Adding dipole contribution and charge contribution to find electric potential

Question

I'm reading a book called Teoria do electromagnetismo by Kleber-Daum-Machado Volume 1. More especifically a section about the effect of electric fields in dielectric materials, in there he describes a dielectric that fills the space and that we are grabbing a small piece $\Delta V$ to see the electric potential generated by it, my confusion arises because he says that there will be two contributions 1 because of the charge enclosed in $\Delta V$ and the other because of the dipoles contained also in $\Delta V$, which would be: $$\Delta \phi(\vec{r})=\frac{1}{4\pi \epsilon_0}\frac{\rho(\vec{r})\Delta V}{|\vec{r}-\vec{r}^{\prime}|}+\frac{1}{4\pi \epsilon_0}\frac{(\vec{P}\Delta V)(\vec{r}-\vec{r}^{\prime})}{|\vec{r}-\vec{r}^{\prime}|^3}$$ Why does he add both terms, my intuition tells me that it would be enough by taking into account the charge inside that volume

Answer 1

The charge inside that volume is taken into account. There is however a field that is due to the electric dipoles. The total potential at any point therefore, will be the sum of the electric field due to the charge in the dielectric plus the field which is due to the electric dipoles in the dielectric.