Electric Dipole:
An electric dipole is a system in which two equal magnitude and opposite point charged particles are placed at a very short distance apart.
Electric Dipole Moment:
The product of magnitude of one point charged particle and the distance between the charges is called the 'electric dipole moment'. It is vector quantity and the direction of electric dipole moment is along the axis of the dipole pointing from negative charge to positive charge.

Electric Dipole 
Let us consider, the two charged particle, which has equal magnitude $+q$ coulomb and $q$ coulomb is placed at a distance of $2l$ in a dipole so the electric dipole moment is →
$\overrightarrow{p}=q\times \overrightarrow{2l}$ 
Unit: $Cm$ Or $Amperemetresec$
Dimension: $[ALT]$
Electric field intensity due to an Electric Dipole:
The electric field intensity due to an electric dipole can be measured at three different points:
 Electric field intensity at any point on the axis of an electric dipole
 Electric field intensity at any point on the equatorial line of an electric dipole
 Electric field intensity at any point on an electric dipole
1. Electric field intensity at any point on the axis of an electric dipole:
Let us consider, An electric dipole $AB$ made up of two charges of $q$ and
$+q$ coulomb are placed in a vacuum or air at a very small distance of $2l$. Let a point $P$ be on the axis of an electric dipole and place it at a distance $r$ from the center point $O$ of the electric dipole. Now put the test charged particle $q_{0}$ at point $P$ for the measurement of electric field intensity due to dipole's charge.

Electric field intensity at any point on the axis of an electric dipole 
So, Electric field intensity (magnitude only) at point $P$ due $+q$ charge of electric dipole
$ E_{+q}=\frac{1}{4\pi\epsilon}\frac{q}{(rl)^{2}} \qquad(1)$
Electric field intensity (magnitude only) at point $P$ due $q$ charge of electric dipole
$ E_{q}=\frac{1}{4\pi\epsilon_{0}}\frac{q}{(r+l)^{2}} \qquad(2)$
The net electric field at point $P$ due to an electric dipole
$ E=E_{+q}E_{q}\qquad(3)$
Subtitute the value of $E_{+q}$ and $E_{q}$ in equation $(3)$, Then the above equation $(3)$ can also be written as follows
$E=\frac{q}{4\pi\epsilon _{0}}\left [ \frac{1}{(rl)^{2}}\frac{1}{(r+l)^{2}} \right ]$
$ E=\frac{q}{4\pi\epsilon _{0}}\left [ \frac{(r+l)^{2}(rl)^{2}}{(r+l)^{2}(rl)^{2}} \right ]$
$ E=\frac{q}{4\pi\epsilon _{0}}\left [ \frac{(r^{2}+l^{2}+2lrr^{2}l^{2}+2lr)}{(r+l)^{2}(rl)^{2}} \right ]$
$ E=\frac{q}{4\pi\epsilon _{0}}\left [ \frac{4rl}{(r^{2}l^{2})^{2}} \right ]\qquad(4)$
Here $l \lt r $ so $l^{2} \lt \lt r^{2}$ therefore neglect the term $l^{2}$ in above equation $(4)$ so we can write above equation
$ E=\frac{q}{4\pi\epsilon _{0}}\left [ \frac{4rl}{(r^{2})^{2}} \right ]$
$E=\frac{1}{4\pi\epsilon _{0}}\left [ \frac{2p}{r^{3}} \right ]\qquad (5)$
This is the equation of electric field intensity at a point on the axis of an electric dipole.
The vector form of the above equation $(5)$ is
$\overrightarrow{E}=\frac{1}{4\pi\epsilon _{0}}\left [ \frac{2\overrightarrow{p}}{r^{3}} \right ]$

2. Electric field intensity at any point on the equatorial line of an electric dipole:
Let us consider, An electric dipole $AB$ made up of two charges of $+q$ and $q$ coulomb are placed in vacuum or air at a very small distance $2l$. Let a point $P$ be on the equatorial line of an electric dipole and place it at a distance $r$ from the center point $O$ of the electric dipole. Now put the test charged particle $q_{0}$ at point $P$ for the measurement of electric field intensity due to dipole's charge.
So, Electric field intensity (magnitude only) at point $P$ due $+q$ charge of electric dipole

