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[Electromagnetics] Dielectric boundary condition

  1. Dec 19, 2015 #1
    http://local.eleceng.uct.ac.za/courses/EEE3055F/lecture_notes/2011_old/eee3055f_Ch4_2up.pdf [Broken]

    (Page 4.4 )

    I am having a trouble with understanding why closed loop line integration is 0 at dielectric boundary.
    As far as I know, closed loop line integration is 0 because electric field is conservative.

    However, if we have a boundary condition, we'd have two electric vector fields where each has different
    permittivity epsilon_1, epsilon_2. and also directional components would be different due to refraction.

    If we break down the line integral into two parts of closed loop line integration, one for the top and the other for the bottom dielectric region, the middle boundary part will cancel out and the result Et1=Et2 makes sense.
    Also if we do the two closed loop line integration for top and bottom dielectric region, we can show that Et1 and Et2 are equal to the electric field at the boundary for any arbitrary closed loop.

    *However, it seems a little conunter intuitive to me because E1t and E2t would be diffrent due to refraction. How can I justify the result physically?
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Dec 19, 2015 #2
    Look at the first equation on 4-2 right-hand side. The surface integral is zero for an infinitesimally small loop since the surface area can't collect any time-varying B field flux.

    A time-varying E-field is not a conservative vector field (it does not have a zero curl). This is in contradiction to electrostatics where the curl of E is always zero. This was one of the major discoveries of Faraday.

    I think you are confusing the reason (boundary conditions) with the result (Snell's law of refraction). Remember tangential E (and tangential H) are always continuous on the boundary of homogeneous media without a surface current. The normal components follow as a result.

    Of course is you are an electromagnetic purist you would say that the normal components of D and B are always conserved across any interface... :rolleyes:
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