Question : Derive an expression for Electric Field intensity at a point on the equatorial line of an electric dipole ( Broad side on Position)
Solution : Consider an electric dipole AB of length $2l$. Let P be the point on the equatorial line at a distance r from the centre O of the dipole.
Step 1. Electric field intensity at P due to +q charge is given by
$\vec{E}_{1} = \frac{1}{4\pi\epsilon_{0}} \frac{q}{BP^2}$ along PD
$\vec{E}_{1}= \frac{1}{4\pi\epsilon_{0}} \frac{q}{(r^2 + l^2)}$ along PD ...(i)
Electric field intensity at P due to $-q$ charge is given by,
$\vec{E}_{2} = \frac{1}{4\pi\epsilon_{0}} \frac{q}{AP^2}$along PC
$\vec{E}_{2}= \frac{1}{4\pi\epsilon_{0}} \frac{q}{(r^2 + l^2)}$along PC ....(ii)
From (1) and (2), $|\vec{E}_{1}| = |\vec{E}_{2}| = \frac{1}{4\pi\epsilon_{0}} \frac{q}{(r^2 + l^2)}$ ...(iii)
$\vec{E}_{1}$ and $\vec{E}_{2}$ are inclined at an angle of $2\theta$, so their resultant can be determined using parallelogram law of vectors addition. This resultant (E, say) is the net electric field intensity due to the electric dipole at point P and given by
$E = \sqrt{E_{1}^2 + E_{2}^2 + 2E_{1}E_{2} \cos 2\theta}$
$(\because E_{1} = E_{2})$
$E = \sqrt{E_{1}^2 + E_{2}^2 + 2E_{1}^2 \cos 2\theta}$
$E= \sqrt{2E_{1}^2 + 2E_{1}^2 \cos 2\theta}$
$= \sqrt{2E_{1}^2 (1 + \cos 2\theta)}$
$(\because 1 + \cos 2\theta = 2\cos^2 \theta)$
$= \sqrt{2E_{1}^2 \times 2\cos^2 \theta}$
$\therefore E = 2E_{1} \cos \theta$
(Using eqn. (3))
$= 2 \times \frac{1}{4\pi\epsilon_{0}} \frac{q}{(r^2 + l^2)} \cos \theta$
Now from $\triangle OAP$,
$\cos \theta = \frac{l}{\sqrt{r^2 + l^2}}$
$E = 2 \times \frac{1}{4\pi\epsilon_{0}} \frac{q}{(r^2 + l^2)} \times \frac{l}{(r^2 + l^2)^{1/2}}$
$E= \frac{q \times 2l}{4\pi\epsilon_{0} (r^2 + l^2)^{3/2}}$
Since
$q \times 2l = p \quad \text{(dipole moment)}$
$E = \frac{p}{4\pi\epsilon_{0} (r^2 + l^2)^{3/2}} \quad \text{along (-) x-axis}$
In vector form,
$\vec{E} = \frac{-\vec{p}}{4\pi\epsilon_{0} (r^2 + l^2)^{3/2}}$
The direction of the electric field $\vec{E}$ is opposite to the direction of the dipole moment $\vec{p}$.
Special Case:
If $r>>>l$ i.e., dipole is short, then $l^2$ can be neglected as compared to $r^2$.hence
$E = \frac{p}{4\pi\epsilon_{0} r^3} \quad \text{along (-) x-axis}$
Important Facts : Angle between the electric dipole moment ($\vec{p}$) and the electric field intensity at any point on its equatorial line is $180^\circ$ or $\pi$ radian.
$E_{axial} = 2 E_{equatorial}$