---
title: "Axial dipole field twice equatorial"
subject: "Engineering Physics"
chapter: 7
key: "semester_1/engineering_physics:dipole-axial-twice-equatorial"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/dipole-axial-twice-equatorial
site: BCT Engineering
---

# Axial dipole field twice equatorial

Engineering Physics · Chapter 7: Electrostatics

- Asked in: 2073 Chaitra, 5 marks; 2070 Chaitra, 5 marks; 2069 Asar, 5 marks (asked 3 times)
- Video: [electric field of dipole for equatorial, axial and general point.](https://www.youtube.com/watch?v=yTAUbhkA8q0) — physics_with_ Nihal_singh
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/dipole-axial-twice-equatorial
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-7

## Question

Prove that the electric field due to a short dipole at a point on the axial line is twice that on the equatorial line.

## Short answer

For a short dipole (r ≫ a), axial field magnitude is 2kp/r³ and equatorial field magnitude is kp/r³. Hence the axial field is twice the equatorial field.

## Answer

A **short dipole** has charges +q and −q separated by 2a, with dipole moment p = 2qa. We find the field at distance r from the centre (r ≫ a), where k = 1/4πε₀.

#### Field on the axial line
```text
   -q          +q           P
    o----------o-----------x---> 
   -a    0     +a     r
```
1. P is on the line joining the charges. Distances from +q and −q are (r − a) and (r + a).
2. Field of +q (away from the charge, along p): kq/(r − a)². Field of −q (towards the charge, opposite to p): kq/(r + a)².
3. E_axial = kq [1/(r − a)² − 1/(r + a)²] = kq · 4ar / (r² − a²)².
4. For r ≫ a: E_axial ≈ 4kqa / r³ = **2kp / r³**, along p.

#### Field on the equatorial line
```text
          P
         /|\
      r1/ | \r2
       /  |r \
   -q o---+---o +q
```
1. P lies on the perpendicular bisector, at a distance r from the centre. Each charge is at distance √(r² + a²).
2. Magnitude of each field: E₁ = E₂ = kq / (r² + a²).
3. Components perpendicular to the dipole axis (along OP) cancel. The components parallel to the axis add up, each with cos θ = a/√(r² + a²), directed from +q towards −q, i.e. opposite to p:
 E_eq = 2 kq/(r² + a²) · a/√(r² + a²) = 2kqa / (r² + a²)^(3/2) = kp / (r² + a²)^(3/2)
4. For r ≫ a: E_eq ≈ **kp / r³**, opposite to p.

#### Ratio
```text
E_axial / E_equatorial = (2kp/r³) / (kp/r³) = 2
```
**Hence the axial field of a short dipole is twice the equatorial field in magnitude (the two have opposite directions relative to p).**

## Check yourself

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

1. On the axial line, why are the fields of +q and -q subtracted?
   A. They are perpendicular so they add in quadrature
   B. They point the same way, so they add
   C. The two fields are equal and cancel completely
   D. They are collinear and opposite in direction; +q is nearer so its field is larger
   Answer: D. E = kq[1/(r - a)^2 - 1/(r + a)^2], net along p. (Key point: Axial combination)

2. On the equatorial line, which components of the two fields cancel and which add?
   A. Components parallel to the axis cancel; those along OP add
   B. All components cancel
   C. Components along OP cancel; components parallel to the dipole axis add
   D. All components add
   Answer: C. The charges are symmetric about the bisector, so the components along OP are opposite while the axis components add. (Key point: Equatorial combination)

3. For a short dipole, what is E_axial / E_equatorial at the same distance?
   A. 1/2
   B. 2 (axial 2kp/r^3, equatorial kp/r^3)
   C. 1
   D. 4
   Answer: B. E_axial ≈ 2kp/r^3 and E_eq ≈ kp/r^3. (Key point: The ratio)

4. How do the directions of the two fields compare relative to p?
   A. Axial field is along p; equatorial field is opposite to p
   B. Both are along p
   C. Both are opposite to p
   D. Axial is opposite to p; equatorial is along p
   Answer: A. On the axis the nearer +q dominates (along p); on the bisector the net field points from +q towards -q, opposite to p. (Key point: Directions)

## Related in Engineering Physics

- [Dipole field on equatorial line](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Adipole-equatorial-field) — 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
- [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
- [Field of long charged rod, equatorial](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Acharged-rod-field-potential-continuous) — Engineering Physics, chapter 7
- [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 potential and resultant field](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Adipole-potential-and-resultant-field) — 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×
- [Define electric dipole and polarization](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Adefine-dipole-and-polarization) — Electromagnetism, chapter 2
- [E due to line charge using Gauss's law](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Aline-charge-field-gauss-law) — Electromagnetism, chapter 2, asked 3×