(a) A circular ring of radius R lying on the x-y plane and centred at the origin, carries a uniform line charge λ. Find the first…

(a) Multipole expansion of a uniformly charged ring

For r > R, V(r,θ) = (1/4πε₀) Σₙ (1/rⁿ⁺¹) ∮ r′ⁿ Pₙ(cosγ) dq′, where γ is the angle between r and the source point. For the ring, r′ = R, dq′ = λR dφ′, Q = 2πλR, and cosγ = sinθ cos(φ′ − φ). Monopole (n = 0): V₀ = (1/4πε₀) Q/r = λR/(2ε₀r). Dipole (n = 1): p = ∮ r′ dq′ = λR² ∫₀²π (cosφ′, sinφ′, 0) dφ′ = 0 by symmetry, so V₁ = 0. Quadrupole (n = 2): V₂ = (1/4πε₀)(λR³/r³) ∫₀²π P₂(sinθ cos(φ′ − φ)) dφ′, with P₂(x) = (3x² − 1)/2. Since ∫cos² dφ′ = π, ∫ P₂ dφ′ = (1/2)[3π sin²θ − 2π] = (π/2)(3 sin²θ − 2) = −π P₂(cosθ). So V₂ = (1/4πε₀)(πλR³/2)(3 sin²θ − 2)/r³ = −(1/4πε₀) QR² P₂(cosθ)/(2r³). Result, for r ≫ R: V(r,θ) = (1/4πε₀)[ Q/r + 0 + (QR²/4)(3 sin²θ − 2)/r³ + … ], Q = 2πλR. Check on the axis (θ = 0): V = (1/4πε₀)(Q/r)(1 − R²/2r²), which matches the binomial expansion of the exact result Q/(4πε₀√(r² + R²)).

(b) Method of images for two charges above a grounded plane

For a grounded plane z = 0, each charge q at height z has an image −q at −z. The z = 0 plane is then at zero potential, which is the required boundary condition for the region z > 0, where P lies. Charges: Q₁ = 3 nC at (0,2,2); Q₁′ = −3 nC at (0,2,−2); Q₂ = 4 nC at (0,−2,4); Q₂′ = −4 nC at (0,−2,−4). Vectors from each charge to P = (3,2,4): Q₁: (3,0,2), r = √13; Q₁′: (3,0,6), r = √45; Q₂: (3,4,0), r = 5; Q₂′: (3,4,8), r = √89. Potential, with 1/4πε₀ = 9×10⁹ N m²/C²: V = 9[3/√13 − 3/√45 + 4/5 − 4/√89] = 9[0.8321 − 0.4472 + 0.8 − 0.4240] = 6.85 V. Field, E = (1/4πε₀) Σ q r/r³, in V/m: Q₁: 9 × 3(3,0,2)/46.87 = (1.728, 0, 1.152) Q₁′: −9 × 3(3,0,6)/301.9 = (−0.268, 0, −0.537) Q₂: 9 × 4(3,4,0)/125 = (0.864, 1.152, 0) Q₂′: −9 × 4(3,4,8)/839.5 = (−0.129, −0.172, −0.343) Sum: E ≈ (2.20 î + 0.98 ĵ + 0.27 k̂) V/m, so |E| ≈ 2.42 V/m.

(c) Molecules of oxygen in a speed interval

Mass of an O₂ molecule: m = 32 × 1.66×10⁻²⁷ = 5.31×10⁻²⁶ kg. Number of molecules: N = 0.1/5.31×10⁻²⁶ = 1.88×10²⁴. T = 273 K, k_B = 1.38×10⁻²³ J/K. Maxwell-Boltzmann speed distribution: f(v) = 4π(m/2πk_BT)³/2 v² exp(−mv²/2k_BT). a = m/(2k_BT) = 7.05×10⁻⁶ s²/m². The prefactor 4π(a/π)³/2 = 4.22×10⁻⁸ s³/m³. The interval is narrow, so take v = 200 m/s and Δv = 10 m/s. Then av² = 0.282 and e^(−0.282) = 0.754, giving f(200) = 4.22×10⁻⁸ × (200)² × 0.754 = 1.27×10⁻³ s/m. ΔN = N f Δv = 1.88×10²⁴ × 1.27×10⁻³ × 10 = 2.4×10²² molecules. Numerical integration over 195-205 m/s gives the same value, 2.40×10²². This is about 1.3% of the gas. The most probable speed is √(2k_BT/m) ≈ 377 m/s, so the interval lies below the peak.