Electric field in matter- Bound Charges why the term $\rho_b$ exist? I am reading the book of Griffiths, Introduction to electrodynamics, and he explains in this chapter about the bound charge densities $\sigma_b$ and $\rho_b$
but I do not understand how is it possible that the term $\rho_b$ can exist. In my head I imagine that in the moment there is an external constant electric field, all the negative charges will be in one direction and the positive in the opposite direction to the negative, wich means that there are no charges inside the body, but only on the surface, so I think there is only term of $\sigma_b$.
I do not understand what is wrong in my assumption.
 A: You have the correct intuition for the case of constant polarization density $\vec{P}$, but not when $\vec{P}$ has a non-zero divergence: recall that $\rho_b = -\vec{\nabla}·\vec{P}$.
As a concrete example consider the 1D problem in which we have a slab in $-a\le x \le a$ with
$$\vec{P}=\hat x P_0\left(1-\frac{x^2}{a^2}\right)$$
$$\rho_b=-\vec{\nabla}·\vec{P}=\frac{2P_0x}{a^2}.$$
The polarization charge is negative for $x<0$ and positive for $x>0$. Why? Because the positive and negative charges in nearby dipoles do not cancel out exactly.
Picture the dipoles as positive and negative charges separated by some fixed distance. Try drawing the dipoles in a diagram to see what I mean. For $x < 0$, a dipole at $x$ has a smaller positive charge than the negative charge of a dipole at $x + \delta x$: the result is a net negative bound charge density for $x < 0$.
A: Your reasoning is correct when the medium is a conductor.  However, in an insulating medium it is possible for there to be a charge density in which the charges cannot move.
