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The potential due to a system of point charges \(q_k\) at position \(\vec{r}_k\) is given by
\begin{align}\label{EQ01} \phi(r) = \sum \frac{q_k}{4\pi\epsilon_0}~\frac{1}{|\vec{r}-\vec{r}_k|}. \end{align} A dipole consist of two equal and opposite charges \(q\) and \(-q\). For a dipole we get \begin{align} \phi(\vec{r})=&\frac{q}{4\pi\epsilon_0|\vec{r}-\vec{r}_1|}-\frac{q}{4\pi\epsilon_0|\vec{r}-\vec{r}_2|}\\\nonumber =&\frac{q}{4\pi\epsilon_0}\left[\frac{1}{(r^2+r_1^2-2\vec{r}\cdot\vec{r}_1)^{1/2}} ~-~\frac{1}{(r^2+r_2^2-2\vec{r}\cdot\vec{r}_2)^{1/2}}\right]\\\label{EQ05} =&\frac{q}{4\pi\epsilon_0}\left[\frac{1}{r}\left(1-\frac{2\vec{r}\cdot\vec{r}_1}{\vec{r}^2} +\frac{r_1^2}{r^2}\right)^{-1/2} - \left(1+\frac{r_2^2}{r^2} - \frac{2\bar{r}\cdot\bar{r}_2}{r^2}\right)^{-1/2}\right]\\\nonumber =&\frac{q}{4\pi\epsilon_0r}\left(1+\frac{\vec{r}\cdot\vec{r}_1}{r^2} - 1 - \frac{\bar{r}\cdot\bar{r}_2}{r^2}\right)+\cdots\\\nonumber \approx&\frac{q}{4\pi\epsilon_0r}~\frac{\bar{r}\cdot(\vec{r}_1-\vec{r}_2)}{r^2}\\\label{EQ02} =&\frac{\vec{p}\cdot\vec{r}}{4\pi\epsilon_0r^3} \end{align}
where \(\vec{p}=q(\bar{r}_1-\bar{r}_2)\) is the dipole moment. If \(\theta\) is the angle between the vectors $\vec{p}$ ~and $\vec{r}$, then
\begin{equation} \phi(\vec{r})=\frac{p\cos\theta}{4\pi\epsilon_0 r^2}. \end{equation}
The above computation can be repeated for potential due to the system of \(n\) charges, as in \eqRef{EQ01}. The potential at large distances will be given again by \eqRef{EQ02} with the dipole moment given by \begin{equation} \vec{p} = \sum_k q_k \vec{r}_k. \end{equation} If the dipole moment is zero, we will need to keep the next non zero term in the Taylor expansion step \eqRef{EQ05}.
