---
title: "Potential of a dipole at any point"
subject: "Engineering Physics"
chapter: 7
key: "semester_1/engineering_physics:dipole-potential-general-point"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/dipole-potential-general-point
site: BCT Engineering
---

# Potential of a dipole at any point

Engineering Physics · Chapter 7: Electrostatics

- Asked in: 2072 Chaitra, 5 marks; 2071 Chaitra, 5 marks (asked 2 times)
- Video: [Potential due to dipole (logical derivation) | Electric potential & cap. | Physics | Khan Academy](https://www.youtube.com/watch?v=utY112E_NIc) — Khan Academy India - English
- 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-general-point
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-7

## Question

Define electric dipole. Determine the electric potential at any point at a distance r from the center of the dipole (also find the potential on the axial line).

## Short answer

An electric dipole is a pair of equal and opposite charges ±q separated by 2a, with dipole moment p = 2qa from −q to +q. For r ≫ a, potential V = p cosθ/(4πε₀ r²). On the axial line, V = ±p/(4πε₀ r²).

## Answer

#### Electric dipole
An electric dipole is a pair of equal and opposite point charges +q and −q separated by a small distance 2a. Its dipole moment is p = q(2a), a vector directed from −q to +q.

#### Potential at a general point
```text
            P
           /|\
        r2/ | \r1
         /  |r  \
       -q---O---+q  
        <-a-> <-a->
        angle θ between OP and p
```
Let P be at distance r from the centre O, with OP making an angle θ with the dipole axis, and r ≫ a. Let k = 1/4πε₀.

1. Distance of P from +q: r₁; from −q: r₂.
2. Potential at P: V = kq (1/r₁ − 1/r₂) = kq (r₂ − r₁) / (r₁ r₂).
3. For r ≫ a, drop a perpendicular from each charge to line OP:
 r₁ ≈ r − a cos θ, r₂ ≈ r + a cos θ
 so r₂ − r₁ ≈ 2a cos θ and r₁r₂ = r² − a² cos² θ ≈ r².
4. Therefore V = kq (2a cos θ) / r², and with p = 2qa:

 **V = p cos θ / (4πε₀ r²) = p·r̂ / (4πε₀ r²)**

#### Special cases
- **Axial line (θ = 0° or 180°):** V = ± p / (4πε₀ r²). Exactly (without approximation), V = q·2a / [4πε₀ (r² − a²)] on the axis on the +q side.
- **Equatorial line (θ = 90°):** V = 0, so the potential is zero everywhere on the perpendicular bisector.

The dipole potential falls as 1/r², unlike the potential of a single charge (1/r).

## Check yourself

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

1. What is the dipole moment of charges +q and -q separated by 2a?
   A. p = 2qa, directed from +q to -q
   B. p = qa, from -q to +q
   C. p = 0
   D. p = 2qa, directed from -q to +q
   Answer: D. p = q(2a), a vector pointing from the negative to the positive charge. (Key point: Dipole moment)

2. For r >> a, which approximations lead to the dipole potential?
   A. r2 - r1 ≈ 2a sin θ and r1 r2 ≈ r^2
   B. r2 - r1 ≈ 0
   C. r2 - r1 ≈ 2a cos θ and r1 r2 ≈ r^2
   D. r2 - r1 ≈ a and r1 r2 ≈ a^2
   Answer: C. Perpendicular drops from each charge to OP give r1 ≈ r - a cos θ and r2 ≈ r + a cos θ. (Key point: Path difference)

3. What is the potential at distance r making angle θ with p?
   A. V = p cos θ/(4πε0 r)
   B. V = p cos θ/(4πε0 r^2)
   C. V = p sin θ/(4πε0 r^2)
   D. V = p/(4πε0 r^3)
   Answer: B. V = kq(2a cos θ)/r^2 with p = 2qa. (Key point: General result)

4. What is the potential on the equatorial line, and how does the dipole potential fall off?
   A. Zero on the equatorial line; falls as 1/r^2
   B. Maximum on the equatorial line; falls as 1/r
   C. Zero on the axial line; falls as 1/r^2
   D. Zero on the equatorial line; falls as 1/r
   Answer: A. At θ = 90°, cos θ = 0 so V = 0, and the 1/r^2 dependence is faster than a single charge's 1/r. (Key point: Special cases)

## Related in Engineering Physics

- [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
- [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
- [Axial dipole field twice equatorial](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Adipole-axial-twice-equatorial) — Engineering Physics, chapter 7, asked 3×
- [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×
- [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

## Related in other subjects

- [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×
- [Potential from point and line charge](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Apotential-point-and-line-charge) — Electromagnetism, chapter 2
- [Distance of point from plane parallel to line](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Adistance-point-from-plane-parallel) — Mathematics, chapter 3
- [Define electric dipole and polarization](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Adefine-dipole-and-polarization) — Electromagnetism, chapter 2