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Gauss' Law in Dielectrics

  1. Nov 16, 2004 #1
    This problem is also giving me a lots and lots of trouble, and I don't even know where to begin.

    A point charge q is imbedded in a solid material of dielectric constant K.

    A) Use Gauss's law as stated in equation \oint{K \vec{E} \cdot \vec{A}} \;=\; \frac{Q_{free}}{\epsilon_{0}} to find the magnitude of the electric field due to the point charge q at a distance d from the charge.

    B) Use your result from part (a) and Gauss's law in its original form as given in equation \oint{\vec{E} \cdot \vec{d A}}\; =\; \frac{Q_{encl}}{\epsilon_{0}} to determine the total charge (free and bound) within a sphere of radius d centered on the point charge q.

    C) Find the total bound charge within the sphere described in part (b).

    I obviously need A before doing B and C, but I don't even know how to get it. I didn't fully understand Gauss law in the first place, so now I am even more confused. Any help appreciated, thanks.
     
  2. jcsd
  3. Nov 16, 2004 #2
    I think you'll have to do a better job with those equations so people can help you without spending all night trying to decode the question.
     
  4. Nov 16, 2004 #3

    dduardo

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    Here is a hint:

    E and D are in the radial direction. That means that there will never be a tangential component. Now all you have to figure out is what the normal component is. This is just a simple boundary condition problem.
     
  5. Nov 16, 2004 #4
    what do you mean a boundary condition problem? I am confused a bit, I'm afraid to say.

    those two equations, which I thought would show up properly, are:
    integral KEA = Q(free)/epsilon_0
    and
    integral EdA = Q(enclosed)/epsilon_0
     
  6. Nov 16, 2004 #5

    dduardo

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    Forget about the formulas and just think for a moment. In the case when you have a charge in free space what is the electric field at some point due to the charge? Once you can answer that question using Gauss's Law you'll be ready to move on to the second part which is: how is an electric field affected by a dielectric boundary.
     
  7. Nov 16, 2004 #6
    The electric field from some point is given by a formula, right? Specifically Kq/(r^2)
     
  8. Nov 16, 2004 #7

    dduardo

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    yes, now what is the relationship between En inside the dielectric and En inside free space.
     
  9. Nov 16, 2004 #8
    the dielectric has a charge built up on the sides of it? And this would effect E?
     
  10. Nov 16, 2004 #9

    dduardo

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    Ok, so what is the charge density?
     
  11. Nov 16, 2004 #10
    I honestly am not sure of the charge density. I know this sounds dumb but I don't know...?
     
  12. Nov 16, 2004 #11

    dduardo

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    [tex]Q = \int \rho dV[/tex] where rho is the charge density
     
  13. Nov 16, 2004 #12

    dduardo

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