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Polarization of a di-electric due to an electric field

  1. Mar 30, 2010 #1
    if a di-electric was introduced to an electric field from the outside it would be polarized and the charge denisties would be as follows :
    **volume bounded charge density ( ρv ) = - div ( P )
    **surface bounded charge density ( ρs ) = P . n
    where P is the polarization vector and n is the unit vector.

    now if the di-electric was originally neutral , should all these bounded charges sum up to zero ?
    ( volume bounded charge density ρv + surface bounded charge density ρs = 0 ) ?

    and what if the source of the electric field was coming from inside the di-electric ( as for example a free point charge ) would it still be the case ?
  2. jcsd
  3. Mar 30, 2010 #2
    We can be certain of it because charge is conserved. We can assume that if polarization causes some distribution of charge then each negative region is balanced by another positive region.

    Is that good enough? It wouldn't make any physical sense to create charges by some special configuration of electric fields.

    If you want a more mathy answer then take a look at Gauss's law in integral form:

    [tex]\oint\oint E \cdot dA = \frac {Q}{\epsilon_0}[/tex]

    E is the electric field and Q is the total charge inside some closed surface over which the integration takes place. The polarization field is a direct function of E so there must be a proof of charge conservation somewhere in Gauss's law. I'll give it some more thought later.
  4. Mar 31, 2010 #3
    I think I have mathematical answer for you. Let's define flux:

    [tex]\Phi = \oint\oint{\vec{V} \cdot \hat{n}}[/tex]

    What this tells us is that some scalar(phi) is equal to the sum of the vectors pointing into and out of a closed surface. For an analogy, V could be the flow of water and the closed surface could be the outline of a lake. If more rivers are flowing into the lake than out, there will be a build up of water in the lake.

    The electrical version is this (Gauss's law):

    [tex]Q = \oint\oint{ \epsilon_0 \vec{E} \cdot \hat{dA}}[/tex]

    This tells us that the sum of electrical vectors pointing into and out of a surface is related to the charge inside the surface. If that charge starts as 0 then we have:

    [tex]\oint\oint{ \epsilon_0 \vec{E} \cdot \hat{dA}} = 0[/tex]

    For linear dielectrics, we know what P is:

    [tex]P = \epsilon_0\chi \vec{E} ; \vec{E} = P / (\epsilon_0\chi)[/tex]

    So, we know right away that the bound surface charges must sum up to zero.

    [tex]\sigma_s = P \cdot \hat{dA}[/tex]

    [tex]\oint\oint{ \epsilon_0 \vec{E} \cdot \hat{dA}} = 0[/tex]

    [tex]\oint\oint{ \epsilon_0 \left( P / (\epsilon_0\chi) \right) \cdot \hat{dA}} = 0[/tex]

    [tex]\oint\oint{ \epsilon_0 \left( P \cdot \hat{dA} \right) \frac{1}{\epsilon_0\chi}} = 0[/tex]

    [tex]\oint\oint{ \epsilon_0 \left( \sigma_s} \right) \frac{1}{\epsilon_0\chi}} = 0[/tex]

    Now, the bound charges inside the dielectric are this:

    [tex] \rho_v = -\nabla \vec{P}[/tex]

    It is a property of vector fields that the following is true:

    [tex]\oint\oint\oint \left( \nabla \vec{V} \right) = \Phi = \oint\oint \vec{V} \cdot \vec{dA} [/tex]

    Therefore, the bound charges must sum up to zero because the flux going into the surface must be zero.

    We could do all that again with a charge inside the surface and find the relationship between bound charges and the free internal charge. Also, I did this analysis for a linear dielectric. I'm not even sure what how to go about describing what would happen in a non-linear material.

    Remember the simple answer? It was the one about how charge must be conserved so charged regions must balance out somehow. That answer is looking better and better don't ya think?
  5. Mar 31, 2010 #4
    That was very helpful really.
    yes now i understand , thanks very much Okefenokee .
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