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Charge density and total charge from D

Short answer

ρ_v = ∇·D = 5 sinθ cosφ / r² C/m³. Total charge Q = ∮ D·dS = 0 because the φ-integral of cosφ over 0 to 2π vanishes. At surface r=2, D = 2.5 sinθ cosφ a_r C/m².

Given the field D = [5 sin(θ) cos(φ)/r] a_r C/m², find: (a) the volume charge density; (b) the total charge contained in the region r < 2 m; (c) the value of D at the surface r = 2.

Given D = (5 sinθ cosφ / r) a_r C/m². Only D_r exists, and D_r depends on r, θ and φ. Use Gauss's law in differential form, ρ_v = ∇·D.

(a) Volume charge density

∇·D = (1/r²) ∂(r² D_r)/∂r
    = (1/r²) ∂(5 r sinθ cosφ)/∂r
    = (1/r²)(5 sinθ cosφ)
ρ_v = 5 sinθ cosφ / r²   C/m³

(The other two terms of the divergence vanish because D_θ = D_φ = 0.)

(b) Total charge in r < 2 m

Using Gauss's law Q = ∮ D·dS over the sphere r = 2 (dS = r² sinθ dθ dφ a_r):

Q = ∫₀²ᵖⁱ ∫₀ᵖⁱ (5 sinθ cosφ / 2)(4 sinθ) dθ dφ
  = 10 ∫₀ᵖⁱ sin²θ dθ · ∫₀²ᵖⁱ cosφ dφ
  = 10 (π/2)(0)
  = 0

Check with the volume integral: Q = ∫ρ_v dv = ∫∫∫ (5 sinθ cosφ/r²) r² sinθ dr dθ dφ. The φ-integral of cosφ over 0 to 2π is zero, so Q = 0. The positive charge for −π/2 < φ < π/2 is cancelled by equal negative charge elsewhere.

Q = 0 C

(c) D at r = 2

D = 5 sinθ cosφ / 2  a_r = 2.5 sinθ cosφ a_r  C/m²

✓ Answer checked by Claude Opus 5.5 (high effort)

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