Given V = 2x²y − 5z, find fields

Given V = 2x²y − 5z (volts) and the point P(−4, 3, 6). Use ε₀ = 8.854×10⁻¹² F/m.

(i) V at P

V = 2(−4)²(3) − 5(6) = 2(16)(3) − 30 = 96 − 30 = 66 V

(ii) E = −∇V

∂V/∂x = 4xy  = 4(−4)(3) = −48
∂V/∂y = 2x²  = 2(16)    = 32
∂V/∂z = −5

E = −(−48 a_x + 32 a_y − 5 a_z) = 48 a_x − 32 a_y + 5 a_z  V/m

(iii) Unit vector a_E

|E| = √(48² + 32² + 5²) = √(2304 + 1024 + 25) = √3353 = 57.9 V/m
a_E = E/|E| = 0.829 a_x − 0.553 a_y + 0.086 a_z

(iv) D = ε₀E

D = 8.854×10⁻¹² (48 a_x − 32 a_y + 5 a_z)
  = 425 a_x − 283 a_y + 44.3 a_z  pC/m²

(v) Volume charge density

ρ_v = ∇·D = −ε₀∇²V.

∇²V = ∂²V/∂x² + ∂²V/∂y² + ∂²V/∂z² = 4y + 0 + 0 = 4(3) = 12
ρ_v = −ε₀(12) = −8.854×10⁻¹² × 12 = −1.06×10⁻¹⁰ C/m³ = −106 pC/m³

Answers: V = 66 V; E = 48 a_x − 32 a_y + 5 a_z V/m; a_E = 0.829 a_x − 0.553 a_y + 0.086 a_z; D = 425 a_x − 283 a_y + 44.3 a_z pC/m²; ρ_v = −106 pC/m³.