---
title: "Field of short linear quadrupole on axis"
subject: "Engineering Physics"
chapter: 7
key: "semester_1/engineering_physics:quadrupole-axis-field"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/quadrupole-axis-field
site: BCT Engineering
---

# Field of short linear quadrupole on axis

Engineering Physics · Chapter 7: Electrostatics

- Asked in: 2067 Asar, 5 marks
- Video: [Electric field intensity due to quadrupole at axial line|Physics BE TU KU PoU Purbanchal University](https://www.youtube.com/watch?v=jpFksUrLbMk) — Civil Engineering Solution Channel
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/quadrupole-axis-field
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-7

## Question

Derive an expression for the electric field at any point on the axis of a short linear quadrupole.

## Short answer

For a short linear quadrupole (+q, −2q, +q) on axis, the net field is E = (1/4πε₀)·6qa²/r⁴ = (1/4πε₀)·3Q/r⁴, with quadrupole moment Q = 2qa². The field decays as 1/r⁴.

## Answer

A **linear quadrupole** is two equal and opposite dipoles placed end to end: charge +q at x = −a, −2q at the origin, and +q at x = +a. The net charge and net dipole moment are zero.

```text
  +q         -2q         +q                 P
  (+)------- (-) ------- (+) ---------------.
  -a          0          +a                 r (on axis)
```

#### Field at P on the axis, distance r from the centre (r > a)
1. Distances of P from the charges: r − a (from +q at +a), r (from −2q), r + a (from +q at −a).
2. All fields lie along the axis, so add them as scalars:

E = (1/4πε0) [ q/(r − a)² − 2q/r² + q/(r + a)² ]

3. Combine terms:

E = (q/4πε0) · [ r²(r + a)² − 2(r² − a²)² + r²(r − a)² ] / [ r²(r² − a²)² ]

The numerator: r²[(r+a)² + (r−a)²] − 2(r² − a²)² = r²(2r² + 2a²) − 2(r⁴ − 2a²r² + a⁴) = 6a²r² − 2a⁴

4. So

E = (q/4πε0) · (6a²r² − 2a⁴) / [ r²(r² − a²)² ]

5. For a short quadrupole (a ≪ r), numerator ≈ 6a²r² and denominator ≈ r⁶:

E = (q/4πε0) · 6a²/r⁴

With quadrupole moment Q = 2qa²:

**E = (1/4πε0) · 3Q / r⁴**

The field of a quadrupole falls as 1/r⁴, faster than a dipole (1/r³) or a point charge (1/r²).

## Check yourself

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

1. A linear quadrupole consists of
   A. +q, −2q, +q equally spaced on a line
   B. +q and −q only
   C. +2q, −q, −q
   D. +q, +q, −2q off-axis
   Answer: A. Two opposite dipoles end to end give +q, −2q, +q with zero net charge and dipole moment. (Key point: Definition)

2. Which expression is the on-axis field for r > a?
   A. (1/4πε0)[q/(r−a)^2 − 2q/r^2 + q/(r+a)^2]
   B. (1/4πε0)[q/(r−a) − 2q/r + q/(r+a)]
   C. (1/4πε0)[q/(r−a)^2 + 2q/r^2 + q/(r+a)^2]
   D. (1/4πε0)[q/r^2 − 2q/r^2 + q/r^2]
   Answer: A. Collinear fields add as scalars with the −2q term opposite in sign. (Key point: Setup of field)

3. For a ≪ r the numerator and denominator become
   A. 2a^4 and r^4
   B. 6a^2 r^2 and r^6
   C. 6a^2 and r^2
   D. a^2 and r^4
   Answer: B. The numerator 6a^2r^2 − 2a^4 ≈ 6a^2r^2, and the denominator ≈ r^6. (Key point: Short approximation)

4. How does the quadrupole field fall with r?
   A. 1/r^2
   B. 1/r^3
   C. 1/r^4
   D. 1/r
   Answer: C. E = 3Q/(4πε0 r^4), faster than a dipole (1/r^3) or point charge (1/r^2). (Key point: Fall-off)

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