# Div D, div E. What rho is it. (electrostatics)

1. Mar 3, 2013

### fluidistic

I know that the D field has to do with free charges. So when we write Maxwell's equation $\vec \nabla \cdot \vec D = \rho$ we mean rho as the free charge density which does not include the bound charges also known as polarized charges.
I know that the E field has to do with both the free charges and bound charges. However if the region I'm concerned about is an isotropic, linear and homogeneous dielectric then $\vec D = \varepsilon \vec E$ and thus $\vec \nabla \cdot \vec E = \frac{\rho }{\varepsilon }$.
Where rho in this case is also the free charge density.
But I'd have thought that $\vec \nabla \cdot \vec E = \frac{\rho _\text{free}+ \rho _{\text{bound}}}{\varepsilon }$. I'm guessing this equation is not false, but it's just that the nature of the dielectric makes that $\vec \nabla \cdot \vec P=0$ (which is also worth $-\rho _\text{bound}$). So that means that in an isotropic, linear and homogeneous dielectric there are no bound charges?

2. Mar 3, 2013

3. Mar 3, 2013

### fluidistic

Yes you are right, I've just rederived it via $\vec D = \varepsilon _0 \vec E - \vec P$, taking the divergence.
Therefore I'm left to see that $\frac{\rho _\text{free}+ \rho _{\text{bound}}}{\varepsilon _0 } = \frac{\rho _\text{free}}{\varepsilon}$.
I can already answer a question in my first post, namely "So that means that in an isotropic, linear and homogeneous dielectric there are no bound charges?", the answer is no, because div P is not worth 0.