--- title: "Verify divergence theorem, D field" subject: "Electromagnetism" chapter: 1 key: "semester_3/electromagnetism:verify-divergence-theorem-dr-theta" source: https://www.bctengineer.com/q/semester_3/electromagnetism/verify-divergence-theorem-dr-theta site: BCT Engineering --- # Verify divergence theorem, D field Electromagnetism · Chapter 1: Introduction - Asked in: 2073 Shrawan, 8 marks - Video: [Electromagnetic fields 47: Divergence Theorem: Example (1)](https://www.youtube.com/watch?v=m_U36N_AV78) — Study with Dr. Hisham أدرس مع د. هشام - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electromagnetism/verify-divergence-theorem-dr-theta - Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-1 ## Question Given the flux density D = (2cosθ/r³) a_r + (sinθ/r³) a_θ C/m², evaluate both sides of the divergence theorem for the region defined by 1 < r < 2, 0 < θ < π/2, 0 < φ < π/2. ## Short answer The volume integral of ∇·D is zero, and the closed surface flux also sums to zero, verifying the divergence theorem. ## Answer **Divergence theorem:** The net outward flux of **D** through a closed surface equals the volume integral of ∇ · D over the enclosed volume: ∮ D · dS = ∫ (∇ · D) dv. **Given:** D = (2cosθ/r³) a_r + (sinθ/r³) a_θ C/m². Region: 1 < r < 2, 0 < θ < π/2, 0 < φ < π/2. #### Right side: volume integral ```text ∇ · D = (1/r²) ∂(r² D_r)/∂r + (1/(r sinθ)) ∂(sinθ D_θ)/∂θ + (1/(r sinθ)) ∂D_φ/∂φ ``` (D has no φ-component, so the last term is zero.) ```text (1/r²) ∂(r² · 2cosθ/r³)/∂r = (1/r²) ∂(2cosθ/r)/∂r = −2cosθ/r⁴ (1/(r sinθ)) ∂(sinθ · sinθ/r³)/∂θ = (1/(r sinθ)) (2 sinθ cosθ / r³) = 2cosθ/r⁴ ∇ · D = −2cosθ/r⁴ + 2cosθ/r⁴ = 0 ``` So ∫ (∇ · D) dv = **0**. #### Left side: closed surface integral The region has five faces. ```text 1. Outer, r = 2: dS = r² sinθ dθ dφ a_r, D_r = 2cosθ/8 Ψ = ∫₀^(π/2)∫₀^(π/2) (2cosθ/8)(4) sinθ dθ dφ = ∫ cosθ sinθ dθ · ∫dφ = (1/2)(π/2) = π/4 2. Inner, r = 1: dS = −r² sinθ dθ dφ a_r, D_r = 2cosθ Ψ = −∫∫ 2cosθ sinθ dθ dφ = −(1)(π/2) = −π/2 3. Face θ = π/2 (part of the xy-plane): outward normal is +a_θ, dS = r sinθ dr dφ a_θ D_θ = 1/r³ at θ = π/2: Ψ = ∫₀^(π/2)∫₁² (1/r³)(r) dr dφ = (1/2)(π/2) = π/4 4. θ = 0: this is only the z-axis (zero area): Ψ = 0 5. φ = 0 and φ = π/2 faces: D has no a_φ component, so D · dS = 0. ``` Total flux: ```text Ψ = π/4 − π/2 + π/4 + 0 = 0 ``` **Answer:** Both sides equal **0**, so the divergence theorem is verified. (D is solenoidal here: the same flux enters through r = 1 as leaves through r = 2 and θ = π/2.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is ∇·D for D = (2cosθ/r^3) a_r + (sinθ/r^3) a_θ? A. 4cosθ/r^4 B. 0, because the two terms cancel (-2cosθ/r^4 + 2cosθ/r^4) C. -2cosθ/r^4 D. 2cosθ/r^4 Answer: B. The radial term gives -2cosθ/r^4 and the θ term gives +2cosθ/r^4, so the sum is zero and the volume integral is zero. (Key point: Divergence value) 2. Which faces of the region contribute flux? A. Only r = 2 B. r = 2, r = 1 and θ = π/2 C. All five faces equally D. Only the φ faces Answer: B. The r = 2, r = 1 and θ = π/2 faces carry flux. The φ faces get none since D has no a_φ component, and θ = 0 is only an axis with no area. (Key point: Surface faces) 3. Why is the flux through r = 1 negative (-π/2)? A. D_r is negative there B. The outward normal there is -a_r, pointing toward the origin C. The area is negative D. Because r is smaller Answer: B. The outward normal of the region at the inner sphere is -a_r, so D·dS = -D_r dS and the flux is negative. (Key point: Inner surface sign) 4. What is the total outward flux? A. π/4 + π/2 + π/4 = π B. π/4 - π/2 + π/4 = 0, equal to the volume integral C. π/2 D. -π/2 Answer: B. Fluxes are π/4 (outer), -π/2 (inner) and π/4 (θ = π/2), summing to 0, which agrees with the divergence integral. 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