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Dipole potential and resultant field

Short answer

The dipole potential is V = (1/4πε₀) p cosθ / r². The radial component is E_r = (1/4πε₀) 2p cosθ / r³, transverse E_θ = (1/4πε₀) p sinθ / r³; resultant magnitude |E| = (p/(4πε₀ r³)) √(1+3cos²θ) and direction tanα = (tanθ)/2.

For a given short electric dipole, show that the electric potential at any point at a distance r is V = P cos θ / r², where θ is the angle made by r to the dipole and P is its dipole moment. Using the above relation find an expression for the resultant electric intensity at that point.

Potential of a short dipole

Charges +q and −q separated by 2a, p = 2aq. Point P at distance r from the centre O, with OP making angle θ with p, and r ≫ a.

              P
             /|
        r2 /  | r1
         /  r |
   -q ---+----O----+--- +q
  1. Distances: r₁ ≈ r − a cosθ (from +q), r₂ ≈ r + a cosθ (from −q).
  2. V = (q/4πε0)(1/r₁ − 1/r₂) = (q/4πε0)(r₂ − r₁)/(r₁r₂) = (q/4πε0)(2a cosθ)/r²

V = (1/4πε0) · p cosθ / r² (in the question's units, V = p cosθ/r²)

Resultant field from V

Use polar components: E_r = −∂V/∂r, E_θ = −(1/r) ∂V/∂θ.

  1. E_r = −∂/∂r [p cosθ / (4πε0 r²)] = 2p cosθ / (4πε0 r³)
  2. E_θ = −(1/r) ∂/∂θ [p cosθ / (4πε0 r²)] = p sinθ / (4πε0 r³)
  3. Resultant:

E = √(E_r² + E_θ²) = (p / 4πε0 r³) √(4cos²θ + sin²θ)

E = (p / 4πε0 r³) √(1 + 3cos²θ)

  1. Its direction makes angle α with the radius vector, where tanα = E_θ/E_r = tanθ / 2.

Checks: θ = 0 gives E = 2p/(4πε0r³) (axial); θ = 90° gives E = p/(4πε0r³) (equatorial).

✓ Answer checked by Claude Opus 5.5 (high effort)

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