(a) In spherical coordinates, V = –25 V on a conductor at r = 2 cm and V = 150 V on another conductor at r = 35 cm. The space…

(a) Let a = 2 cm = 0.02 m and b = 35 cm = 0.35 m. For spherical symmetry, Laplace's equation gives

V(r) = A/r + B.

Using the boundary values: V(a) = A/a + B = –25 V, V(b) = A/b + B = 150 V.

Subtracting, A(1/b – 1/a) = 175. Now 1/b – 1/a = 1/0.35 – 1/0.02 = 20/7 – 50 = –330/7 m⁻¹. Thus A = –175 × 7/330 = –245/66 V m. Then B = –25 – A/a = –25 + (245/66)/0.02 = 5300/33 V.

Hence V(r) = –245/(66r) + 5300/33 V, and the radial electric field is E_r = –dV/dr = A/r² = –245/(66r²) V/m. The field is negative, i.e. directed radially inward.

For a conductor, the surface charge density is σ = D·n = ε₀εᵣE·n, where n is outward from the conductor into the dielectric.

At the inner conductor r = a, n = +r̂: σ_inner = ε₀εᵣE_r(a) = ε₀εᵣA/a². With ε₀ = 8.854×10⁻¹² F/m, εᵣ = 3.12, a = 0.02 m: σ_inner = –2.56×10⁻⁷ C/m².

At the outer conductor r = b, n = –r̂: σ_outer = –ε₀εᵣE_r(b) = –ε₀εᵣA/b². Thus σ_outer = +8.37×10⁻¹⁰ C/m².

Final: σ_inner ≈ –2.56×10⁻⁷ C/m², σ_outer ≈ +8.37×10⁻¹⁰ C/m².

(b) Let the square loop have side L and carry current I. Use the Biot–Savart law for one side. The magnetic flux density due to a finite straight wire at perpendicular distance d is

B = μ₀I/(4πd)(sinθ₁ + sinθ₂).

At the centre of the square, for any one side, d = L/2. The two end angles are equal: θ₁ = θ₂ = 45°, so sinθ₁ = sinθ₂ = 1/√2. Therefore

B_side = μ₀I/(4π(L/2))(2/√2) = √2 μ₀I/(2πL).

All four sides produce magnetic fields in the same direction at the centre by the right-hand rule. Hence

B_total = 4B_side = 2√2 μ₀I/(πL).

Since H = B/μ₀ in free space, H = 2√2 I/(πL).

Final: H = 2√2 I/(πL) A/m at the centre, directed normal to the plane of the loop according to the right-hand rule.

(c) Treat the Earth as a conducting sphere of radius R = 6.37×10⁶ m. By Gauss's law, outside a conducting sphere,

E = Q/(4π ε₀ R²),

where Q is the total net surface charge. The field magnitude is 150 N/C and is directed radially inward, so Q must be negative. Using magnitudes:

|Q| = 4π ε₀ R² E.

Substitute:

|Q| = 4π(8.854×10⁻¹²)(6.37×10⁶)²(150) C.

Now R² = (6.37×10⁶)² = 4.05769×10¹³ m². Thus

|Q| = 4π(8.854×10⁻¹²)(4.05769×10¹³)(150) ≈ 6.77×10⁵ C.

Because E is inward, Q = –6.77×10⁵ C.

Final: total net surface charge of the Earth ≈ –6.77×10⁵ C.

(d) Let W be the work done by the gas. For a quasi-static adiabatic process, dQ = 0. By the first law,

dQ = dU + dW,

so dW = –dU.

For a perfect gas, internal energy is U = nC_VT. Since C_V = R/(γ – 1) and PV = nRT, U = PV/(γ – 1).

Therefore, for a finite adiabatic change, W = Uᵢ – U_f = (PᵢVᵢ – P_fV_f)/(γ – 1).

Now use the adiabatic relation for a perfect gas: PV^γ = constant. Thus P_fV_f^γ = PᵢVᵢ^γ, which gives V_f = Vᵢ(Pᵢ/P_f)^(1/γ).

Hence P_fV_f = P_fVᵢ(Pᵢ/P_f)^(1/γ) = PᵢVᵢ(P_f/Pᵢ)^(1 – 1/γ) = PᵢVᵢ(P_f/Pᵢ)^((γ – 1)/γ).

Substituting into W: W = PᵢVᵢ/(γ – 1)[1 – (P_f/Pᵢ)^((γ – 1)/γ)].

Final: W = PᵢVᵢ/(γ – 1)[1 – (P_f/Pᵢ)^((γ – 1)/γ)], valid for a perfect gas undergoing a quasi-static adiabatic expansion.

(e) For sodium, one free electron per atom. The electron number density is

n = ρN_A/M.

Given ρ = 0.97 g/cm³ = 970 kg/m³, M = 23 g/mol = 0.023 kg/mol, and N_A = 6.022×10²³ mol⁻¹:

n = (970 × 6.022×10²³)/0.023 ≈ 2.54×10²⁸ m⁻³.

The Fermi energy is given by the free-electron formula

E_F = (ħ²/(2mₑ))(3π²n)^(2/3).

Using ħ = 1.055×10⁻³⁴ J s and mₑ = 9.109×10⁻³¹ kg: 3π²n = 3π²(2.54×10²⁸) ≈ 7.52×10²⁹ m⁻³. Thus k_F = (3π²n)^(1/3) ≈ 9.10×10⁹ m⁻¹.

Therefore E_F = ħ²k_F²/(2mₑ) ≈ (1.055×10⁻³⁴)²(9.10×10⁹)²/(2 × 9.109×10⁻³¹) J ≈ 5.05×10⁻¹⁹ J.

Converting to electron-volt, E_F = 5.05×10⁻¹⁹/1.602×10⁻¹⁹ eV ≈ 3.15 eV.

Final: Fermi energy of sodium ≈ 3.15 eV.