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Dielectric boundaries and symmetry

  1. Jul 21, 2003 #1
    I have been reviewing the method of images for electrostatic problems and noticed the following:

    Textbooks point out that to use Gauss' Law successfully there must be geometric symmetry that makes the Efield constant over some reasonable surface. I thought this was just to make the math tractable but then I noticed the following issue.

    1. Point charge inside a thin spherical shell of a different dielectric constant. You can calculate the Efield inside without any knowledge of the shell using Gauss' Law.

    2. Point charge near a planar dielectric boundary. You must know the distance from the charge to the boundary and both dielectric constants to calculate the Efield on the charge side of the boundary.

    In one case you only need to know about the region of interest, in the second case you need to know everything. Is symmetry alone the difference?

    For the spherical shell I can see that the field due to any polarization charge will 'cancel out' inside the shell and so the Efield is the same as if there were no shell. What about other cases? How about an infinite planar charge and a parallel dielectric boundary?
     
  2. jcsd
  3. Jul 21, 2003 #2
    Inquiry

    Hi mmwave:

    "...point charge inside a thin spherical shell of a different dielectric constant. You can calculate the Efield inside ..."

    Allow me give my explanation of why the E field inside a sphere is zero. It is essentially because the E field scales as 1/r^2 whereas the "area", or "solid angle" you are exposed to scales as r^2. Let me try and clarify.

    {
    When inside a spherical shell, draw any cone (both nappes) centered on your position. The nappe of the cone intersecting the surface closest to your position will enclose a small area, and the nappe opposite to that will enclose a larger area. The E field generated at your position due to both these areas will be such that they will EXACTLY cancel. So, symmetry (specifically, the spherical symmetry of the shell (r^2) and the spherical symmetry of Gauss' Law (1/r^2) is responsible for the net effect.
    }

    "...how about an infinite planar charge and a parallel dielectric boundary"

    The electric field of an infinite plane is a constant. It does not depend on the distance to the plane. If the charge is in some dielectric medium (i.e. epsilon != epsilon_0), simply replace epsilon_0 with epsilon.

    "...the second case [infinite plane] you need to know everything."

    Not quite. Please clarify.

    ---
    I am sure I have misunderstood your question. Your question seems to be asking a deeper question that I am not getting. Please try and re-phrase it (P.S. make quotes from Griffiths or Jackson if that is what you are reading).
     
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