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Dipole field on equatorial line

Short answer

An electric dipole is a pair of equal opposite charges ±q separated by 2a, with dipole moment p = 2aq. On the equatorial line, the field is E = (1/4πε₀) p/(r²+a²)^(3/2), antiparallel to p; for r ≫ a, E = (1/4πε₀) p/r³.

What is electrical dipole and dipole moment? Derive an expression of the electric field intensity at a point due to dipole at equatorial line.

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

Field on the equatorial (perpendicular bisector) line

Let P be at distance r from the centre O along the perpendicular bisector.

              P
              .
            / | \
        r1 /  |  \ r2
          /   | r  \
   -q ---/----O-----\--- +q
        <-a->  <-a->
  1. Each charge is at distance r' = √(r² + a²) from P.
  2. Magnitudes are equal: E₊ = E₋ = (1/4πε0) q/(r² + a²).
  3. E₊ points away from +q; E₋ points towards −q. Each makes angle θ with the line parallel to the dipole axis, where cosθ = a/√(r² + a²).
  4. Components perpendicular to the axis (along OP) cancel; components parallel to the axis add. Both point from +q to −q, opposite to p:

E = 2 E₊ cosθ = 2 · (1/4πε0) · q/(r² + a²) · a/√(r² + a²)

  1. With p = 2aq:

E = (1/4πε0) · p / (r² + a²)^(3/2), directed antiparallel to p.

  1. For a short dipole (r ≫ a):

E = (1/4πε0) · p / r³

This is half the magnitude of the axial-line field (2p/4πε0r³) at the same distance, and in the opposite direction.

✓ Answer checked by Claude Opus 5.5 (high effort)

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