Derive an expression for the intensity of the electric field at a point on the equatorial plane of an electric dipole.

Explanation

(1) Consider an electric dipole consisting of two charge -q and +q separated by distance '2a' with centre at 'O'.
(2) We shall calculate electric field E at P on equatorial line of dipole and at a distance `OP=r`.
(3) Let `E_(1)` and `E_(2)` be the electric fields at P due to charges +q and -q respectively.
(4) The `bot^(r)` components (`E_(1) sin theta` and `E_(2) sin theta`) cancel each other because they are equal and opposite. The `II^(el)` components (`E_(1) cos theta` and `E_(2) cos theta`) are in the same direction and hence add up.
(5) The resultant field intensity at point P is given by `E=E_(1) cos theta+E_(2) cos theta`
But `E_(1)=E_(2)=1/(4pi epsi_(0))xxq/(AP^(2))=1/(4 pi spri_(0))xxq/((r+a))`
`E=1/(4pi epsi_(0))xxq/((r+a)) [cos theta+cos theta]`
`=1/(4pi epsi_(0))xx(2q cos theta)/((r+a))`
(6) From figure, `cos theta=a/((r^(2)+a^(2))^(1//2))`
`:. E=1/(4pi epsi_(0))xx(2aq)/((r^(2)+a^(2))^(3//2))`
(7) If `r gt gt a`, then `a^(2)` can be neglected in comparison to `r^(2)`. Then
`E=P/(4pi epsi_(0)) xx1/r^(3)`
In vector form `vec(E)=vec(P)/(4 pi epsi_(0) r^(3))`