- #1
fluidistic
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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?
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?