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Differential form of Gauss' theorem with dielectrics

  1. May 23, 2014 #1
    Hi all,
    I'm stuck on this incompatibility within the differential form of Gauss' thearem (or Maxwell's first equation) with dielectrics.

    [itex]\vec{\nabla}\cdot\vec{E}=\frac{\rho_{free}+\rho_{bound}}{\epsilon_{0}}[/itex]

    [itex]\rho_{bound}=-\vec{\nabla}\cdot\vec{P}[/itex]





    But with a linear, homogeneous, isotropic dielectric we have

    [itex]-\vec{\nabla}\cdot\vec{P}=0=\rho_{bound}[/itex]




    And therefore we get

    [itex]\vec{\nabla}\cdot\vec{E}=\frac{\rho_{free}}{\epsilon_{0}}[/itex] (1)




    But using the general formula we have

    [itex]\vec{\nabla}\cdot\epsilon_{0}\vec{E} + \vec{\nabla}\cdot\vec{P}=\rho_{free}[/itex]

    [itex]\vec{P}=\epsilon_0\cdot\chi_0\vec{E}[/itex]



    So ([itex]1+\chi_e=\epsilon_r[/itex], [itex]\epsilon_0\cdot\epsilon_r=\epsilon[/itex])

    [itex]\vec{\nabla}\cdot\epsilon\vec{E}=\rho_{free}[/itex],

    which means

    [itex]\vec{\nabla}\cdot\vec{E}=\frac{\rho_{free}}{\epsilon}[/itex]

    and is incompatible with (1).


    Where is the mistake?

    Thank you in advance,
    Ocirne
     
  2. jcsd
  3. May 23, 2014 #2

    TSny

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    Hi, Ocirne94. Welcome to PF!

    I don't think there is any mistake. But you did assume that ##\rho_{bound} = 0##. You have essentially proven an interesting fact about linear, homogeneous, and isotropic dielectrics whenever ##\rho_{bound} = 0##. There is only one way for your two equations for ##\vec{\nabla} \cdot \vec{E}## to be compatible. What is it?
     
  4. May 23, 2014 #3
    where did [itex]-\vec{\nabla}\cdot\vec{P}=0=\rho_{bound}[/itex] come from?
     
  5. May 23, 2014 #4
    Hi,
    thank you for your answers.

    There are two ways for those two equations to be compatible inside the dielectric: the first is that
    [itex]\epsilon = \epsilon_0[/itex]
    which I would interpret as saying that the dielectric isn't there.

    The other one is that the divergence is zero in both cases, which implies
    [itex]\rho_{free}=0[/itex]
    inside any linear, homogenous and isotropic dielectric.
    In other words, no LIH dielectric can have free electrons inside its volume.
    Now this is somewhat disturbing (so much that in three hours of messing around the problem I have never considered this possibility), but eventually consistent with the fact that
    [itex]\vec{E}\propto\vec{P}[/itex]
    and I have started from
    [itex]\vec{\nabla}\cdot\vec{P}=0[/itex]

    Thank you very much for your illuminating answer! :smile:
    Ocirne
     
  6. May 23, 2014 #5
    You still haven't explained why you think the divergent of P is zero.
     
  7. May 23, 2014 #6

    TSny

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    Keep in mind that you assumed ##\rho_{bound} = 0## in your analysis. So, what you showed is that if ##\rho_{bound} = 0## then ##\rho_{free} = 0## also.

    You should be able to show the converse.

    So, in a "Class A" dielectric (i.e., linear, homogeneous, and isotropic), ##\rho_{free} = 0## ⇔ ##\rho_{bound} = 0##.

    But, as dauto is suggesting, it is not necessarily true that these charge densities are both zero. But if one is zero, the other is zero. If one is nonzero, then the other is nonzero. You could certainly imagine that some free charge density is embedded inside a dielectric, so that ## \rho_{free} \neq 0##. Then you would also have ##\rho_{bound} \neq 0##.

    You might try to show ##\rho_{bound} = - \frac{\epsilon_r-1}{\epsilon_r} \rho_{free}##.
     
  8. May 24, 2014 #7
    Hi,
    I was taught that (as a rough model with some approximations) within a LIH dielectric the dipoles get a uniform orientation: this implies that no net bound charge can be found inside the dielectric's volume, and only surface charge is present.


    I definitely agree that the maths allows nonzero charge densities, but in this case I face a problem with definitions: (quoting from my and many other textbooks), "in dielectrics all the charges are bound to specific atoms and molecules".
    So considering a free charge density within a dielectric, hence a bound charge density, looks like violating the definition: this likely enters the domain of the microscopic structure of matter and of the particles' interaction, but I would say that the volume taken by a free charge density within a dielectric cannot be considered dielectric.
     
  9. May 24, 2014 #8

    TSny

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    Free charge in a dielectric is just additional charge added into the material. For example, you could imagine an electron beam that sends electrons into a dielectric material and that these electrons become entrapped in the material with some volume charge density ρf that might vary with position inside the material. These electrons would be considered "free charge" that is not due to polarization of the molecules of the dielectric. Admittedly, this is not a typical situation for dielectric materials. But in such a situation I think you would consider the free charge and dielectric as occupying the same region.

    If you embed small, but macroscopic-sized charged particles into a dielectric, then I think your are right that you could consider these particles as not occupying the same region as the dielectric but, rather, as particles that are surrounded by the dielectric material.

    Anyway, you will see problems in standard textbooks where you have dielectrics that contain some specified free volume charge density.
     
  10. May 24, 2014 #9
    That's only true if the electric field is uniform. If the electric field varies from spot to spot in the dielectric - that is it is a function of the coordinates - than there will be non-zero net bound charges in the dielectric. if there are embedded "free" charges in the dielectric those free charges produce no uniform electric field that induce a net bound charge in the dielectric that partially shields the free charge.
     
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