(N/A) Let a dipole consist of charges $+q$ and $-q$ separated by a distance $2a$. Let $P$ be a point on the equatorial plane at a distance $r$ from the center $O$ of the dipole.
The distances of point $P$ from $+q$ and $-q$ are equal:
$r_{+} = r_{-} = \sqrt{r^{2} + a^{2}}$
The magnitudes of the electric fields due to the two charges are equal:
$E_{+q} = E_{-q} = \frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2} + a^{2}}$
Resolving the electric field vectors into components:
$1$. The components perpendicular to the dipole axis $(E \sin \theta)$ are equal in magnitude and opposite in direction,so they cancel each other out.
$2$. The components parallel to the dipole axis $(E \cos \theta)$ are in the same direction (opposite to the dipole moment direction $\hat{p}$).
The total electric field at $P$ is:
$E = -(E_{+q} \cos \theta + E_{-q} \cos \theta) \hat{p}$
$E = -2 \left( \frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2} + a^{2}} \right) \cos \theta \hat{p}$
From the geometry,$\cos \theta = \frac{a}{\sqrt{r^{2} + a^{2}}}$. Substituting this:
$E = -\frac{2aq}{4 \pi \varepsilon_{0} (r^{2} + a^{2})^{3/2}} \hat{p}$
Since the dipole moment is $p = q(2a)$:
$E = -\frac{p}{4 \pi \varepsilon_{0} (r^{2} + a^{2})^{3/2}} \hat{p}$
For a short dipole $(r \gg a)$:
$E \approx -\frac{p}{4 \pi \varepsilon_{0} r^{3}} \hat{p}$