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From wikipedia, the electric displacement field is $\vec D=\epsilon_0 \vec E+ \vec P$ and it satisfies $\nabla \cdot D=\rho-\rho_b$ where $\rho_b$ is "the density of all those charges that are part of a dipole, each of which is neutral".

If these equations works at macroscopic level (by which i mean that every quantity is the result of an average over a small volume containing a huge number of particles) then why $\rho_b$ is not zero?

If any particle is neutral then every small volume contains zero net charge and so $\rho_b \triangleq dQ/dV=0$

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  • $\begingroup$ You have to take into account that any volume, whether finite or infinitesimal, carrying effective volume charge density $\rho_b = - \nabla \cdot P$ also has a corresponding surface charge density $\sigma_b = n \cdot P$, where $n$ is the normal to the boundary of the volume. When integrating total charge, these two contributions always cancel by Gauss theorem. $\endgroup$
    – PolyPhys
    Commented Apr 25, 2022 at 22:17

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As you stated:

$\rho_b$ is "the density of all those charges that are part of a dipole, each of which is neutral".

This means that the dipole as a whole is neutral.

But there can still be a charge distribution with non-zero values internally for the dipole. For example, two charges of equal magnitude but opposite value (e.g., $q$ and $-q$) separated by a distance $d$.

Why ρb is not zero? If any dipole is neutral then at macroscopic level it should always be zero

The dipole is not zero (not even "at macroscopic level"). The total charge of the dipole is zero, but the dipole is not "zero"--it is a dipole.

Update, based on chat:

It is true that you could can often choose some volume where the total charge is zero, but that is not the only thing that matters. For example, suppose you have a whole bunch of dipoles lined up in a row (which you actually can/do have in macroscopic matter). Inside the material there are a lot of different volumes that you could choose where the charge inside the volume is zero.

But, nevertheless, at the surfaces, there must be bound charge (think about a whole line of dipoles all in a row facing the same direction, at the very ends of the line there is uncompensated charge). (In a similar polarized volume this would mean that the bound charge is vanishing except for at the surface.)

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  • $\begingroup$ Since every single particle is made by a positive and a negative part that equals each other, then every small volume containing a bunch of particles should be neutral. What is wrong with this reasoning? $\endgroup$
    – SimoBartz
    Commented Apr 25, 2022 at 19:48
  • $\begingroup$ Sure, every "small volume" that contains an equal number of positive and negative charges contains net zero charge. So what? The distribution of charge still matters, even though the total charge in the "small volume" is zero. $\endgroup$
    – hft
    Commented Apr 25, 2022 at 19:51
  • $\begingroup$ so the macroscopic charge density should be zero, it's basically $dQ/dV$ $\endgroup$
    – SimoBartz
    Commented Apr 25, 2022 at 19:52
  • $\begingroup$ No it is not and it should not. Consider, for example, an infinitely thin dipole layer. If you take any "small volume" that contains a single dipole, there will be zero net charge, and there will be no MONOPOLE MOMENT. But there will be a DIPOLE MOMENT. Sure, it is very small, but in a dipole layer, those very many very small dipole moments matter. $\endgroup$
    – hft
    Commented Apr 25, 2022 at 19:56
  • $\begingroup$ if with a single dipole you mean a single particle then you can't take a small volume containing just one particle, it would be a microscopic approach. Here i think we are treating equations at macroscopic level $\endgroup$
    – SimoBartz
    Commented Apr 25, 2022 at 20:00

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