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Homework Help: Infinite sheet of dielectric

  1. Dec 2, 2011 #1
    Here is the problem from I.E.Irodov's Basic laws of electromagnetism.

    An infinitely large plate made of homogeneous linear isotropic dielectric with dielectric constant
    [itex]\epsilon[/itex] is uniformly charged by an extraneous charge (see footnote) with volume density [itex]\rho > 0[/itex].The thickness of the plate is 2a. Find the magnitude of [itex]\mathbf{E}[/itex] and the potential [itex]\varphi[/itex] as functions of distance x from the middle of the plate (assume that the potential is zero at the middle of the plate).

    Now the author has given the solution in the book . I am just trying to understand it. He says, "From symmetry considerations it is clear that E=0 in the middle of the plate, while at all other points vectors [itex]\mathbf{E}[/itex] are perpendicular to the surface of the plate"

    I am trying to see this from more mathematical arguments using the typical equations involved here.

    [tex]\vec{\nabla}\times \vec{E} =0[/tex]

    [tex]\vec{\nabla}\times \vec{D}=\vec{\nabla}\times \vec{P}[/tex]

    [tex]\vec{\nabla}\bullet \vec{D}=\rho_f [/tex]

    How do I proceed ?

    footnote: Extraneous charges are frequently called free charges , but this term is not
    convenient in some cases since extraneous charges may be not free ( this is author's footnote and I don't really understand what he means by it)
  2. jcsd
  3. Dec 3, 2011 #2

    Simon Bridge

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    Do you understand what the "symmetry considerations" here are?

    Free charges means not polarization charges.
    You want equation 3.
    Choose a coordinate system and write it out.
    Solve the resulting differential equations for each region (you have three). D has to be continuous across the boundaries but E does not.

    Consider: how would you go about this problem if you replace the dielectric with an insulator?

    "free charge" means "not polarization charges".
  4. Dec 3, 2011 #3
    I can see that sheet is symmetric about the axis going through the middle. But other than
    that I don't get symmetry considerations as the author says.

    So equation 3 would be kind of a Poisson's equation where [itex]\rho_f[/itex] is constant
    inside the dielectric, right ?

    Isn't an insulator a form of a dielectric ?

    so does it mean that extraneous charge may or may not be polarization charge ?
  5. Dec 4, 2011 #4

    Simon Bridge

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    Try sketching the situation with the field lines.

    The slab has "rectangular symmetry" which means you pick Cartesian coordinates.
    If you try it, you'll find that two of directions have vanishing contributions, so it turns into a 1D problem.

    I take it you have not done this for an insulator then?
    Usually these are given as an exercise before you do dielectrics.
    It normally goes: single-charge, charges on a conductor, uniform charge density through an insulator, then dielectrics.

    Charges don't move in an insulator, but simple insulators don't have polarization charges to speak of. It's a way of getting students to consider things other than a thin shell of charge.

    Suggest you go see how these are done - go back over your class notes.

    Have a look at these notes - scroll down for example 2.4.
    Or this problem.

    Once you know how to do it just for E, you can do it for D.
    Last edited: Dec 4, 2011
  6. Dec 5, 2011 #5
    Simon, is it possible to use symmetries about the laws of physics in general. For example, if I choose the axis going through the middle as z axis, and if we rotate this sheet about this z axis, then since the charge distribution around this z axis is symmetric, the field lines should be symmetric too. That would eliminate lot of weird lines. Also we can talk about the reflection in the mirror located at z=0. That should also keep the field lines invariant, since after the reflection, the charge distribution is symmetric.

    I had read somewhere about such arguments in physics. I don't know what they are called
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