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Homework Help: How do I get this interface condition?

  1. Jan 10, 2018 #1
    1. The problem statement, all variables and given/known data
    I have an infinitely long cylinder of a dielectric material, surrounded by another dielectric material and coated with graphene which has surface conductivity [itex]\sigma[/itex], implying it has a superficial current. The sheet of graphene is very thin, and the dielectrics are asumed to be isotropic. There's an incident p-polarised monochromatic plane wave in the x direction, and I have to show the interface conditions are:
    [tex]\frac{1}{\epsilon_1} \frac{dF_1}{dr} = \frac{1}{\epsilon_2} \frac{dF_2}{dr} [/tex]

    [tex]F_2 - F_1 = \frac{4i \pi}{\omega \epsilon_1} \sigma \frac{dF_1}{dr}[/tex]

    where F is the H field in the z direction, the subindex 1 indicates the material inside the cylinder and 2 outside the cylinder. I'm having trouble with the second one.

    2. Relevant equations
    The boundary conditions given is that the tangential component of E (tangential to the normal vector of the surface of the cylinder) is continuous, and that the tangential component of H is proportional to the surface conductivity density.

    I have tried using Maxwell equations: [tex]\nabla \times H = \frac{4 \pi}{c}J - \frac{i \omega}{c} \epsilon E[/tex]

    3. The attempt at a solution
    The [itex]\phi[/itex] component is: [tex]-\frac{dH_z}{dr}= \frac{4 \pi}{c}J_{\phi} - \frac{i \omega \epsilon}{c} E_{\phi}[/tex]

    If I do the integral between R- and R+ (R is the radius of the cylinder) then the left side would look like [itex]F_2 - F_1[/itex], which is what I'm looking for, but what about the right side? I feel like I have to use the boundary conditions given, but I'm not sure how. Please help! Thank you!
  2. jcsd
  3. Jan 16, 2018 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
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