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Continuity of the first Maxwell equation.

  1. Nov 3, 2007 #1
    Suppose that we will proof the continuity of the first maxwell equation:

    So we have [tex]div(\vec{E})=\frac{1}{\epsilon _0} \rho [/tex] than [tex]\iiint \ div(\vec{E}) = \oint_v \vec{E} d\vec{s}=\iiint \frac{1}{\epsilon _0 } \rho [/tex]
    than follewed [tex]E_{y1} l -E_{y2}l=Q [/tex]

    Therefore E must continue is this a good proof? Thanks.
  2. jcsd
  3. Nov 3, 2007 #2


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    You have not defined what "y" direction is. So your y components are ambiguous.

    Only the parallel component of the E-field (parallel to the surface) is continuous.

  4. Nov 3, 2007 #3
    Thanks if I define my y component as the parallel one. Is than the proof oké?
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