Electric field intensity at a point of the equatorial line of an electric dipole 
$E_{+q}=\frac{1}{4\pi\epsilon_{0}}\left [\frac{q}{(r^{2}+l^{2})} \right ] \qquad(1)$
Electric field intensity (magnitude only) at point $P$ due to $q$ charge of electric dipole
$E_{q}=\frac{1}{4\pi\epsilon_{0}}\left [\frac{q}{(r^{2}+l^{2})} \right ] \qquad(2)$
The net electric field at point $P$ due to an electric dipole
$E=E_{+q}\:cos\theta + E_{q}\: cos\theta \qquad(3)$
Put the value of $E_{+q}$ and $E_{q}$ in the above equation $(3)$, So equation $(3)$ can also be written as follows
$E=2\left [ \frac{q}{4\pi\epsilon_{0} }\frac{1}{(r^{2}+l^{2})} \right ]cos\theta \qquad (4)$
From figure, In $\Delta \: POB$,
$cos\:\theta=\frac{l}{\sqrt{(r^{2}+l^{2})}}$
Put the value of $cos\theta$ in equation $(4)$, so equation $(4)$ can also be written as follows
$E=\frac{1}{4\pi\epsilon_{0}}\left [ \frac{q\times2l}{(r^{2}+l^{2})^{3/2}} \right ] \qquad (5)$
Here $l \lt r$ so $l^{2} \lt \lt r^{2}$ so neglect the term $l^{2}$ in above equation $(5)$ so we can write above equation
$ E=\frac{1}{4\pi\epsilon_{0}}\left [ \frac{q\times2l}{r^{3}} \right ] \qquad (6)$
$E=\frac{1}{4\pi\epsilon_{0}} \frac{p}{r^{3}} \qquad (7)$
This is the equation of electric field intensity at a point on the equatorial line of an electric dipole.
The vector form of the above equation $(7)$ is
$\overrightarrow{E}=\frac{1}{4\pi\epsilon_{0}} \frac{\overrightarrow{p}}{r^{3}}$

3. Electric field intensity at any point of an electric dipole:
Let us consider, An electric dipole $AB$ of length $2l$ consisting of the charge $+q$ and $q$. Let's take a point $P$ in general and its position vector is $\overrightarrow{r}$ from the center point $O$ of the electric dipole AB.

Electric field intensity at any point of an electric dipole 
The electric dipole moment is a vector quantity that has a direction from $q$ charge to $+q$ charge. So the electric dipole moment's direction is resolved in twocomponent one is along the vector position $\overrightarrow{r}$ i.e
pcosθ and the other is normal to vector position $\overrightarrow{r}$ i.e $psinθ$. So
Electric field intensity due to dipole moment of component
$p\:cos\theta$ {Electric field along the Axial Line }
$ \overrightarrow{E_{\parallel }}=\frac{1}{4\pi\epsilon_{0}} \frac{2pcos\theta}{r^{3}} \qquad(1)$
Electric field intensity due to dipole moment of component
$p\:sin\theta$ {Electric field along the Equatorial Line}
$ \overrightarrow{E_{\perp}}=\frac{1}{4\pi\epsilon_{0}} \frac{psin\theta}{r^{3}} \qquad(2)$
The resultant electric field vector $E$ at point $P$
$ \overrightarrow{E}=\sqrt{E_{\perp}^{2}+E_{\parallel}^{2}+2 E_{\perp}E_{\parallel}cos90^{\circ}}$
From figure, The angle between $E_{\perp}$ and $E_{\parallel}$ is $90^{\circ}$. So
$\overrightarrow{E}=\sqrt{E_{\perp}^{2}+E_{\parallel}^{2} }\qquad (3)$
Now substitute the value of equation $(1)$ and equation $(2)$ in equation $(3)$. Then
$ \overrightarrow{E}=\frac{1}{4\pi\epsilon_{0}}\frac{p}{r^{3}}\sqrt{(sin^{2}\theta+4cos^{2}\theta)}$
$ \overrightarrow{E}=\frac{1}{4\pi\epsilon_{0}}\frac{p}{r^{3}}\sqrt{(1+3cos^{2}\theta)}$

This is the equation of electric field intensity at any point due to an electric dipole.
The direction of the resultant electric field intensity vector $\overrightarrow{E}$ from the axial line is
$tan\alpha =\frac{\overrightarrow{E}_{\perp }}{\overrightarrow{E_{\parallel }}} \qquad(4)$
Put the value of $\overrightarrow{E_{\perp}}$ and $\overrightarrow{E_{\parallel}}$ in equation (4). we get
$tan\alpha =\frac{sin\theta}{2cos\theta}$
$tan\alpha =\frac{1}{2}tan\theta$

Here $\alpha$ is the angle between the resultant electric field intensity $\overrightarrow{E}$ and the axial line.