Let an electric dipole consist of two equal and opposite point charges – q at A and +q at b ,separated by a small distance AB =2a ,with centre at O.

The dipole moment                 p=q*2a

We will calculate potential at any point P,where

OP=r and  angle BOP= θ

Let                                   AP=r₁ and BP=r₂

Draw AC perpendicular PQ and BD perpendicular PO

In ∆AOC.                                     Cosθ=OC/OA =OC/a

OC= a cos θ

Similarly,                                      OD= a cos θ

Potential at P due to               +q=1/4πε₀q/r₂

And  Potential at P due to     -q= -1/4πε₀ q/r₁

Net potential at P due to the dipole

V=q/4πε₀r₂ –q/4πε₀r₁

V= q/4πε₀[1/r₂-1/r₁]

Now ,               r₁=AP=CP

=OP+OC

=r+a cos θ

And                   r₂=BP=DP

=OP –OD

=r- a cos θ

V=q/4πε₀[1/r-a cos θ – 1/r+a cos θ]

= q/4πε₀[r+a cos θ –r + a cos θ/r²-a² cos²θ]

V= q/ 4πε₀[2a cos θ/r²-a² cos² θ]

i.e.                              V=p cosθ/4πε₀(r²-a² cos² θ)                             (p=q*2a)

Special cases:-

(i)                 When the point P lies on the axial line of the dipole ,θ=0⁰

Cos θ=cos 0⁰ =1

V=p/4πε₀(r²-a²)

If                             a<<r.then V=q/4πε₀r²

Thus due to an electric dipole ,potential, V∞ 1/r²

(ii)               When the point P lies on the equatorial line of the dipole ,θ=90⁰

Cos θ =cos 90⁰=0

i.e electric potential due to an electric dipole is zero at every point on the equatorial line of the dipole.