Dipole moment in an electric system when the charges are not symmetrically distributed

Question

Suppose we have two -q charges and +2q charge ate the corners of an equilateral triangle. Now to calculate the dipole moment suppose we assume that the +2q charge is 2 charges of +q magnitude. My doubt arises here. If we assume that, then shouldn't we have 4 dipole pairs.(each -q charge forms two pairs with the +q charges). Hence the dipole moment calculated will be different compared to calculating the dipole moment by finding the "centre of charge" (Like com) .

Answer 1

Say the three charges are at points $\mathbf{x}_1=(0,d)$, $\mathbf{x}_2=(\sqrt{3}~d/2,d/2)$ and $\mathbf{x}_3=(-\sqrt{3}~d/2,d/2)$, where $d$ is the distance from the centroid of the triangle to each vertex. By definition, the electric dipole moment is \begin{align} \mathbf{P} &= 2 q~\mathbf{x}_1 -q~\mathbf{x}_2-q~\mathbf{x}_3\\ &= (0,qd). \end{align} Now say I split the $2q$ charge into two, each with charge $+q$. The position of the new charge (call it $\mathbf{x}_0$), would be the same as the original $2q$ charge, i.e. $\mathbf{x}_0 = \mathbf{x}_1 = (0,d)$. So \begin{align} \mathbf{P} &= q~\mathbf{x}_0 + q~\mathbf{x}_1 -q~\mathbf{x}_2-q~\mathbf{x}_3\\ &= q~2\mathbf{x}_1 -q~\mathbf{x}_2-q~\mathbf{x}_3\\ &= (0,qd), \end{align} which is the same, as expected.

Note that the connection of the dipole moment to the centroid of the geometry (for equal charges) is only true if the geometry is non-degenerate, i.e. all vertices of the polygon formed by the point charges have to be distinct. Another way to think about it is this: you can perturb the position of the extra $+q$ charge in the second case by some small constant $\epsilon$ in any direction, i.e. $\mathbf{x}_0 = (\epsilon,d) \neq \mathbf{x}_1$. Now you have a non-degenerate quadrilateral with a centroid \begin{align} \bar{\mathbf{x}} = \frac14 \sum_{i=0}^3\mathbf{x_i} = \frac14 (\epsilon,3 d), \end{align} which will be parallel to $\mathbf{P}$ if you now take the $\epsilon\rightarrow 0$ limit.