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Electric field inside a cavity in a dielectric

  1. Aug 8, 2017 #1
    upload_2017-8-8_20-7-32.png



    2. Relevant equations


    3. The attempt at a solution

    A)
    In the spherical cavity, let me have a Gaussian Spherical Surface around the center.
    Applying Gauss' s Law for dielectrics,
    ##\oint _S\vec D⋅ d \vec a = Q_{f_{en}} ##
    Due to the symmetry of the problem and ## Q_{f_{en}} =0## , ## \vec D = 0 ##.
    Since, ## \vec P = 0 ##, ##\vec E ## inside the cavity, too, is zero.
    Is this correct?
     

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  3. Aug 8, 2017 #2

    Charles Link

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    ## D ## is the quantity that stays uniform everywhere. ## \nabla \cdot D=0 ## since there are no free electrical charges. In general, ## D=\epsilon_o E+P ##. But ## P=0 ## inside the cavity. This allows you to get an answer for ## E ## inside the cavity in terms of ## D ##. ## \\ ## One additional question you might have is what is ## P ## in the material? That depends upon the polarizability ## \alpha ## of the material. In general ## P= \alpha E_o ##, but they don't tell you what ## \alpha ## is for the material. They simply tell you that ## P ## has some known value. ## \\ ## And one additional input: I believe the assumption that ## D ## is perfectly uniform is an approximation. For the case of a dielectric sphere in a uniform electric field, that approximation is exact. When the material is uniform, except for a (spherical) cavity, I believe just outside the spherical cavity, the electric field is not perfectly uniform for the exact solution to this problem, but the assumption is made, (and it is accurate to a very good approximation), that the electric field is indeed uniform throughout the material, including the region just outside the cavity.
     
    Last edited: Aug 8, 2017
  4. Aug 9, 2017 #3
    A) There is something wrong in this argument. I can't assume symmetry as I don't know the nature of ##\vec E_0##.

    Considering the cavity as the superposition of original dielectric material and a sphere of polarization - ## \vec P ##.
    Electric field due to this sphere inside the cajvity is ## \vec E _{sp} = \vec P/3ε_0 ##
    The electric field inside the cavity is ## \vec E_{ca} = \vec E_0 + \{\vec E _{sp} = \vec P/3ε_0\}##
    ## \vec D = ε_0 \vec E_0 + 0 ## ##= ε_0 \vec E_0 + \vec P - 2 \vec P /3 ##
    ## = \vec D_0 - \frac 2 3 \vec P
    ##

    B) Considering the cavity as the superposition of original dielectric material and a needle of polarization - ## \vec P ##.
    There will be surface charge density only on the two ends of the needle. Assuming that the total charge is so small that the electric field due to it is negligible compared to ##\vec E_0##.
    So,
    ## \vec E_{ca} = \vec E_0 ##
    ## \vec D = \vec D_0 - \vec P##

    C) Considering the cavity as the superposition of original dielectric material and a wafer of polarization - ## \vec P ##.

    upload_2017-8-9_14-15-49.png
    The positive surface charge density is on the upper surface and the negative charge density is on the lower plate.
    So, electric field due to this wafer ## \vec E_{wa} = \frac {-\vec P} {ε_0}
    \\ \vec E = \vec E_0 + \frac {-\vec P} {ε_0} ##
    ## \vec D = \vec D_0 - 2 \vec P##

    Is this correct so far?

    The solution means that depending on the shape of the cavity, the electric field and the displacement are different inside cavities in a large dielectric with uniform polarization, electric field, displacement.
     
  5. Aug 9, 2017 #4
    The charge distribution in the above figure is wrong. Actually I thought the wafer as a || plate capacitor and the external electric field due to which the polarization occurs is in the direction of polarization of wafer. Because of this, the positive charge of the upper plate goes to the lower surface of the upper plate and the negative charge of the lower plate goes to the upper surface of the lower plate.

    But, this is wrong. What happens is : Due to the external electric field, there devlops Polarization in the dielectrics.
    Now, the surface charge density on the upper and the lower surface is given by: ##σ_{up} = \vec P' ⋅ \hat z = - P \\ σ_{down} = \vec P' . \left(- \hat z \right ) =P##
    upload_2017-8-9_15-31-23.png

    So, electric field due to this wafer
    ##\vec E_{wa} = \frac {\vec P} {ε_0} \\ \vec E = \vec E_0 + \frac {\vec P} {ε_0}

    \\ \vec D = \vec D_0 ##

    Is this correct so far?
     
  6. Aug 9, 2017 #5

    BvU

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    Correct on three counts. Well done.
     
  7. Aug 9, 2017 #6

    Charles Link

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    Yes, @Pushoam very good. My original suggestion, that ## \vec{D} ## is continuous (and uniform) only works for case C. You correctly superimposed a material of polarization "-P" for the other two cases.
     
  8. Aug 9, 2017 #7

    Charles Link

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    One additional comment=I am somewhat puzzled by the solution of the superposition of the P's because essentially it implies that ## \vec{P} ## remains constant around the cavity. Meanwhile, if you were to compute the electric field outside the cavity caused by the polarization of "-P" and its surface polarization charge ## \sigma_p=\vec{P} \cdot \hat{n} ##, it is not uniform, so that it would cause the assumed uniform ## \vec{P} ## in the material just outside the cavity to change. In any case, you solved it correctly, but it appears the assumption/approximation is made that ## \vec{P} ## remains uniform.
     
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