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Homework Help: Field and Displacement Inside Carvities Within A Dielectric

  1. Oct 30, 2012 #1
    1. The problem statement, all variables and given/known data

    Suppose the field inside a large piece of dielectric is [itex]\vec{E}_{0}[/itex], so that the electric displacement is [itex]\vec{D}_{0}=\epsilon_{0}\vec{E}+\vec{P}[/itex].

    a. Now a small spherical cavity is hollowed out of the material. Find the field at the center of the cavity in terms of [itex]\vec{E}_{0}[/itex] and [itex]\vec{P}[/itex]. Also find the displacement at the center of the cavity in terms of [itex]\vec{D}_{0}[/itex] and [itex]\vec{P}[/itex].

    b. Do the same for a long needle-shaped cavity running parallel to [itex]\vec{P}[/itex].

    c. Do the same for a thin wafer shaped cavity perpendicular to [itex]\vec{P}[/itex]


    2. Relevant equations

    [itex]\vec{D}_{0}=\epsilon_{0}\vec{E}[/itex] when [itex]\vec{P}=0[/itex] (inside the cavity).


    3. The attempt at a solution

    For each part, I found the E of the cavity as if it were a solid piece, and then subtracted that from [itex]\vec{E}_{0}[/itex] to account for the change in [itex]\vec{E}[/itex]. Then, because [itex]\vec{D}_{0}=\epsilon_{0}\vec{E}[/itex] inside the cavity, [itex]\vec{D}_{0}=\epsilon_{0}\vec{E}_{new}[/itex].

    a. This part was simple. Inside a polarized sphere, [itex]\vec{E}=\frac{-1}{3\epsilon_{0}} \vec{P}[/itex], so [itex]\vec{E}_{new}=\vec{E}_{0}+\frac{1}{3\epsilon_{0}} \vec{P}[/itex], and [itex]\vec{D}=\vec{D}_{0}-\frac{2}{3}\vec{P}[/itex].


    b. Here it gets more tricky. The length of the needle is parallel to [itex]\vec{P}[/itex], so the [itex]\vec{E}[/itex] is pointed into and outside the ends. There would be a - charge and + charge at opposite ends, creating a field between them. But the field is negligible because the distance between them is large compared to the dipole? In that case, [itex]\vec{E}_{new}=\vec{E}_{0}[/itex]. Once we neglect the field inside, I can move forward, but can anyone explain better why exactly we ignore that field.

    c. Now we have a thin wafer-shaped cavity. I tried to use potential to calculate the electric field within the wafer.

    [itex]V=\frac{1}{4\pi\epsilon_{0}}\oint_{S}\frac{1}{r} \vec{P}\cdot d\vec{a}[/itex].

    (I did not use the second half of this eqn because we are assuming uniform polarization, which means the gradient of polarization is 0.)

    From here I find
    [itex]\vec{E}=-\vec{\triangledown}V=\frac{-1}{2\epsilon_{0}}\vec{P}[/itex].

    I then multiplied this by 2 to account for the top and bottom sides of the wafer and found

    [itex]\vec{E}_{new}=\vec{E}+\frac{1}{\epsilon_{0}}\vec{P}[/itex].

    The answer is right, but I'm not sure about my work/reasoning.
     
    Last edited: Oct 30, 2012
  2. jcsd
  3. Oct 30, 2012 #2
    I'm fixing the code currently
     
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