Electric Field of a Dipole on Its Equatorial Line

Now consider a point $P$ on the dipole's equatorial line -- the perpendicular bisector of the line joining $+q$ and $-q$ -- at distance $r$ from the centre $O$. By the geometry of a right triangle, the distance from either charge to $P$ is the same, equal to $\sqrt{r^2+a^2}$, so the magnitudes of the two individual fields at $P$ are also equal:

The two fields, however, point in different directions: $\vec{E}_+$ points away from $+q$ (along the line from $+q$ to $P$), and $\vec{E}_-$ points toward $-q$ (along the line from $P$ to $-q$). Resolving each field into a component parallel to the dipole axis and a component perpendicular to it (i.e. along the equatorial line itself): by the symmetry of the figure, the two perpendicular (equatorial-direction) components are equal and opposite, so they cancel exactly; the two components parallel to the axis, however, both point in the same direction -- opposite to $\vec{p}$ -- and so they add.

Each field's component along the axis has magnitude $E_+\cos\theta$ (or $E_-\cos\theta$), where $\theta$ is the angle each line ($+q$-to-$P$ or $P$-to-$-q$) makes with the equatorial line, and $\cos\theta = a/\sqrt{r^2+a^2}$ from the right triangle. So the net field is

For a short dipole, $r\gg a$, so $(r^2+a^2)^{3/2}\approx r^3$, giving

$E_{\text{eq}} \approx \frac{kp}{r^3}$ …