Derive an expression for the intensity of the electric field at a point on the axial line of an electric dipole.

Explanation

Electric field at a point on the axis of a dipole:
(1) Consider an electric dipole consisting two charge -q and +q separated by a adistance '2a' with centre 'O'.
(2) We shall calculate electric field E at point P on the axial line of dipole, and at a distance `OP=r`.
(3) Let `E_(1)` and `E_(2)` be the intensities of electric field at P due to charge +q and -q respectively.
(4) Therefore, `E_(1)=1/(4 pi epsi_(0))xxq/((AP)^(2))=1/(4 pi epsi_(0))xxq/((r-a)^(2))` along AP
and `E_(2)=1/(4 pi epsi_(0))xxq/((BP)^(2))=1/(4 pi epsi_(0))xxq/((r+a)^(2))` along PB
The resultant intensity at P is `E=E_(1)-E_(2)` [`:'` They are opposite and `E_(1) gt E_(2)`]
`implies E=1/(4pi epsi_(0))xxq/((r-a)^(2))-1/(4pi epsi_(0))xxq/((r+a)^(2))=q/(4pi epsi_(0))[1/((r-a)^(2))-1/((r+a)^(2))]`
`=q/(4pi epsi_(0))[((r+a)^(2)-(r-a)^(2))/((r^(2)-a^(2))^(2))]`
`=q/(4 pi epsi_(0))xx(4 ra)/((r^(2)-a^(2))^(2))`
`:. E=1/(4 pi epsi_(0)) (2pr)/((r^(2)-a^(2))^(2))` where `P=2` aq
If `r gt gt a` then `a^(2)` can be neglected in comparision to `r^(2)`.
`:. E=1/(4 pi epsi_(0))xx(2 Pr)/r^(4)=1/(4pi epsi_(0))xx(2P)/r^(3)`
In vector form, `vec(E)=(2vec(P))/(4pi epsi_(0) r^(3))`