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

  1. Mar 3, 2013 #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?
     
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  3. Mar 3, 2013 #2

    atyy

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  4. Mar 3, 2013 #3

    fluidistic

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    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.
     
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