**Let us calculate**

**Potential At A Point Due To An Electric Dipole:-**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=r_{1} and BP=r_{2}

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=1/4πε_{0}q/r_{2}

And Potential at P due to -q= -1/4πε_{0} q/r_{1}

Net potential at P due to the dipole

V=q/4πε_{0}r_{2} –q/4πε_{0}r_{1}

V= q/4πε_{0}[1/r_{2}-1/r_{1}]

Now , r_{1}=AP=CP

=OP+OC

=r+a cos θ

And r_{2}=BP=DP

=OP –OD

=r- a cos θ

V=q/4πε_{0}[1/r-a cos θ – 1/r+a cos θ]

= q/4πε_{0}[r+a cos θ –r + a cos θ/r^{2}-a^{2} cos^{2}θ]

V= q/ 4πε_{0}[2a cos θ/r^{2}-a^{2} cos^{2} θ]

i.e. V=p cosθ/4πε_{0}(r^{2}-a^{2} cos^{2} θ) (p=q*2a)

**Special cases:- **

(i) When the point P lies on the axial line of the dipole ,θ=0^{0}

Cos θ=cos 0^{0} =1

V=p/4πε_{0}(r^{2}-a^{2})

If a<<r.then V=q/4πε_{0}r^{2}

Thus due to an electric dipole ,potential, V∞ 1/r^{2}

(ii) When the point P lies on the equatorial line of the dipole ,θ=90^{0}

Cos θ =cos 90^{0}=0

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

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