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: Electromagnetism - Boundary conditions for Polarization field at interface

  1. Oct 30, 2011 #1
    Not actually a homework question, this is a question from a past exam paper (second year EM and optics):

    1. The problem statement, all variables and given/known data

    Use a Gaussian surface and Amperian loop to derive the electrostatic boundary conditions for a polarization field P at an interface between media 1 and 2 with relative permittivities εr1 and εr2.

    2. Relevant equations

    D = ε0E + P,

    [itex]\nabla[/itex].P = -ρb

    where D is the displacement field, E the electric field and ρb the bound volume charge density.

    3. The attempt at a solution

    Lack of a solution is the reason for posting?

    Thanks in advance.
  2. jcsd
  3. Oct 31, 2011 #2
    Ok, after obtaining a copy of Feynman's lectures and referring to Volume 2, Chapter 33-3, I think I have found a solution.

    Treating the boundary as a separate region (media 3 with a relative permittivity that begins at εr1 and changes continuously to εr2) then since the P field is different in each region, in region 3 (the boundary) there is a [itex]\delta[/itex]Px/[itex]\delta[/itex]x where Px is the P field in the x direction. So;

    d/dx (Dx) = d/dx ( (ε0Ex) + (P)) ... (partial derivatives with respect to x [can't use latex very well]

    D doesn't change in materials so:

    -d/dx (Exε0) = d/dx (Px)

    Integrating each side with respect to x over region 3 and letting P2 be the polarisation in εr2 region and P1.. :

    Px2-Px1 = -ε0(Ex2 - Ex1)

    Also there is no B divergence from Maxwell so:
    B1 = B2 (for all directions)

    Also from Maxwell:

    curl(E) = -dB/dt (again partials)

    Which gives the following set:

    dEz/dy -dEy/dz = -dBx/dt
    dEz/dx - dEx/dz = -dBy/dt
    dEy/dx - dEx/dy = -dBz/dt

    The E field only changes in the direction so only the following is considered:

    dEz/dx = -dBy/dt
    dEy/dx = -dBz/dt

    Now if E were to change the right hand side of the equation would have to be balanced by a change of B with respect to time which does not happen so:

    Ez1 = Ez2 and Ey1 = Ey2

    So I have the boundary conditions of a dielectric interface. The boundary conditions on P are then just found by substituting P = E(εr-1)ε0:

    (1) Px2-Px1 = -ε0(Ex2 - Ex1)
    Px2 - Px1 = (-Px2/(εr2-1))+(Px1/(εr1-1))
    Px1εr1 / (εr1-1) = Px2εr2 / (εr2-1)

    (2) B1 = B2

    (3) (P1/(εr1-1))y = (P2/(εr2-1))y

    (4) (P1/(εr1-1))z = (P2/(εr2-1))z

    Hoping someone can verify this for me. Fairly sure about the derivation of boundary conditions; it is essentially Feynman's derivation modified for two materials (he does it for vacuum to material).

    However, not sure if the substitution of P is the right thing to do - does this form constitute an answer to the question or is it asking something else?

    Again thanks in advance.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook