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

Explanation

Consider an electric dipole consisting of two point charges +q and -q separated by a small distance 2a.
We have to find the electric field intensity at p on the equatorial line of the dipole when `thetap=r`.
`E_1` is electric field intensity at P due to charge +q is ,
`|E_1|=1/(4piepsilon_0)q/(AP^2)` where `AP^2=a^2+r^2`
`|E_1|=1/(4piepsilon_0)q/(a^2+r^2)`
The magnitude of the electric field intensity at P due to charge -q is
`|E_2|=1/(4piepsilon_0)q/(a^2+r^2)`
The direction of `vecE_1` is away from the charge +q and the direction of `vecE_2` is towards the each other charge -q as shown `vecE_1` and `vecE_2` can be resolved into two components . The components normal to the dipole axis cancel away . The components along the dipole axis add up.
`therefore vecE_1=(E_1 cos theta + E_2 cos theta )-hatP`
Here the direction of `vecE` is opposite to the dipole monent P
`therefore -hatP` is a unit vector .
`vecE=(E_1+E_2)cos theta -hatP " " because |E_1|=|E_2|`
`vecE=(1/(4piepsilon_0) q/((a^2+r^2)) + 1/(4piepsilon_0)q/((a^2+r^2))) cos theta -hatP`
From the figure , cos `theta =1/((a^2+r^2)^(1//2))-hatP`
`vecE=1/(4piepsilon_0)q/(a^2+r^2)[1+1]a/((a^2+r^2)^(1//2))-hatP`
`=2 1/(4piepsilon_0)(qa)/((a^2+r^2)^(3//2))-hatP " " because P=qxx2a`
`=1/(4piepsilon_0)P/((a^2+r^2)^(3//2))-hatP`
`vecE=1/(4piepsilon_0)vecP/((a^2+r^2)^(3//2)) " " because hatP=vecP/P`
a < < < r So, a can be neglected .
`vecE=1/(4piepsilon_0). vecP/((r^2)^(3//2))-vecP/P`
`vecE=-1/(4piepsilon_0)vecP/r^3`