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Boundary condition of EM field

  1. Dec 31, 2008 #1


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    On the boundary (surface) of two regions, the tangential components of electric fields on above and below surface are continuous. I wonder if it is also true for displacement [tex]\vec{D}[/tex] and polarization [tex]\vec{P}[/tex]? That is, can I say:
    the tangential component of [tex]\vec{D}[/tex] or [tex]\vec{P}[/tex] on above and below surface are continuous?

    For magnetic field, the statement of the magnetic field about [tex]\vec{B}[/tex] is:

    [tex](\vec{B}_{above} - \vec{B}_{below} )\cdot\hat{n} = 0[/tex]
    [tex](\vec{B}_{above} - \vec{B}_{below} )\times \hat{n} = \mu_0\vec{K}[/tex]

    I wonder if [tex]\vec{K}[/tex] means the free current surface density? What is the boundary conditions for [tex]\vec{H}[/tex]?
  2. jcsd
  3. Dec 31, 2008 #2
    Do the two regions have the same dielectric constant? Think about the formula that relates D and E...
  4. Jan 2, 2009 #3


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    That is the question. In the text, it said the tangential components of the electric fields on the boundary are continuous. But it doesn't tell if the tangential components of the displacement or polarization are also continuous or not. So if the dielectric constants in these two regions not the same, does it mean they will not be continuous even along the tangential direction?

    By the way, in some text, it reads

    [tex](\vec{P}_2-\vec{P}_1)\cdot\hat{n} = -\sigma_p[/tex]

    and [tex]\sigma_p[/tex] is what we call the density of polarized charges. I wonder if this is the same name as bound charges which is used in other text?
    Last edited: Jan 2, 2009
  5. Jan 3, 2009 #4
    I was just saying that by looking at the formula that relates D and E, you will see the answer to that question. The dielectric constant is the constant of proportionality between D and E, so if E is continuous, but the dielectric constant changes, what is going to happen to D? See what I mean?
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