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Dipole field on axial line

Short answer

An electric dipole consists of equal and opposite charges ±q separated by 2a; dipole moment p = 2qa from −q to +q. On the axial line at distance r ≫ a, E = (1/4πε₀)·2p/r³, so E ∝ 1/r³.

What is an electric dipole and dipole moment? Show that the electric field for a short dipole drops inversely to the cube of the distance at any point from the dipole on an axial line.

An electric dipole is a pair of equal and opposite point charges +q and −q separated by a small distance 2a. The dipole moment is the vector p = q(2a), of magnitude 2aq, directed from −q to +q. Its unit is C·m.

Field on the axial line

Let O be the centre of the dipole and P a point on the axis at distance r from O (r > a).

   -q          O          +q                P
 --(-)---------+---------(+)------------------.-->
   <--- a --->  <--- a --->
   <------------------ r ------------------->
  1. Distance of P from +q is (r − a); from −q is (r + a).
  2. Field of +q at P (away from +q, along p): E₁ = (1/4πε0) q/(r − a)²
  3. Field of −q at P (towards −q, opposite to p): E₂ = (1/4πε0) q/(r + a)²
  4. E₁ is larger and the fields are collinear, so the net field is along p:
E = E₁ − E₂ = (q/4πε0) [ 1/(r−a)² − 1/(r+a)² ]
  = (q/4πε0) · 4ar / (r² − a²)²
  = (1/4πε0) · 2pr / (r² − a²)²      (p = 2aq)
  1. For a short dipole, a ≪ r, so r² − a² ≈ r²:

E = (1/4πε0) · 2p / r³

Hence E ∝ 1/r³: the field of a short dipole falls off inversely as the cube of the distance, faster than the 1/r² of a single charge, because the fields of +q and −q nearly cancel at large distance.

✓ Answer checked by Claude Opus 5.5 (high effort)

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