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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.

imbAF
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1 Answers1

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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?

mike stone
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