(N/A) The electric potential $V$ at a point due to an electric dipole is given by: $V = k \left( \frac{\vec{p} \cdot \hat{r}}{r^2} \right) = \frac{k p \cos \theta}{r^2}$,where $k = \frac{1}{4 \pi \epsilon_0}$,$\vec{p}$ is the dipole moment,and $\theta$ is the angle between the position vector $\vec{r}$ and the dipole moment $\vec{p}$.
Important Features:
$(i)$ The potential due to an electric dipole depends on the angle $\theta$ between the position vector $\vec{r}$ of the point and the dipole moment $\vec{p}$.
$(ii)$ The electric dipole potential decreases as $\frac{1}{r^2}$ with distance,unlike the potential of a point charge which decreases as $\frac{1}{r}$.
Special Cases:
$(1)$ For a point on the axial line: $\theta = 0^\circ$ or $180^\circ$,hence $V_a = \pm \frac{k p}{r^2}$.
$(2)$ For a point on the equatorial plane: $\theta = 90^\circ$,hence $V_e = 0$.