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Charged wire inside a linear dielectric

  1. May 10, 2014 #1
    1. The problem statement, all variables and given/known data

    see attachment 2

    2. Relevant equations

    a.


    3. The attempt at a solution

    see atachment 1


    i am stuck at problem a. the question is to write down the boundary conditions for that the E field has to satisfy infinetly far away from the wire and at the boundary of the dielectric z = 0.

    I'm not sure how to do this, any help is welcome

    thanks

    Jenny
     

    Attached Files:

  2. jcsd
  3. May 10, 2014 #2
    Dat mag niet. Groetjes B. N.
     
  4. May 10, 2014 #3

    vanhees71

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    You have the following equations, valid everywhere
    [tex]\vec{\nabla} \times \vec{E}=0, \quad \vec{\nabla} \cdot \vec{D}=\rho, \quad \vec{D}=\epsilon \vec{E}.[/tex]
    In your case you have
    [tex]\epsilon=\Theta(z)+\epsilon_{r} \Theta(-z).[/tex]
    To get the boundary conditions you just apply Stokes's theorem to the first equation with an arbitrary rectangular surface perpendicular to the surface. This leads to the condition that the tangential components of the electric field must be continuous across the surface, i.e., in your case
    [tex]\vec{e}_z \times [\vec{E}(x,y,0^+)-\vec{E}(x,y,0^-)]=0,[/tex]
    because [itex]\vec{e}_z[/itex] is the normal vector of the dielectric's boundary surface.

    Further you use an infinitesimal cube parallel to the boundary on the 2nd equation. This yields the boundary condition
    [tex]\vec{e}_z [\vec{D}(x,y,0^+)-\vec{D}(x,y,0^-)] = \sigma.[/tex]
    Here [itex]\sigma[/itex] is the surface charge density on the surface of the dielectric.
     
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