Define electric fied intensity and derive an expression for it at a point on the neuttral axis of a dipole. Also determine its direction.

Explanation

Electric field intensity at a point in the electric field is defined or the force experienced by a unit positive charge placed at that point.
Equatorial lines : A line passing through the centre of an electric dipole and perpendicular to it is known as an equatorial line.
Consider an electric dipole AB of dipole length 2a. Let P be the any point of the equatorial line at a distance r from the centre .O. of the dipole.
In right angled `DeltaAOP `
`AP^2 = AO^2 + OP^2`
`AP^2 = a^2 + r^2 = r^2 + a^2`
Also `BP^2 = AP^2 = r^2 + a^2`
`implies BP = AP = sqrt(r^2 + a^2)`
Let `E_1` = Electric field at point P due to charge `– q` lying at A.
`E_1 = 1/(4 pi in_0) (q_1)/(AP^2)` (along PA)
`E_1 = 1/(4pi in_0) (q)/((r^2 + a^2)) `(along PA)
Let `E_2` = Electric field due to charge + q lying at B.
`E_2 = 1/(4 pi in_0) q/(BP^2)` (along PD)
`E_2 = 1/(4pi in_0) q/((r^2 - a^2))` (along PD)
`implies` Magnitude of `E_1` and `E_2` is same. We can find the resultant of E, and E, by completing the parallelogram PCMD. The resultant of `E_1` and `E_2` is represented by the diagonal of the parallelogram. So by triangle law of vector addition, we have
`E/(BA) = (E_1)/(PA) = (E_2)/(BP)`
`implies E = E_1 (BA)/(PA)`
`E = 1/(4pi in_0) q/((r^2 + a^2)) (2a)/((r^2 + a^2)^(1//2)) `( along PM)
`E = 1/(4pi in_0) q/((r^2 + a^2)^(3//2))` where `p = q(2a)`
= dipole moment of the dipole
When dipole is of very small length :
Then `r gt gt a`
and hence `a^2` can be neglected as compared to the `r^2`.
So `E = 1/(4pi in_0) P/(r^3)` (along PM)