---
title: "Field components from potential V = 15x−3x²y+12yz²"
subject: "Engineering Physics"
chapter: 7
key: "semester_1/engineering_physics:potential-xyz-field-components"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/potential-xyz-field-components
site: BCT Engineering
---

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

Engineering Physics · Chapter 7: Electrostatics

- Asked in: 2072 Chaitra, 5 marks
- Video: [How to get electric potential from electric field](https://www.youtube.com/watch?v=Ls8FG-Rj9Ts) — Physics with Professor Matt Anderson
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/potential-xyz-field-components
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-7

## Question

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?

## 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.

## Answer

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.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is E found from V(x, y, z)?
   A. E = +∇V
   B. E = V/r
   C. E = -∇V, so Ex = -∂V/∂x etc.
   D. E = ∫V dx
   Answer: C. The field is the negative gradient of potential. (Key point: Relation)

2. What is Ex for V = 15x - 3x^2 y + 12 y z^2?
   A. 15 - 6xy
   B. -15 + 6xy
   C. -15 - 6xy
   D. -3x^2 + 12z^2
   Answer: B. ∂V/∂x = 15 - 6xy; Ex is its negative. (Key point: Ex)

3. What is Ey at P(1, 0, -2)?
   A. -45 V/m
   B. +45 V/m
   C. -15 V/m
   D. 0
   Answer: A. Ey = 3x^2 - 12z^2 = 3 - 48 = -45. (Key point: Ey at P)

4. What is |E| at P?
   A. 60 V/m
   B. 45 V/m
   C. 15 V/m
   D. About 47.4 V/m
   Answer: D. |E| = sqrt(15^2 + 45^2 + 0^2) = sqrt(2250) ≈ 47.4. (Key point: Magnitude)

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