Obtain an expression for the electric field intenstiy at a point on the equatorial line of an electric dipole.

Explanation

Let q be the magnitude of charge on the dipole. Let 2a be the dipole length. Let r be the distance of a point from the centre of the dipole. Let PO be perpendicular to AB. The directin of `vecE_(-q)` and `vecE_(+q)` are as shown in the fig.
Resolving `vecE_(q)` and `vecE_(-q)` along two perpendicular directions, we not the `sin theta` components of `E_(+q)` and `E_(-q)` cancel each other and `cos theta` components reinforce each other at that point.
Sicne `|vecE_(+q)|=|vecE_(-q)|` and `vecE_(B)=2Ecos theta (-hatp)`
`:.E_(B)=2 E cos theta`
i.e. `E_(b)=2=(1/(4pi epsilon_(0)))(q/(AP^(2)))(a/(AP))`
where `AP=(r^(2)+a^(2))^(1/2)`
`E_(B)=(p/(4 pi epsilon_(0)))(q/((r^(2)+a^(2))^(3/2)))`
where `p=2qa=` dipole moment.
Since the direction of `E_(B)` is opposite to `vecp`
`vecE_(B)=(p/(4 p epsilon_(0)))(1/((r^(2)+q^(2))^(3/2)))hatp`
where `p hatp =vecp`
`vecE_(B)=(-vecp)/((4pi epsilon_(0))(r^(2)+a^(2))^(3//2))`