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Electrostatic boundary conditions

  1. Mar 8, 2006 #1
    Im having trouble following how this is derived: The normal component of the electric field is discontinuous by an amount sigma/epsilon_0 at any boundary (when you cross a continuous surface charge). They talk about taking a little box so that the surface integral E dot da = 1/epsilon_0 * sigma * A (where A is area parallel to surface charge) and making its width perpendicular to the surface charge very small. Somehow they get that this implies E_perpendicualAbove -E_perpendicularBelow = 1/epsilon_0 * sigma. How's this? And also, they go on saying that in cases like the surface of a uniformly charged solid sphere this doesnt apply because there is no surface charge, but I dont get this...what about the edge of the sphere, its still charged. So please any clarification will help, as i have a test tomorrow.
     
  2. jcsd
  3. Mar 8, 2006 #2

    Galileo

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    It's just an application of Gauss' law applied to a small pillbox straddling the surface. Looking only at the perpendicular component of E, the sides contribute nothing to flux. The flux through the pillbox is [itex](E_\perp^{above} -E_{\perp}^{below})A[/itex]. The enclosed charge is sigma*A, so by Gauss' law you get the result.

    In a continuous volume charge distribution the E field is always continuous.
     
  4. Mar 8, 2006 #3

    Meir Achuz

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    The Gaussian pillbox has infintesimal thickness \Delta t, so the charge enclose at the surface of a uniformly charged sphere goes to zero
    with \Delta t.
     
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