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Holy Dielectric

  1. Oct 24, 2009 #1
    Holy Dielectric!!!

    1. The problem statement, all variables and given/known data
    A large block of dielectric contains small cavities of various shapes that may be assumed not to disturb appreciably the polarization. Show that, inside a needle-like cavity parallel to P, E is the same as in the dielectric


    2. Relevant equations

    [tex]\oint_{S}\vec{E}\cdot \vec{da}=\frac{Q_{enc}}{\epsilon_0}[/tex]

    3. The attempt at a solution
    Not really sure how to attack this problem....

    I can create a Gaussian object inside the dielectic and using Gauss's law....

    Taking the Gaussian surface to be cylindrical, i get that [tex]E=\frac{\pi p_{b}}{2\epsilon_0}[/tex], where [tex]p_{b}[/tex] is the bound charge density inside the dielectric.

    How would I calculate E inside of the cavity? Same method? Doesn't [tex]p_{b}=0[/tex]?
     
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  3. Oct 24, 2009 #2

    gabbagabbahey

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    Re: Holy Dielectric!!!

    What makes you think the electric field is cylindrically symmetric? Is the dielectric linear?

    Reading your problem statement, I would assume that the polarization [itex]\textbf{P}[/itex] inside the dielectric is approximately uniform, and that the dielectric need not be linear (which means [itex]\textbf{E}[/itex] and [itex]\textbf{D}[/itex] need not be uniform).

    I think you'll just want to calculate the electric field due to a uniformly polarized needle with polarization [itex]-\textbf{P}[/itex] and then use thew superposition principle.
     
  4. Oct 24, 2009 #3
    Re: Holy Dielectric!!!

    So the needle can be treated as a line, length d, say. But Griffiths leave out the discussion of such a line in Ch.4...
     
  5. Oct 24, 2009 #4

    gabbagabbahey

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    Re: Holy Dielectric!!!

    You shouldn't need Griffiths to tell you how to calculate such a field....where are the bound charges in such a needle? How large/small of a field would you expect these charges to create and why?
     
  6. Oct 25, 2009 #5
    Re: Holy Dielectric!!!

    The bound charges exist at the ends of the needle. Treating this as a dipole of ends -q and +q, i get:

    still calculating....

    nevermind dipole approach seems to messy...
     
    Last edited: Oct 25, 2009
  7. Oct 25, 2009 #6
    Re: Holy Dielectric!!!

    Then again... treating this as a dipole yields...

    [tex]V=\frac{\vec{p}\cdot \hat{r}}{4\pi\epsilon_0\cdot r^{2}}[/tex]

    and therefore...

    [tex]E=\frac{p}{4\pi\epsilon_0\cdot r^{3}}\cdot (2cos(\theta)\cdot \hat{r} + sin(\theta)\cdot \hat{\theta})[/tex]
     
  8. Oct 25, 2009 #7

    gabbagabbahey

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    Re: Holy Dielectric!!!

    But for a skinny needle, how much surface area will there be at the ends....will the resulting surface bound charge be large or small?...If the needle is long these charges will be far apart....doesn't that reduce the field even more?

    I would say [itex]\textbf{E}_{\text{needle}}\approx0[/itex].:wink:
     
  9. Oct 25, 2009 #8
    Re: Holy Dielectric!!!

    Makes sense, but seemed like way too easy of an answer x] so now I've got to prove that E is zero in the rest of the dielectric
     
  10. Oct 25, 2009 #9
    Re: Holy Dielectric!!!

    umm does it make sense that E inside the dielectric is zero? The bound charge density is zero because P is uniform, leaving the bound surface charge.

    Bound surface charge is also zero because it only exists at the surface and I'm looking at a gaussian surface inside?


    Therefore potential is zero inside the dielectric and so is E?
     
  11. Oct 25, 2009 #10

    gabbagabbahey

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    Re: Holy Dielectric!!!

    No you don't.In fact, you should be able to argue that E is much much larger inside the dielectric, than the field of the needle---which is why you can neglect the field of the needle entirely. You only need to use the superposition principle (superimpose a needle of negative polarization onto the slab to create the needle shaped cavity);

    [tex]\textbf{E}_{\text{cavity}}=\textbf{E}_{\text{slab}}+\textbf{E}_{\text{needle}}[/tex]
     
  12. Oct 25, 2009 #11

    gabbagabbahey

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    Re: Holy Dielectric!!!

    No, the bound surface charge is very large, and doesn't necessarily create a symmetric field; so just because the net flux is zero inside, doesn't mean the field is zero. (Gauss' Law requires symmetry to be useful!)
     
  13. Oct 26, 2009 #12
    Re: Holy Dielectric!!!

    So I've got to prove that [tex]E_{cavity}=E_{slab}+E_{needle}=E_{no cavity}[/tex] because the original problem statements is that [tex]E_{cavity}=E_{no cavity}[/tex]?
     
  14. Oct 29, 2009 #13
    Re: Holy Dielectric!!!

    The only way I can see this working is if [tex]\textbf{E}_{\text{cavity}}=\textbf{E}_{\text{slab} }+\textbf{E}_{\text{needle}}[/tex] and [tex]\textbf{E}_{\text{needle}}[/tex] is neglible (say zero), then [tex]\textbf{E}_{\text{slab}}[/tex] is also zero. Therefore, [tex]\textbf{E}_{\text{cavity}}[/tex] is zero.

    Giving end result [tex]\textbf{E}_{\text{cavity}}=0=\textbf{E}_{\text{needle}}[/tex]
     
  15. Oct 29, 2009 #14

    gabbagabbahey

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    Re: Holy Dielectric!!!

    You need to understand that [itex]\textbf{E}_{\text{cavity}}[/itex] represents the total field inside the cavity; [itex]\textbf{E}_{\text{slab}}[/itex] represents the field inside the cavity, due to just the dielectric slab; and [itex]\textbf{E}_{\text{needle}}[/itex] represents the field inside the cavity due to just a uniformly polarized needle (w/ polarization [itex]-\textbf{P}[/itex])....
     
  16. Oct 29, 2009 #15
    Re: Holy Dielectric!!!

    [itex]\textbf{E}_{\text{slab}}[/itex] is the electric field from the whole chunk of dielectric??
     
  17. Oct 29, 2009 #16

    gabbagabbahey

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    Re: Holy Dielectric!!!

    You tell me....what does the superposition principle say?
     
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