Derive an expression for the intensity of the electric field at a point on the axial line of an electric dipole.

Consider an electric dipole consisting charges –q and q, seperated by a distance 2a. Let p be a point on the axial line of a dipole at a distance r from centre of dipole as shown in figure
Electric field due to charge –q at P is

${\overrightarrow{\mathrm{E}}}_{-\mathrm{q}}=\frac{1}{4{\mathrm{\pi ϵ}}_0}\frac{\mathrm{q}}{(\mathrm{r}+\mathrm{a}{)}^2}\text{ towards left }$
Electric field due to charge +q at P is
$\overrightarrow{\mathrm{E}}+\mathrm{q}=\frac{1}{4\mathrm{\pi }{\in }_0}\frac{\mathrm{q}}{(\mathrm{r}-\mathrm{a}{)}^2}\text{ towards right }$
Resultant electric field at point P is
$\begin{array}{l}{\overrightarrow{\mathrm{E}}}_{\mathrm{axial}}=\frac{1}{4\mathrm{\pi }{\in }_0}\frac{\mathrm{q}}{(\mathrm{r}-\mathrm{a}{)}^2}-\frac{1}{4\mathrm{\pi }{\in }_0}\frac{\mathrm{q}}{(\mathrm{r}+\mathrm{a}{)}^2}\\ {\overrightarrow{\mathrm{E}}}_{\mathrm{axial}}=\frac{\mathrm{q}}{4\mathrm{\pi }{\in }_0}\left[\frac{1}{(\mathrm{r}-\mathrm{a}{)}^2}-\frac{1}{(\mathrm{r}+\mathrm{a}{)}^2}\right]\end{array}$
$\begin{array}{l}{\overrightarrow{\mathrm{E}}}_{\mathrm{axial}}=\frac{\mathrm{q}}{4\mathrm{\pi }{\in }_0}\left[\frac{(\mathrm{r}+\mathrm{a}{)}^2-(\mathrm{r}-\mathrm{a}{)}^2}{(\mathrm{r}-\mathrm{a}{)}^2(\mathrm{r}+\mathrm{a}{)}^2}\right]\\ {\overrightarrow{\mathrm{E}}}_{\text{axial }}=\frac{\mathrm{q}}{4\mathrm{\pi }{\in }_0}\frac{4\mathrm{ra}}{{\left({\mathrm{r}}^2-{\mathrm{a}}^2\right)}^2}\\ {\overrightarrow{\mathrm{E}}}_{\text{axial }}=\frac{\mathrm{q}}{4\mathrm{\pi }{\in }_0}\frac{2\left(2\mathrm{a}\right)\left(\mathrm{q}\right).\mathrm{r}}{{\left({\mathrm{r}}^2-{\mathrm{a}}^2\right)}^2}\\ {\overrightarrow{\mathrm{E}}}_{\mathrm{axial}}=\frac{2}{4{\mathrm{\pi ϵ}}_0}\frac{\mathrm{Pr}}{{\left({\mathrm{r}}^2-{\mathrm{a}}^2\right)}^2}\end{array}$
if a<<<r then 'a' is negligible
$\begin{array}{r}{\overrightarrow{\mathrm{E}}}_{\text{axial }}=\frac{2}{4{\mathrm{\pi ϵ}}_0}\frac{\mathrm{Pr}}{{\mathrm{r}}^4}\\ {\overrightarrow{\mathrm{E}}}_{\text{axial }}=\frac{1}{4{\mathrm{\pi ϵ}}_0}\frac{2\mathrm{P}}{{\mathrm{r}}^3}\end{array}$
The direction of $\overrightarrow{\mathrm{E}}$ is in the direction of dipole moment