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Potential of electric quadrupole on axis

Short answer

An electric quadrupole is a charge arrangement with zero net charge and dipole moment, e.g., +q, −2q, +q on a line. Its axial potential at distance r from centre is V = 2qa²/[4πε₀ r(r²−a²)], which ≈ Q/(4πε₀ r³) for large r, where quadrupole moment Q = 2qa².

What is an electric quadrupole? Calculate the electric potential of a linear quadrupole (of separation 2a) at an axial distance r from its center.

Electric quadrupole

An electric quadrupole is a system of charges with zero net charge and zero net dipole moment. A linear quadrupole consists of two equal and opposite dipoles placed end to end: charges +q, −2q, +q on a line with equal spacing.

   +q          -2q          +q
    o-----a-----o-----a-----o----------> P
   x=-a         0          x=+a     r

Potential on the axis

Take the −2q at the origin and +q at x = ± a (distance between the two end charges = 2a). Point P is on the axis at distance r from the centre (r > a).

  1. The distances of P from the charges are (r − a), r and (r + a).
  2. By superposition: V = (1/4πε₀) [ q/(r − a) − 2q/r + q/(r + a) ]
  3. Combine the first and third terms: q/(r − a) + q/(r + a) = 2qr / (r² − a²)
  4. V = (q/4πε₀) [ 2r/(r² − a²) − 2/r ] = (q/4πε₀) · [2r² − 2(r² − a²)] / [r (r² − a²)] = (q/4πε₀) · 2a² / [r (r² − a²)]

V = 2qa² / [4πε₀ r (r² − a²)]

  1. For r ≫ a, r² − a² ≈ r²:

V ≈ 2qa² / (4πε₀ r³) = Q / (4πε₀ r³), where Q = 2qa² is the quadrupole moment.

The potential falls off as 1/r³, faster than a dipole (1/r²) and a monopole (1/r).

✓ Answer checked by Claude Opus 5.5 (high effort)

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