Let $N$ identical electric charges be arranged arbitrarily in space, in such a way that each of them has a fixed position $\vec{r}_i$, a charge $q$ and a mass $m$. If I were to calculate the total dipole moment of said distribution ($\vec{p}_T$), then I could proceed as follows, taking into account dipole moments are additive magnitudes:
$$\vec{p}_T=\displaystyle\sum_{i=1}^N \vec{p}_i=\displaystyle\sum_{i=1}^N q_i\vec{r}_i$$
Since the charges are, by hypothesis, identical, they have the same charge and mass, so I can take $q$ out of the summation. Also, I'm interested in writing this sum in terms of the position of the center of mass of the distribution:
$$\vec{p}_T=q\left(\frac{1}{M_T}\displaystyle\sum_{i=1}^N m\vec{r}_i\right)\frac{M_T}{m}$$
Where $M_T$ is the total mass of the distribution. The term inside the parenthesis can easily be identified as the system's center of mass ($\vec{r}_{CM}$), so I can write this expression more compactly as follows:
$$\vec{p}_T=q\ \vec{r}_{CM}\frac{M_T}{m}$$
since the particles are identical, they all contribute the same to the system's total mass, so evidently we have: $M_T=Nm$. Substituting this into our equation:
$$\vec{p}_T=Nq\ \vec{r}_{CM}$$
And if we define $\vec{p}_{CM}\equiv q\vec{r}_{CM}$ as the dipole moment possessed by a particle of charge $q$ such as those present in the system and placed in the center of mass, we then get:
$$\vec{p}_T=N\vec{p}_{CM}$$
Therefore, although each charge in my system is randonly oriented, the total dipole moment is identical to another system with the same number of particles in which all of the dipoles are oriented in the same direction. What are the implications of this?
