Energy stored for given potential field

Given: Region 2 mm < r < 3 mm, 0 < θ < 90°, 0 < φ < 90°, free space (ε₀ = 8.854 × 10⁻¹² F/m). Energy from the field: W_E = (1/2) ε₀ ∫ |E|² dv, with E = −∇V and dv = r² sinθ dr dθ dφ.

(a) V = 200/r

E = −∇V = −(dV/dr) a_r = (200/r²) a_r,     E² = 4 × 10⁴ / r⁴

W_E = (1/2) ε₀ ∫₀^(π/2) ∫₀^(π/2) ∫_(0.002)^(0.003) (4×10⁴ / r⁴) r² sinθ dr dθ dφ
    = (1/2) ε₀ (4×10⁴) [ ∫ r⁻² dr ] [ ∫ sinθ dθ ] [ ∫ dφ ]

∫ r⁻² dr = 1/0.002 − 1/0.003 = 500 − 333.33 = 166.67
∫₀^(π/2) sinθ dθ = 1,    ∫₀^(π/2) dφ = π/2 = 1.5708

W_E = 0.5 (8.854×10⁻¹²)(4×10⁴)(166.67)(1)(1.5708)
    = 4.636 × 10⁻⁵ J

W_E = 46.4 μJ

(b) V = (300/r²) cosθ

E_r = −∂V/∂r = 600 cosθ / r³
E_θ = −(1/r) ∂V/∂θ = 300 sinθ / r³
E_φ = 0

E² = (3.6×10⁵ cos²θ + 9×10⁴ sin²θ) / r⁶

W_E = (1/2) ε₀ ∫ ∫ ∫ (3.6×10⁵ cos²θ + 9×10⁴ sin²θ) r⁻⁶ r² sinθ dr dθ dφ

Radial part: ∫ r⁻⁴ dr = (1/3)(r₁⁻³ − r₂⁻³) = (1/3)(1.25×10⁸ − 3.704×10⁷) = 2.932 × 10⁷

Angular part (θ): ∫₀^(π/2) (3.6×10⁵ cos²θ + 9×10⁴ sin²θ) sinθ dθ = 3.6×10⁵ (1/3) + 9×10⁴ (2/3) = 1.2×10⁵ + 0.6×10⁵ = 1.8×10⁵

Angular part (φ): π/2.

W_E = 0.5 (8.854×10⁻¹²)(1.8×10⁵)(1.5708)(2.932×10⁷)
    = 36.7 J

Answer: (a) W_E = 46.4 μJ; (b) W_E ≈ 36.7 J.

(The large value in (b) comes from the extremely strong field of this 1/r³ distribution at millimetre distances; such a potential would be unrealistic physically, but the working above follows the given V.)