(a) Using Maxwell's equations, obtain Poisson's equation and Laplace's equation. The region -(π)/2 < z/(z₀) < (π)/2 has a charge…

(a) From Gauss's law and E = −∇V: ∇·E = ρ/ε₀, so ∇·(−∇V) = ρ/ε₀. Hence ∇²V = −ρ/ε₀ (Poisson's equation). If ρ = 0, this becomes ∇²V = 0 (Laplace's equation).

Here ρ = ρ₀ cos(z/z₀), ρ₀ = 10⁻⁸ C/m³, inside |z| < πz₀/2; elsewhere ρ = 0. By translational symmetry in x and y, V = V(z). Poisson's equation is d²V/dz² = −(ρ₀/ε₀) cos(z/z₀).

Integrating once: dV/dz = −(ρ₀z₀/ε₀) sin(z/z₀) + C₁. Integrating again: V = (ρ₀z₀²/ε₀) cos(z/z₀) + C₁z + C₂.

Symmetry about z = 0 gives C₁ = 0. Taking V(0) = 0 gives C₂ = −ρ₀z₀²/ε₀. Thus for |z| < πz₀/2: V = (ρ₀z₀²/ε₀)[cos(z/z₀) − 1] volt, E = −(dV/dz) ẑ = (ρ₀z₀/ε₀) sin(z/z₀) ẑ V/m.

Outside the slab, ρ = 0, so Laplace's equation gives V = Az + B, E_z = −A. At z = a = πz₀/2, continuity of E gives A = −ρ₀z₀/ε₀. Matching V at z = a gives B = (ρ₀z₀²/ε₀)(π/2 − 1). Therefore for z > πz₀/2: V = (ρ₀z₀²/ε₀)(π/2 − 1 − z/z₀) volt, E = (ρ₀z₀/ε₀) ẑ V/m. For z < −πz₀/2, by symmetry: V = (ρ₀z₀²/ε₀)(π/2 − 1 + z/z₀) volt, E = −(ρ₀z₀/ε₀) ẑ V/m. Numerically ρ₀/ε₀ = 10⁻⁸/8.854×10⁻¹² ≈ 1.13×10³ V/m², so |E| outside ≈ 1.13×10³ z₀ V/m (z₀ in metres). An arbitrary constant may be added to V; E is unchanged.

(b) (i) Anomalous dispersion is the region of dispersion in which the refractive index decreases as frequency increases, equivalently dn/dλ > 0, usually near an absorption band. Normal dispersion has dn/dλ < 0. In the resonance (Sellmeier) picture, n rises steeply just below a natural frequency of the bound electrons and falls sharply through the absorption band; Le Roux first observed this in iodine vapour. In a prism, the refractive index depends on wavelength: violet light has larger n than red in normal dispersion, so violet is deviated more and red less. Hence white light emerges as a spectrum of separated colours.

(b) (ii) Use scalar magnetic potential Φ_m with H = −∇Φ_m. Since ∇·B = 0 and B = μ₀(H + M), ∇²Φ_m = 0 inside and outside because M is uniform inside. At r = a, σ_m = M·r̂ = M₀ cosθ, and ∂Φ_in/∂r − ∂Φ_out/∂r = M₀ cosθ.

The source is cosθ, so retain only the l = 1 term: Φ_in = A r cosθ, Φ_out = B cosθ/r². Continuity at r = a gives A a = B/a², so B = A a³. Now ∂Φ_in/∂r = A cosθ, and ∂Φ_out/∂r = −2B cosθ/a³ = −2A cosθ. Therefore ∂Φ_in/∂r − ∂Φ_out/∂r = A cosθ + 2A cosθ = 3A cosθ = M₀ cosθ. Thus A = M₀/3. Hence for r < a, Φ_m = (M₀ r cosθ)/3 = M₀ z/3, with M₀ in A/m and Φ_m in A. Then H = −M₀ ẑ/3 A/m, and B = (2μ₀M₀/3) ẑ T.

(c) Internal energy U is the total energy associated with the microscopic motions and interactions of the constituents of the system, excluding bulk kinetic and potential energy. For a closed system with only PV work, the first law gives dU = T dS − P dV. Helmholtz function: F = U − TS. Enthalpy: H = U + PV. Gibbs potential: G = U − TS + PV = H − TS = F + PV.

From dU = T dS − P dV, U = U(S,V), so (∂T/∂V)_S = −(∂P/∂S)_V. … (1)

From F = U − TS, dF = −S dT − P dV. Thus (∂S/∂V)_T = (∂P/∂T)_V. … (2)

From H = U + PV, dH = T dS + V dP. Thus (∂T/∂P)_S = (∂V/∂S)_P. … (3)

From G = U − TS + PV, dG = −S dT + V dP. Thus (∂S/∂P)_T = −(∂V/∂T)_P. … (4)

The four Maxwell thermodynamic relations are therefore: (∂T/∂V)_S = −(∂P/∂S)_V, (∂S/∂V)_T = (∂P/∂T)_V, (∂T/∂P)_S = (∂V/∂S)_P, (∂S/∂P)_T = −(∂V/∂T)_P. They are valid for a closed system of fixed composition with only reversible PV work.