= Solution
The <electric dipole moment> is $\mathbf p=2aq\hat{\mathbf z}$. By the <superposition principle>, the exact <electric potential> is
$$
V(\mathbf r)=\frac{q}{4\pi\epsilon_0}\left[\frac1{\sqrt{r^2+a^2-2ar\cos\theta}}-\frac1{\sqrt{r^2+a^2+2ar\cos\theta}}\right].
$$
For $a/r\ll1$, the <binomial expansion> gives each reciprocal <distance> as $r^{-1}(1\pm(a/r)\cos\theta+O(a^2/r^2))$. The leading $1/r$ terms cancel because the total <electric charge> is zero. The next even terms also cancel between the two signs. Consequently
$$
V(\mathbf r)=\frac{2aq}{4\pi\epsilon_0}\frac{\cos\theta}{r^2}+O\left(\frac{qa^3}{4\pi\epsilon_0r^4}\right),\qquad \boxed{\alpha=\frac{2aq}{4\pi\epsilon_0}=\frac{p}{4\pi\epsilon_0}.}
$$
This is the leading <electric dipole> term of the <multipole expansion>; it describes the far field, not the <electric potential> arbitrarily close to either <point charge>.
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