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Dicontinuity of electric displacement normal to a boundary

  1. Jun 19, 2010 #1
    1. Hi i need help with this question,

    Show, using maxwell's equations as a starting point, that the discontinuity in the component of the electric displacement normal to a boundary between different media is equal to the free surface charge density on a boundary.

    i have tried by using the integral form and integrating over a cylindrical pill box (define top lid= A1, bottom lid = A2 cylinder wall = A3) to on the surface (a flat plane with constant charge density)
    [tex]\oint E . dA[/tex] = [tex]\int[/tex] E . dA1 + [tex]\int[/tex] E . dA2 + [tex]\int[/tex] E . dA3

    last integral is zero as the electric field is perpendicular to the area

    but i am not sure as to whether i can re-use maxwell's equations as an expression for the electric field that doesn't seem rigorous and
     
  2. jcsd
  3. Jun 19, 2010 #2
    You should apply this equation: [tex]\oint \vec{D}d\vec{A}=Q_{free}[/tex]. The problem is about electric displacement, right?
    Why is the electric field perpendicular to the area? Does the problem state this?
    Oh and notice that the height of the cylinder is very small compared to the dimenions of the lid.
     
  4. Jun 19, 2010 #3
    Take the displacement above to be D1 and below to be D2 and use Gauss law.

    Take a rectangular pill box (that's up to you) and apply Gauss law in the limit that the height of the box approaches zero. (You want the difference in D just above the surface and D just below the surface)
     
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