(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 between the conductors is a dielectric for which εᵣ = 3·12. Find the surface charge densities on the conductors. (10 marks)

Question

(b) Find the magnetic field strength (H) at the centre of a square current loop of side L. (10 marks)

(c) The magnitude of the average electric field normally present in the Earth's atmosphere just above the surface of the Earth is about 150 N/C, directed radially inward, toward the centre of the Earth. What is the total net surface charge carried by the Earth? Assume the Earth to be a conductor. (The radius of the Earth is 6·37×10⁶ m) (10 marks)

(d) Prove that the work done by a perfect gas during a quasi-static adiabatic expansion is given by

W = (Pᵢ Vᵢ)/(γ - 1)[1 - ((P_f)/(Pᵢ))^((γ-1)/(γ))]

where γ is the ratio of specific heats. (10 marks)

(e) Calculate the Fermi energy in electron-volt for sodium assuming that it has one free electron per atom. The density of sodium = 0.97 gm/cc and the atomic weight of sodium is 23. (10 marks)

UPSC Answer Check · Accepted Answer

(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.

(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…

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