ from a to d−a dr/(d − r) ]. Both integrals give ln((d − a)/a). Hence Φ_ext = (μ₀ I/π) ln((d − a)/a). Therefore the external inductance per unit length is L_ext = Φ_ext/I = (μ₀/π) ln((d − a)/a).

Now include internal inductance. Inside one wire, Ampere’s law gives B(r) = μ₀ I r/(2π a²), 0 ≤ r ≤ a. The magnetic energy per unit length inside one wire is U_int,one = ∫ from 0 to a of B²/(2μ₀) × 2π r dr = μ₀ I²/(16π). Since U = 1/2 L I², the internal inductance of one wire is L_int,one = μ₀/(8π). For two wires, L_int = 2 × μ₀/(8π) = μ₀/(4π). Hence the total inductance per unit length is L = (μ₀/π) [ ln((d − a)/a) + 1/4 ] H/m. For d ≫ a this reduces to L ≈ (μ₀/π) [ ln(d/a) + 1/4 ] H/m. This assumes no skin effect; at high frequencies the internal term is modified.

(c) Maxwell’s differential equations in the presence of charge density ρ and current density J are

• ∇·E = ρ/ϵ₀
• ∇·B = 0
• ∇×E = −∂B/∂t
• ∇×B = μ₀ J + μ₀ ϵ₀ ∂E/∂t.

Take the divergence of the Ampere–Maxwell law: ∇·(∇×B) = μ₀ ∇·J + μ₀ ϵ₀ ∂/∂t (∇·E). But the divergence of a curl is identically zero: ∇·(∇×B) = 0. Using Gauss’s law ∇·E = ρ/ϵ₀, 0 = μ₀ ∇·J + μ₀ ϵ₀ ∂/∂t (ρ/ϵ₀) = μ₀ ( ∇·J + ∂ρ/∂t ). Since μ₀ is nonzero, ∇·J + ∂ρ/∂t = 0. This is the continuity equation expressing conservation of electric charge. In integral form, ∮ J·da + d/dt ∫ ρ dV = 0. Thus the continuity equation is embedded in Maxwell’s equations. The displacement-current term μ₀ϵ₀ ∂E/∂t is essential; without it, taking the divergence would give ∇·J = 0, which fails for time-varying charge density.

(d) Two systems are in thermal equilibrium when, after being connected by a diathermal wall, their macroscopic state variables no longer change with time. The Zeroth law states: if system A is in thermal equilibrium with system C, and system B is also in thermal equilibrium with C, then A and B are in thermal equilibrium with each other. Therefore thermal equilibrium is transitive and behaves like an equivalence relation. All states that are mutually in thermal equilibrium can be placed in one equivalence class. We assign a scalar label θ to each equivalence class. This label is called empirical temperature. Two systems have the same temperature if and only if they are in thermal equilibrium.

Operationally, choose a thermometer C with a measurable thermometric property X, such as volume, pressure, resistance, or emf. Define θ by a suitable calibration, for example θ = 100 (X − X_ice)/(X_steam − X_ice) °C, where X_ice and X_steam are the values at ice point and steam point. When C is in thermal equilibrium with a system, the system is assigned the temperature θ read by C.

An isotherm is the locus of states of a system having the same temperature. For a simple system described by pressure P and volume V, the equation of state f(P,V,θ) = 0 defines isotherms θ = constant. For an ideal gas, PV = nRθ, so isotherms are hyperbolas on a P–V diagram.

To draw isotherms of two different systems A and B:

• Use the same thermometer C and choose fixed readings θ₁, θ₂, θ₃, …
• For system A, keep C at θ₁ and adjust the state of A until A and C are in thermal equilibrium. Mark that state. Repeat for many states of A while C remains at θ₁. Join these points to obtain the θ₁ isotherm of A. Repeat for θ₂, θ₃, …
• Do the same for system B using the same thermometer and the same θ values. By the Zeroth law, any state on the θ₁ isotherm of A and any state on the θ₁ isotherm of B would be in mutual thermal equilibrium if connected by a diathermal wall. Thus isotherms of different systems can be labelled by a common empirical temperature, although their shapes may differ because their equations of state differ.

(e) The Fermi–Dirac distribution gives the average occupation number for fermions: f_FD(E) = 1/[ exp((E − μ)/(k_B T)) + 1 ]. The Bose–Einstein distribution gives the average occupation number for bosons: f_BE(E) = 1/[ exp((E − μ)/(k_B T)) − 1 ], with the condition E > μ; for photons μ = 0, so f_BE(E) = 1/[exp(E/k_B T) − 1].

For the Fermi–Dirac case, at T = 0 the distribution is a step function: f_FD = 1 for E < μ and f_FD = 0 for E > μ. At T > 0 the step is smoothed over an energy width of order k_B T. At E = μ, f_FD = 1/2. For E − μ ≫ k_B T, f_FD ≈ exp(−(E − μ)/(k_B T)), while for μ − E ≫ k_B T, f_FD ≈ 1.

For the Bose–Einstein case, f_BE diverges as E → μ⁺. For E − μ ≫ k_B T, it also tends to exp(−(E − μ)/(k_B T)). For photons with μ = 0, f_BE diverges at E = 0 and decreases monotonically to zero as E increases.

Schematic plots versus energy E are:

Fermi–Dirac: `` f(E) 1 | ________ | ___ 1/2| • at E = μ | ____ 0 | ____ +--------|--------------> E μ ``

Bose–Einstein: `` f(E) ^ | diverges as E → μ⁺ | \ | ___ | ____ | ______ +--------|--------------> E μ ``

For both distributions, when E − μ ≫ k_B T, the exponential term dominates and both reduce to the classical Maxwell–Boltzmann form f ≈ exp(−(E − μ)/(k_B T)). The difference is that Fermi–Dirac suppresses occupation because of the Pauli exclusion principle, while Bose–Einstein enhances occupation at low energy.