(a) A neutral atom consists of a point nucleus +q surrounded by a uniformly charged spherical cloud (-q) of radius r. Show that…

(a) Atomic polarizability of the nucleus-cloud model

Let the nucleus +q be displaced by x from the centre of the uniformly charged cloud of total charge −q and radius r, in a weak field E. By Gauss's law, the cloud's field at distance x < r from its centre comes only from the charge enclosed, −q(x/r)³: E_cloud · 4πx² = −q x³/(ε₀ r³), so E_cloud = −q x/(4πε₀ r³). The restoring force on the nucleus is F = qE_cloud = −q²x/(4πε₀ r³), a linear (spring-like) force. Equilibrium: the force qE from the external field balances the restoring force: qE = q²x/(4πε₀ r³), so x = 4πε₀ r³E/q. The induced dipole moment is p = qx = 4πε₀ r³E, so α = p/E = 4πε₀ r³ = 3ε₀V, where V = (4/3)πr³ is the volume of the cloud. Hence α ∝ V. Validity: the field is weak, so x ≪ r and the response is linear. Check: α has the dimensions of ε₀ × volume. A larger cloud holds the charges farther apart and so is more easily polarised.

(b) Piston-cylinder: heating air against stops

State 1: P₁ = 150 kPa, T₁ = 300 K, V₁ = 0.400 m³. The piston rests on the stops, so the first stage is at constant volume until P = 350 kPa (the pressure needed to lift the piston). The second stage is at constant P = 350 kPa until V₃ = 2V₁ = 0.800 m³. Mass of air: m = P₁V₁/(RT₁) = (150 × 0.400)/(0.287 × 300) = 0.697 kg, and mR = 0.200 kJ/K. (i) Final temperature: T₃ = P₃V₃/(mR) = (350 × 0.800)/0.200 = 1400 K (1127°C). At the end of stage 1, T₂ = T₁P₂/P₁ = 700 K. (ii) Work: only stage 2 does work, W = P₂(V₃ − V₂) = 350 × (0.800 − 0.400) = 140 kJ. In the P-V diagram this is the area under the horizontal line at 350 kPa; the vertical stage adds none. (iii) Heat: with the given values, ΔU = m(U_final − U₃₀₀K) = 0.697 × (1113 − 214) = 626.5 kJ. By the first law, Q = ΔU + W = 626.5 + 140 = 766.5 kJ transferred to the air.

(c) Vector potential of a spinning charged shell

The surface current density is K = σv = σ(ω × r′), with r′ on the shell (|r′| = R). The vector potential is A(r) = (μ₀/4π) ∮ K da′/ℛ, with ℛ = |r − r′|, da′ = R² sinθ′ dθ′ dφ′. Choose the z-axis along the field point, r = r ẑ, and ω = ω(sinψ x̂ + cosψ ẑ). With r′ = R(sinθ′cosφ′ x̂ + sinθ′sinφ′ ŷ + cosθ′ ẑ), (ω × r′)_y = ωR(cosψ sinθ′cosφ′ − sinψ cosθ′). ℛ does not depend on φ′, so the φ′-integral removes the cosφ′ term, and the x and z components of K integrate to zero. Only A_y remains: A_y = −(μ₀σωR³ sinψ/2) ∫₀^π cosθ′ sinθ′ dθ′/ℛ, with ℛ = √(R² + r² − 2Rr cosθ′). Expand 1/ℛ = Σ (r_<^l/r_>^l+1) Pₗ(cosθ′). Only l = 1 survives against cosθ′, and ∫₋₁¹ u² du = 2/3, so the integral is (2/3)(r_</r_>²): r < R: 2r/(3R²); r > R: 2R/(3r²). Since ω × r = −ωr sinψ ŷ, this gives A = (μ₀σR/3)(ω × r) for r < R, A = (μ₀σR⁴/3r³)(ω × r) for r > R. Outside, A is that of a dipole of moment m = (4π/3)σωR⁴, with A = (μ₀/4π)(m × r)/r³. Inside, B = ∇ × A = (2μ₀σRω/3) is uniform.