# Electrostatic field in a Dielectric equations misbehaving

Equation 1: $\int{\vec{E}.\vec{ds}} = \int\frac{\rho_{free} + \rho_{bound}}{\epsilon_0} dv$ (Gauss's Law)

Equation 2: $\int{\vec{D}.\vec{ds}} = \int\rho_{free} dv$ (Gauss's Law)

, but $\vec{D} = \epsilon_0 \epsilon_r \vec{E}$ , so equation 2 becomes $\:\:\:\epsilon_0 \ \epsilon_r\int{\vec{E}.\vec{ds}} = \int\rho_{free} dv$

or, $\int{\vec{E}.\vec{ds}} = \int\frac{\rho_{free}}{\epsilon_0 \epsilon_r} dv$

But if we compare this last equation with Equation 1 , we get $\rho_{free} = \epsilon_r(\rho_{free} + \rho_{bound})$

which doesn't make sense because there are situations where there is only a "frozen in " volume bound charge and no free charges. Whereas the last relation I derived says there cannot be a bound charge without a free charge.

What am I doing wrong?

You're not doing anything wrong. To make the most of your equation, $\rho_\text{free}=\epsilon_r(\rho_\text{free}+\rho_\text{bound})$, it is best to rearrange it as $$\rho_\text{bound}=\frac{1-\epsilon_r}{\epsilon_r}\rho_\text{free}= -\frac{\chi_\text e}{1+\chi_\text e}\rho_\text{free}.$$ This equation expresses the fact that a free charge in a dielectric will polarize the medium around it and this will cause a concentration of bound charge around the free charge which will make it produce an effectively smaller electric field (E field) than it would have done in vacuum. This is expressed very well by Fig. 4.22 in Griffiths:
A positive point charge $q$ will attract the negative ends of the dipoles in the dielectric, which will congregate around it and create a 'wall' of charge which will make the effective charge (as seen by the E field) smaller. The size of this wall of charge will depend on how easy it is to polarize the medium but it will never be quite enough to cancel out the original charge.
Regarding your edit, it is important to note that the constitutive relation $\mathbf D=\epsilon \mathbf E$ only holds for linear dielectrics, and that for these the absence of free charges means there is no E field, and therefore no polarization, bound charges, or D field. Certain materials, known as ferroelectrics, can indeed have frozen-in polarization, so that they have nonzero bound charges and a nonzero electric field even without free charges. Evidently, these are not linear dielectrics.