An electric dipole has dipole moment p. If r W s, where

s is the separation between the charges, show that the electric

potential of the dipole can be written

$V=\frac{1}{4\mathrm{\pi }\in \circ }\frac{p\cos \theta }{r^2}$

where r is the distance from the center of the dipole and u is the

angle from the dipole axis.

It is a finite distance s that comprises both fair and opposite charges. by using the expression of the charge that is potential the calculation of dipole charge at the point of P.

the expression of the charge is=

$V=\frac{1}{4\mathrm{\pi }\in \circ }\frac{q}{r}$

here free space is presented as $\in \circ$, charge is presented as q and r is presented as distance between the points.

the figure indicates each chargeqthat is separated by distances.

here, O represents the origin here distance is represented as randrfrom the point P. the distance from the dipole Pis r.

The potential of the dipole at the point of P has pair of charges which is,

$$\begin{gathered} v=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r_{+}}+\frac{1}{4\pi {\epsilon }_0}\frac{(-q)}{r_{-}} \\ =\frac{q}{4\pi {\epsilon }_0}(\frac{1}{r_{+}}+\frac{1}{r_{-}}) \end{gathered}$$

From Geometry $r_{+}$represented as;

$r_{+}=\sqrt{\left(y^2-sx+{\left(\frac{s}{2}\right)}^2+x^2\right)}$

From Geometry $r_{-}$represented as;

$r_{-}=\sqrt{\left(y^2+sy+{\left(\frac{s}{2}\right)}^2+x^2\right)}$

Combining all the aspects of the given aspect;

localid="1648114549617" $$\begin{gathered} v=\frac{q}{4\pi {\epsilon }_0}(\frac{1}{r_{+}}+\frac{1}{r_{-}}) \\ =\frac{q}{4\pi {\epsilon }_0}(\frac{1}{\sqrt{\left(y^2-sx+{\left(\frac{s}{2}\right)}^2+x^2\right)}}+\frac{1}{\sqrt{\left(y^2+sy+{\left(\frac{s}{2}\right)}^2+x^2\right)}}) \end{gathered}$$

As the s<<r , thus this equation can be written as;

localid="1648114771256" $$\begin{gathered} =\frac{q}{4\pi {\epsilon }_0}(\frac{1}{\sqrt{\left(y^2-sy+{\left(\frac{s}{2}\right)}^2+x^2\right)}}+\frac{1}{\sqrt{\left(y^2+sy+{\left(\frac{s}{2}\right)}^2+x^2\right)}}) \\ =\frac{q}{4\pi {\epsilon }_0}(\frac{1}{\sqrt{\left(y^2+x^2-sy\right)}}+\frac{1}{\sqrt{\left(y^2+x^2+sy\right)}}) \end{gathered}$$

As per the Given Geometry;

$x^2+y^2=r^2$

Hence;

localid="1648115010294" $$\begin{gathered} V=\frac{q}{4\pi {\epsilon }_0}(\frac{1}{\sqrt{\left(r^2-sy\right)}}+\frac{1}{\sqrt{\left(r^2+sy\right)}}) \\ =\frac{q}{4\pi {\epsilon }_{0}r}\left({\left(1-\frac{sy}{r^2}\right)}^{-\frac{1}{2}}-{\left(1+\frac{sy}{r^2}\right)}^{-\frac{1}{2}}\right) \end{gathered}$$

Through using Binomial expression to expand, equation can be written as;

$$\begin{gathered} V=\frac{q}{4\pi {\epsilon }_{0}r}\left(\left(1+\frac{sy}{2r^2}\right)-\left(1+\frac{sy}{2r^2}\right)\right) \\ =\frac{q}{4\pi {\epsilon }_0}\left(\frac{sy}{r^3}\right) \end{gathered}$$

From Geometry it can be stated that;

$y=r\cos \theta$

Applying this within sum of potential from step 3;

$$\begin{gathered} =\frac{q}{4\pi {\epsilon }_0}\left(\frac{sr\cos \theta }{r^3}\right) \\ =\frac{1}{4\pi {\epsilon }_0}\left(\frac{qs\cos \theta }{r^2}\right) \\ =\frac{1}{4\pi {\epsilon }_0}\left(\frac{p\cos \theta }{r^2}\right)\left[As;p=qs\right] \end{gathered}$$

Therefore , it can be stated that the Potential due to Dipole when s<<r an be stated as;$=\frac{1}{4\pi {\epsilon }_0}\left(\frac{p\cos \theta }{r^2}\right)$