---
title: "Dipole potential and resultant field"
subject: "Engineering Physics"
chapter: 7
key: "semester_1/engineering_physics:dipole-potential-and-resultant-field"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/dipole-potential-and-resultant-field
site: BCT Engineering
---

# Dipole potential and resultant field

Engineering Physics · Chapter 7: Electrostatics

- Asked in: 2068 Shrawan, 5 marks
- Video: [Physics 36  The Electric Field (5 of 18) The Electric Dipole](https://www.youtube.com/watch?v=PSlnlXjaHA4) — Michel van Biezen
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/dipole-potential-and-resultant-field
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-7

## Question

For a given short electric dipole, show that the electric potential at any point at a distance r is V = P cos θ / r², where θ is the angle made by r to the dipole and P is its dipole moment. Using the above relation find an expression for the resultant electric intensity at that point.

## Short answer

The dipole potential is V = (1/4πε₀) p cosθ / r². The radial component is E_r = (1/4πε₀) 2p cosθ / r³, transverse E_θ = (1/4πε₀) p sinθ / r³; resultant magnitude |E| = (p/(4πε₀ r³)) √(1+3cos²θ) and direction tanα = (tanθ)/2.

## Answer

#### Potential of a short dipole
Charges +q and −q separated by 2a, p = 2aq. Point P at distance r from the centre O, with OP making angle θ with p, and r ≫ a.

```text
              P
             /|
        r2 /  | r1
         /  r |
   -q ---+----O----+--- +q
```

1. Distances: r₁ ≈ r − a cosθ (from +q), r₂ ≈ r + a cosθ (from −q).
2. V = (q/4πε0)(1/r₁ − 1/r₂) = (q/4πε0)(r₂ − r₁)/(r₁r₂) = (q/4πε0)(2a cosθ)/r²

**V = (1/4πε0) · p cosθ / r²**  (in the question's units, V = p cosθ/r²)

#### Resultant field from V
Use polar components: E_r = −∂V/∂r, E_θ = −(1/r) ∂V/∂θ.

1. E_r = −∂/∂r [p cosθ / (4πε0 r²)] = **2p cosθ / (4πε0 r³)**
2. E_θ = −(1/r) ∂/∂θ [p cosθ / (4πε0 r²)] = **p sinθ / (4πε0 r³)**
3. Resultant:

E = √(E_r² + E_θ²) = (p / 4πε0 r³) √(4cos²θ + sin²θ)

**E = (p / 4πε0 r³) √(1 + 3cos²θ)**

4. Its direction makes angle α with the radius vector, where tanα = E_θ/E_r = tanθ / 2.

Checks: θ = 0 gives E = 2p/(4πε0r³) (axial); θ = 90° gives E = p/(4πε0r³) (equatorial).

## Check yourself

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

1. For r ≫ a, the distance from +q to P is approximately
   A. r + a cosθ
   B. r − a cosθ
   C. r − a
   D. r cosθ
   Answer: B. The +q charge is nearer to P by a cosθ, so r1 ≈ r − a cosθ and r2 ≈ r + a cosθ. (Key point: Distance approximation)

2. The dipole potential is
   A. p cosθ/(4πε0 r^2)
   B. p sinθ/(4πε0 r^2)
   C. p cosθ/(4πε0 r^3)
   D. p/(4πε0 r)
   Answer: A. V = (q/4πε0)(r2−r1)/(r1r2) = p cosθ/(4πε0 r^2). (Key point: Potential)

3. E_r = −∂V/∂r gives
   A. p sinθ/(4πε0 r^3)
   B. p cosθ/(4πε0 r^3)
   C. 2p cosθ/(4πε0 r^3)
   D. p cosθ/(4πε0 r^2)
   Answer: C. Differentiating 1/r^2 gives 2/r^3, so E_r = 2p cosθ/(4πε0 r^3). (Key point: Radial component)

4. The resultant field magnitude is
   A. (p/4πε0 r^3) sqrt(1 + 3cos^2θ)
   B. (p/4πε0 r^3) sqrt(1 + 3sin^2θ)
   C. (p/4πε0 r^3)(2cosθ + sinθ)
   D. (2p/4πε0 r^3) cosθ
   Answer: A. sqrt(4cos^2θ + sin^2θ) = sqrt(1 + 3cos^2θ); this gives 2p/(4πε0 r^3) on axis and p/(4πε0 r^3) at θ = 90°. (Key point: Resultant)

## Related in Engineering Physics

- [Potential of a dipole at any point](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Adipole-potential-general-point) — Engineering Physics, chapter 7, asked 2×
- [Dipole field on equatorial line](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Adipole-equatorial-field) — Engineering Physics, chapter 7
- [Dipole potential; no work along bisector](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Adipole-potential-no-work-perpendicular) — Engineering Physics, chapter 7
- [Dipole field on axial line](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Adipole-axial-field-inverse-cube) — Engineering Physics, chapter 7
- [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×
- [Axial field of dipole and quadrupole](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Adipole-quadrupole-axial-field-conclusion) — Engineering Physics, chapter 7

## Related in other subjects

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- [Potential gradient gives electric field](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Apotential-gradient-gives-electric-field) — Electromagnetism, chapter 2
- [Dipole moment and Clausius-Mossotti equation](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Adipole-moment-clausius-mossotti) — Electrical Engineering Material, chapter 3, asked 2×
- [Retarded potential importance and infinitesimal dipole](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Aretarded-potential-importance-dipole) — Propagation and Antenna, chapter 1