1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Boundary between dielectrics

  1. Mar 13, 2008 #1
    1. The problem statement, all variables and given/known data
    Say I have a boundary between two dielectrics then it's easy to show using a gaussian pillarbox that:
    D(1)-D(2)=free surface charge density=s
    where D(1) is the component of the first medium normal to the surface.
    But suppose that there's nothing else apart from two infinite dielectrics with a constant free charge density between then, how would I work out what the actual values of D(1) and D(2) are rather than just the difference?

    3. The attempt at a solution
    It seems reasonable that the D field should be the same on both sides so that D=s/2 but I'm not sure how I'd prove this?

    Any help appreciated.
  2. jcsd
  3. Mar 13, 2008 #2
    I'm having a hard time interpreting what you are asking. Could you draw it?
  4. Mar 13, 2008 #3
    I think my question was badly worded/ ive probably misused some terminology.

    I suppose my actual qustion is:
    Its obvious from the symmetry that for an uncharged nonconducting non-dielectric that if you put a surface charge density s on it the field above is the same magnitude as the field below: D=s/2.

    But suppose that the material is a dielectric then I'm not sure how you prove what the D field is above the surface (in air) and below the surface (in the dielectric) (under the same condition of the dielectric being infinite with surface charge s and no other free charge anywhere)

    Hope this is a bit clearer, please say if it isn't.
  5. Mar 13, 2008 #4
    Oh, well D is going to be the same for both air and the dielectric, but E will be different because E is D/(epsilon) which will change between the two (epsilon_0 for air and dielectric epsilon in the dielectric). Is this what you are asking about?
  6. Mar 13, 2008 #5
    That's it.

    Its probably really obvious but how do you know that D has to be the same on both sides of the boundary?
  7. Mar 13, 2008 #6
    It comes from the way D is defined. After its whole redefinition then you just get a Gauss's Law for materials which goes as

    [tex]\iint \mathbf{D} \cdot d\mathbf{a} = \sigma_f[/tex]

    and that means that the only thing that you care about is the free charge, which will be the same on both sides.
  8. Mar 13, 2008 #7
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook