---
title: "Field and electron acceleration from potential"
subject: "Engineering Physics"
chapter: 7
key: "semester_1/engineering_physics:potential-field-point-electron-acceleration"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/potential-field-point-electron-acceleration
site: BCT Engineering
---

# Field and electron acceleration from potential

Engineering Physics · Chapter 7: Electrostatics

- Asked in: 2069 Chaitra, 5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/potential-field-point-electron-acceleration
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-7

## Question

What is the magnitude of the electric field at the point (3,2) m if the electric potential is given by V = 2x + 5xy + 3y² volts. What acceleration does an electron experience in the x-direction?

## Short answer

The electric field components are E_x = −12 V/m, E_y = −27 V/m; magnitude |E| = √(144+729) ≈ 29.5 V/m. An electron experiences force opposite to E, so acceleration along +x is a_x = e|E_x|/m ≈ 2.11×10¹² m/s².

## Answer

Given V = 2x + 5xy + 3y² (volts), point (3, 2) m.

#### Field components
E = −∇V.
1. Eₓ = −∂V/∂x = −(2 + 5y) = −(2 + 10) = −12 V/m
2. E_y = −∂V/∂y = −(5x + 6y) = −(15 + 12) = −27 V/m

|E| = √(Eₓ² + E_y²) = √(144 + 729) = √873 = 29.5 V/m

Direction: tanφ = E_y/Eₓ = 27/12, so φ = 66.0° below the −x axis (components both negative).

#### Acceleration of an electron along x
Force on the electron: F = −eE, so Fₓ = −e Eₓ = −(1.6 × 10⁻¹⁹)(−12) = +1.92 × 10⁻¹⁸ N.

aₓ = Fₓ/m = 1.92 × 10⁻¹⁸ / 9.11 × 10⁻³¹ = 2.11 × 10¹² m/s²

**Answer: |E| ≈ 29.5 V/m; the electron's acceleration in the x-direction is ≈ 2.11 × 10¹² m/s², along +x (opposite to Eₓ because the charge is negative).**

## Check yourself

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

1. How is E found from V?
   A. E = −∇V (negative partial derivatives)
   B. E = ∇V
   C. E = V/r
   D. E = ∫V dx
   Answer: A. The field is the negative gradient of potential: Ex = −∂V/∂x, Ey = −∂V/∂y. (Key point: Field from potential)

2. What is Ex at (3, 2) for V = 2x + 5xy + 3y^2?
   A. −12 V/m
   B. +12 V/m
   C. −27 V/m
   D. −17 V/m
   Answer: A. Ex = −(2 + 5y) = −(2 + 10) = −12 V/m. (Key point: Ex value)

3. What is |E| at (3, 2)?
   A. 15 V/m
   B. 39 V/m
   C. 29.5 V/m
   D. 873 V/m
   Answer: C. Ey = −(5x + 6y) = −27, so |E| = sqrt(144 + 729) = 29.5 V/m. (Key point: Magnitude)

4. Why is the electron's x-acceleration along +x although Ex is negative?
   A. Ex is actually positive
   B. The electron's charge is negative, so F = −eE
   C. Electrons ignore electric fields in x
   D. Acceleration is always along +x
   Answer: B. For a negative charge F is opposite to E, giving ax = 2.11 × 10^12 m/s^2 along +x. (Key point: Electron direction)

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