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In Wikipedia https://en.wikipedia.org/wiki/Electric_dipole_moment it is said:

More generally, for a continuous distribution of charge confined to a volume V, the corresponding expression for the dipole moment is:

$\vec p(\vec r)=\int_V \rho (\vec r_0)(\vec r_0 -\vec r)dV_0$

where r locates the point of observation and d3r0 denotes an elementary volume in V.

From a mathematical point of view, I know that the we derive the above expression from the multipole expansion of a charge distribution. I am trying to have physical understanding of it.

In the easiest case for an electric dipole, we have :

$\vec p = q \vec a$ where $\vec a$ is a vector that points from the negative to the positive charge. As you can see here we have no dependency of $\vec p$ from $\vec r$.

In the expression for the continuous charge we do have the electric dipole dependent from $\vec p$. Does this vector $\vec r$ points in an arbitrary location in the volume where the charge distribution is located? Which would imply that for different $\vec r$ values we will have different values of the electric dipole? By different I mean different magnitude and different direction.

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The electric dipole moment of a charge distribution about a point ${\bf R}$ is independent of ${\bf R}$ only when the total charge is zero. In that case $$ p({\bf R})\equiv \int ({\bf r}-{\bf R}) \rho({\bf r})d^3r = \int {\bf r} \rho({\bf r})d^3r-\int {\bf R} \rho({\bf r})d^3r\\ = \int {\bf r} \rho({\bf r})d^3r-{\bf R}\int \rho({\bf r})d^3r\\ = \int {\bf r} \rho({\bf r})d^3r. $$ This means that only neutral objects have well-defined electric dipole moments. For anything else (an electron for example) you have to explain what point ${\bf R}$ you are using.

Electric dipole moment of electron: about what point is the moment taken?

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  • $\begingroup$ So if you have a continuous charge distribution over some volume, and the total net charge is zero, then you can have an electric dipole moment of the whole charge distribution, and you can depict it the same way you depict the dipole moment from 2 equally but opposite charges? $\endgroup$
    – imbAF
    Commented Nov 19, 2021 at 14:42
  • $\begingroup$ Yes. Exactly!... $\endgroup$
    – mike stone
    Commented Nov 19, 2021 at 16:58
  • $\begingroup$ Mike, thanks for the confirmation. May I ask one more question. I am currently studying dielectrics and in my script the following is said : $\rho_{total} = \rho_{external} + \rho_pol$ where $\rho_{total}$ is responsible for the electric field, (in the dielectric I assume), $\rho_pol$ is the source of the polarisation field and $\rho_{external}$ is the source of the electric displacement. What exactly is $\rho_{external}$, where does it come from? What it is meant with external charge density, is it outside the dielectric ? $\endgroup$
    – imbAF
    Commented Nov 19, 2021 at 17:09
  • $\begingroup$ Without access to your book I am not sure what $\rho_{external}$ means in this context, but most likely it is all charges other than those included in $\rho_{pol} =- \nabla\cdot {\bf P}$ arising from from the dielectric itself. $\endgroup$
    – mike stone
    Commented Nov 19, 2021 at 17:36
  • $\begingroup$ other charges, in the dielectric? or outside of it? How can a dielectric have external/free charges? $\endgroup$
    – imbAF
    Commented Nov 19, 2021 at 18:01

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