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Field of short linear quadrupole on axis

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.

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

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.

  +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 ]

  1. 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

  1. So

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

  1. 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).

✓ Answer checked by Claude Opus 5.5 (high effort)

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