Two equal charges $‘q’$ of opposite sign are separated by a small distance $‘d’$. The electric intensity $‘E’$ at a point on the straight line passing through the two charges at a very large distance $‘r’$ from the midpoint of two charges is :

As shown in the fig there are two electric fields $E_{+q}$ and $E_{-q}$ due to $+q$ and $-q$ charges respectively.

Resultant of this two oppositely directed electric fields gives the net electric field at that point.

$\therefore$ Electric field at that point is given by,
$E=E_{+q}-E_{-q}=Kq/(r-d/2{)}^2-Kq/(r+d/2{)}^2$

                            $=Kq[\frac{1}{(r^2-rd+d^4/4)}-\frac{1}{(r^2+rd+d^4/4)}]$

                            

                            $=Kq[\frac{1}{(r^2-rd)}-\frac{1}{(r^2+rd)}]$ ......($d^2/4\approx 0$)

                            

                           $=Kq[\frac{(r^2+rd)}{r^4-r^2{d}^2}-\frac{(r^2-rd)}{r^4-r^2{d}^2}]$

                             

                           $=Kq[\frac{(2rd)}{r^2(r^2-d^2)}]$

                           

                          $=Kq\frac{(2rd)}{r^4}$  ....$(d^2\approx 0)$

                        $\therefore E=\frac{1}{4\pi {ϵ}_o}\frac{2qd}{r^3}$