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Field components from potential V = 15x−3x²y+12yz²

Short answer

E_x = −15+6xy, E_y = 3x²−12z², E_z = −24yz V/m. At P(1,0,−2): E_x = −15, E_y = −45, E_z = 0 V/m; magnitude |E| = √2250 ≈ 47.4 V/m.

Over a certain region of space the electric potential is V = 15x − 3x²y + 12yz². Find the expression for the x, y and z components of the electric field over this region. What is the magnitude of the field at the point P that has coordinates (1, 0, −2) m?

The field is the negative gradient of the potential: E = −∇V, so Eₓ = −∂V/∂x, E_y = −∂V/∂y, E_z = −∂V/∂z.

V = 15x − 3x²y + 12yz²

Components

  1. Eₓ = −∂V/∂x = −(15 − 6xy) = −15 + 6xy
  2. E_y = −∂V/∂y = −(−3x² + 12z²) = 3x² − 12z²
  3. E_z = −∂V/∂z = −(24yz) = −24yz

(All in V/m.)

At P(1, 0, −2) m

  • Eₓ = −15 + 6(1)(0) = −15 V/m
  • E_y = 3(1)² − 12(−2)² = 3 − 48 = −45 V/m
  • E_z = −24(0)(−2) = 0

|E| = √[(−15)² + (−45)² + 0²] = √(225 + 2025) = √2250 = 47.4 V/m

Answer: E = (−15 + 6xy) i + (3x² − 12z²) j − 24yz k; at P, E = −15 i − 45 j V/m, |E| ≈ 47.4 V/m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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