---
title: "Field from potential V = a + bx²"
subject: "Engineering Physics"
chapter: 7
key: "semester_1/engineering_physics:potential-region-x-squared-field"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/potential-region-x-squared-field
site: BCT Engineering
---

# Field from potential V = a + bx²

Engineering Physics · Chapter 7: Electrostatics

- Asked in: 2072 Kartik, 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-region-x-squared-field
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-7

## Question

The potential in a region between x = 0m and x = 6m is V = a + bx² where a = 10 and b = −7V/m. Determine (i) the potentials at x = 0m, 3m and 6m and (ii) the magnitude and direction of electric fields at x = 0m, 3m and 6m.

## Short answer

Potentials: at x = 0 m, 10 V; at 3 m, −53 V; at 6 m, −242 V. E = −dV/dx = 14x; magnitude and direction: at 0 m, 0; at 3 m, 42 V/m (+x); at 6 m, 84 V/m (+x).

## Answer

Given V = a + bx² with a = 10 V and b = −7 V/m², so V = 10 − 7x². (b is given in V/m; dimensionally it is V/m².)

#### (i) Potentials
- x = 0: V = 10 − 0 = **10 V**
- x = 3 m: V = 10 − 7(9) = 10 − 63 = **−53 V**
- x = 6 m: V = 10 − 7(36) = 10 − 252 = **−242 V**

#### (ii) Electric fields
E = −dV/dx = −d(10 − 7x²)/dx = 14x (V/m)

| x (m) | E = 14x (V/m) | Direction |
|---|---|---|
| 0 | 0 | none |
| 3 | 42 | +x |
| 6 | 84 | +x |

The field is positive, so it points along +x, the direction in which the potential decreases, as expected.

**Answer: V = 10 V, −53 V, −242 V; E = 0, 42 V/m, 84 V/m (along +x) at x = 0, 3 m, 6 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 = 10 - 7x^2?
   A. E = +dV/dx = -14x
   B. E = V/x
   C. E = -dV/dx = 14x
   D. E = -7x^2
   Answer: C. E is the negative gradient of potential. (Key point: Field from potential)

2. What is the potential at x = 3 m?
   A. -63 V
   B. -53 V
   C. -21 V
   D. +53 V
   Answer: B. V = 10 - 7(9) = -53 V. (Key point: Potential values)

3. Why does E point along +x at x = 3 and 6 m?
   A. E = 14x is positive, and E points in the direction in which V decreases
   B. V increases along +x, and E points toward higher V
   C. E = 0 everywhere
   D. E always points along -x
   Answer: A. V falls from 10 to -242 V, and E points toward lower potential. (Key point: Direction)

4. What is the field at x = 0?
   A. 14 V/m
   B. 10 V/m
   C. -7 V/m
   D. Zero, since E = 14x
   Answer: D. Substituting x = 0 gives zero even though V = 10 V. (Key point: Field at x = 0)

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