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Axial field of dipole and quadrupole

Short answer

On the axis, dipole field E = (1/4πε₀)·2p/r³; linear quadrupole field E = (1/4πε₀)·3Q/r⁴ with Q=2qa². Conclusion: field decays faster for higher multipoles — point charge 1/r², dipole 1/r³, quadrupole 1/r⁴.

Calculate the electric field at any point at an axial distance due to a dipole and a quadrupole. What conclusion can you draw from your results?

Dipole on its axis

Charges +q and −q separated by 2a, centre at origin; point P at distance r on the axis.

E = (q/4πε0)[1/(r − a)² − 1/(r + a)²] = (q/4πε0) · 4ar/(r² − a²)²

For r ≫ a, with p = 2aq:

E_dipole = (1/4πε0) · 2p / r³

Linear quadrupole on its axis

Charges +q at x = −a and x = +a, and −2q at the origin (net charge zero, dipole moment zero).

  +q          -2q          +q               P
  (+)-------- (-) -------- (+) -------------.
 -a            0            +a               r
  1. E = (q/4πε0)[1/(r − a)² + 1/(r + a)² − 2/r²]
  2. Expand for a ≪ r: 1/(r ∓ a)² = (1/r²)(1 ± 2a/r + 3a²/r² + …)
  3. Sum: 2/r² + 6a²/r⁴ + … − 2/r² = 6a²/r⁴

With quadrupole moment Q = 2qa²:

E_quad = (1/4πε0) · 3Q / r⁴

Conclusion

SourceField falls as
Point charge1/r²
Dipole1/r³
Quadrupole1/r⁴

As charges are arranged so that their individual fields cancel more completely (net charge zero, then dipole moment zero), the field decays faster with distance. Each step multiplies the field by a factor of about a/r, so higher multipoles are important only close to the source.

✓ Answer checked by Claude Opus 5.5 (high effort)

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