Dipole potential; no work along bisector

Potential due to a dipole

Dipole: charges +q and −q, separation 2a, moment p = 2aq. Take point P at distance r from the centre O, where OP makes angle θ with the dipole axis, and r ≫ a.

                           P
                        .' |
                 r   .'    |
                  .'       | r1 (from +q)
               .'  θ       |
   -q ---------O-----------+q ------ axis
    <---- a ---><---- a --->
   r1 ≈ r − a cosθ  (from +q)
   r2 ≈ r + a cosθ  (from −q)

1. V = (q/4πε0)(1/r₁ − 1/r₂) = (q/4πε0)(r₂ − r₁)/(r₁r₂)
2. From the geometry (for r ≫ a): r₂ − r₁ ≈ 2a cosθ and r₁r₂ ≈ r².
3. V = (q/4πε0)(2a cosθ)/r²

V = (1/4πε0) · p cosθ / r²

Special cases: on the axis (θ = 0), V = p/(4πε0 r²); on the equatorial line (θ = 90°), V = 0.

No work along the perpendicular bisector

1. Every point on the perpendicular bisector has θ = 90°, so cos θ = 0 and V = 0 at all those points, including at infinity (V = 0 there).
2. Work done in bringing a test charge q₀ from infinity to a point P on the bisector:

W = q₀ (V_P − V_∞) = q₀ (0 − 0) = 0

Equivalently, the forces from +q and −q on the test charge have equal magnitudes, and their components along the bisector cancel; only a component parallel to the axis remains, which is perpendicular to the path, so it does no work.

Hence no work is done in bringing a charge from infinity to the dipole along its perpendicular bisector; the bisector is an equipotential at zero potential.