Let two charges $+q$ and $-q$ be separated by a distance $l$. Then electric potential at a point at distance $r>>l$ from this dipole,
$$
\varphi(r)=\frac{+q}{4 \pi \varepsilon_{0} r_{+}}+\frac{-q}{4 \pi \varepsilon_{0} r_{-}}=\frac{q}{4 \pi \varepsilon_{0}}\left(\frac{r_{-}-r_{+}}{r_{+} r_{-}}\right)
$$
But
From
where
and Now,
So $\quad E=\sqrt{E_{r}^{2}+E_{0}^{2}}=\frac{p}{4 \pi \varepsilon_{0} r^{3}} \sqrt{4 \cos ^{2} \theta+\sin ^{2} \theta}$