= Solution
Take $\mathbf d$ from the negative to the positive <point charge>. Their <electric dipole moment> is $\mathbf p=q\mathbf d$. Place them at $\pm\mathbf d/2$ and use the <superposition principle> for the <electric potential>:
$$
V(\mathbf s)=\frac{q}{4\pi\epsilon_0}
\left(\frac1{|\mathbf s-\mathbf d/2|}-\frac1{|\mathbf s+\mathbf d/2|}\right).
$$
For $s\gg d$, the first-order <Taylor expansion> of the reciprocal distances gives
$$
\boxed{V(s,\theta)=\frac{\mathbf p\cdot\hat{\mathbf s}}{4\pi\epsilon_0s^2}
=\frac{p\cos\theta}{4\pi\epsilon_0s^2}}
$$
to leading <electric dipole> order. In <spherical coordinates> aligned with $\mathbf p$, the <gradient> gives $E_s=-\partial_sV$, $E_\theta=-s^{-1}\partial_\theta V$ and $E_\phi=0$. Thus
$$
\boxed{\mathbf E(s,\theta)=\frac{p}{4\pi\epsilon_0s^3}
\left(2\cos\theta\,\hat{\mathbf s}+\sin\theta\,\hat{\boldsymbol\theta}\right).}
$$
Equivalently, the <electric dipole field> is $[3(\mathbf p\cdot\hat{\mathbf s})\hat{\mathbf s}-\mathbf p]/(4\pi\epsilon_0s^3)$. The omitted terms vanish relative to these leading expressions as $d/s\to0$.
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