Electromagnetic Theory (v2) #30 | Problem Solving #07: Divergence Theorem & Gauss's Law

Instructor: Dr. Süleyman Burak ÇELİK

Seventh problem-solving episode of the EMT (v2) series. Today we go from local to global. The divergence theorem says the flux of any vector field through a closed surface S equals the volume integral of its divergence over the region V enclosed by S. ∮ A · dS = ∫ ∇·A dV. The intuition is beautiful: cut V into millions of tiny boxes, every interior face is shared with a neighbor and cancels in the sum, only the outermost surfaces survive — that is exactly the boundary S. So summing every local source equals the total flux through the outer skin. Cube example — clean verification. Take the unit cube with the polynomial field we have been carrying since PS05: A = x²y x̂ + y²z ŷ + z²x ẑ. Volume side: ∫(2xy+2yz+2zx) dV over [0,1]³. Each of the three terms separates into a product of three single integrals — 2 · (1/2) · (1/2) · 1 = 1/2 per term — and three terms give 3/2. Surface side: six faces, but on x=0, y=0, z=0 the field component is zero, so those vanish. On x=1: ∫y dy dz = 1/2. On y=1: ∫z dz dx = 1/2. On z=1: ∫x dx dy = 1/2. Total: 3/2. Both sides match — the theorem is verified, no tricks. Sphere example — the lesson lives in the singularity. Same field as PS06: A = (1/r²) r̂. Compute the divergence-theorem integrand inside a sphere of radius R. Off the origin, ∇·A = 0 (PS05), so a naive volume integral gives zero. Surface integral: A · dS = (1/R²) · R² sinθ dθ dφ = sinθ dθ dφ. Integrate sinθ from 0 to π and dφ from 0 to 2π: 2 · 2π = 4π. Mismatch! 0 ≠ 4π. What broke? Nothing — the divergence is not zero everywhere; it is a delta function at the origin: ∇·(r̂/r²) = 4π δ³(r). Integrate that over V and you get 4π — exactly matching the surface side. The theorem is restored when we are honest about the singularity. And here is the payoff. Multiply both sides by q/(4π ε₀). The inverse-square radial field becomes the electric field of a point charge q. The surface integral becomes the flux of E. The volume side becomes Q_enc/ε₀. That is Gauss's law in integral form. Every Gaussian surface calculation in electrostatics — cylindrical shell, charged plane, coaxial cable — rides on this theorem plus the delta-function resolution of the singularity at the source. Topics: 0:00 Cover 0:03 Theorem statement + interior cancellation intuition 1:39 Cube worked example: both sides equal 3/2 4:23 Sphere worked example: 0 vs 4π → delta → Gauss's law 7:05 Summary Key equations: • Divergence theorem: ∮ A · dS = ∫ ∇·A dV • ∇·(r̂/r²) = 4π δ³(r) (singular distributional identity) • Gauss's law: ∮ E · dS = Q_enc / ε₀ Answers: Cube ∮ = ∫ = 3/2. Sphere ∮ = 4π = ∫ (with delta at origin). → Gauss's law for any closed surface. Reference: Sadiku, "Elements of Electromagnetics" 7th Ed, Chapter 3 (divergence theorem); Griffiths Ch 1.