---
title: "Potential gradient gives electric field"
subject: "Electromagnetism"
chapter: 2
key: "semester_3/electromagnetism:potential-gradient-gives-electric-field"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/potential-gradient-gives-electric-field
site: BCT Engineering
---

# Potential gradient gives electric field

Electromagnetism · Chapter 2: Electric field

- Asked in: 2067 Shrawan, 5 marks
- Video: [Electric Potential: Basics, Definition, and Its Relationship with Electric Field](https://www.youtube.com/watch?v=zl3Uj18eHek) — Engineering Funda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/potential-gradient-gives-electric-field
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-2

## Question

Deduce how the potential gradient can be used to determine the electric field intensity.

## Short answer

The electric field E is the negative gradient of the potential V: E = −∇V. E points in the direction of the fastest decrease of potential and is perpendicular to equipotential surfaces.

## Answer

**Idea.** The potential difference between two nearby points gives the work per unit charge. Since the electric field is the force per unit charge, E is the (negative) space rate of change of V.

#### Derivation
The potential difference V_AB = −∫E·dL (from B to A). For a small displacement dL, the change in potential is

```text
dV = −E·dL                                   (1)
```
In Cartesian coordinates, V = V(x, y, z), so by the chain rule

```text
dV = (∂V/∂x)dx + (∂V/∂y)dy + (∂V/∂z)dz       (2)
```
and with dL = dx a_x + dy a_y + dz a_z,

```text
E·dL = E_x dx + E_y dy + E_z dz              (3)
```
Comparing (1), (2), (3) for arbitrary dx, dy, dz:

```text
E_x = −∂V/∂x ,  E_y = −∂V/∂y ,  E_z = −∂V/∂z

E = −(∂V/∂x a_x + ∂V/∂y a_y + ∂V/∂z a_z) = −∇V
```

#### Interpretation
- The gradient ∇V is the vector pointing in the direction of the maximum increase of V, with magnitude equal to the maximum rate of change dV/dn. E is the negative of it, so **E points in the direction of the fastest decrease of potential**, from high to low potential.
- E is perpendicular to the equipotential surfaces (since along them dV = 0, so E·dL = 0).
- In other coordinates: cylindrical E = −(∂V/∂ρ a_ρ + (1/ρ)∂V/∂φ a_φ + ∂V/∂z a_z); spherical E = −(∂V/∂r a_r + (1/r)∂V/∂θ a_θ + (1/(r sinθ))∂V/∂φ a_φ).

#### Example
For a point charge, V = Q/(4πε₀r), so E = −∂V/∂r a_r = Q/(4πε₀r²) a_r, which is Coulomb's law.

## Check yourself

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

1. Which relation is compared with the chain-rule form of dV?
   A. dV = E·dL
   B. dV = −E·dL
   C. dV = ∇×E
   D. dV = −∮E·dL
   Answer: B. dV = −E·dL; comparing with dV = (∂V/∂x)dx + ... gives E components. (Key point: Starting relation)

2. The comparison gives:
   A. E = ∇V
   B. E = −∇V
   C. E = ∇×V
   D. E = −∇·V
   Answer: B. E_x = −∂V/∂x etc., so E = −∇V. (Key point: Result)

3. E points in the direction of:
   A. Maximum increase of V
   B. Fastest decrease of V, perpendicular to equipotentials
   C. Along the equipotential
   D. Zero change
   Answer: B. ∇V points toward increasing V, so E = −∇V points toward fastest decrease and is perpendicular to equipotential surfaces. (Key point: Direction of E)

4. For V = Q/(4πε0 r), E = −∂V/∂r a_r gives:
   A. Q/(4πε0 r) a_r
   B. Q/(4πε0 r²) a_r
   C. −Q/(4πε0 r²) a_r
   D. Q r/(4πε0)
   Answer: B. Differentiating 1/r gives −1/r², and the minus sign cancels it, yielding Coulomb's law. (Key point: Example)

## Related in Electromagnetism

- [Electric dipole potential and field](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Adipole-potential-and-field-derive) — Electromagnetism, chapter 2, asked 4×
- [Energy stored for given potential field](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Aenergy-stored-potential-field) — Electromagnetism, chapter 2
- [H of infinite filament via vector potential](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Ah-infinite-filament-vector-potential) — Electromagnetism, chapter 3
- [Field between coaxial conductors](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Acoax-cylinders-field-and-line-charge) — Electromagnetism, chapter 2
- [Parallel plate sheets, field in regions](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Aparallel-plate-sheets-three-regions) — Electromagnetism, chapter 2
- [Potential V=100xz/(x²+4), charge on conductor](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Apotential-v-100xz-conductor-charge) — Electromagnetism, chapter 2

## Related in other subjects

- [Ring on axis: potential then field](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Aring-axial-potential-then-field) — Engineering Physics, chapter 7, asked 2×
- [Dipole potential and resultant field](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Adipole-potential-and-resultant-field) — Engineering Physics, chapter 7
- [Potential of electric quadrupole on axis](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Aquadrupole-axis-potential) — Engineering Physics, chapter 7, asked 3×
- [pn junction 10¹⁷/10¹⁶: potential, widths, field](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Apn-1e17-1e16-depletion) — Electrical Engineering Material, chapter 6, asked 3×