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Gauss's (hard to understand) Law

  1. Jul 13, 2005 #1
    ok here the problem im having...
    a point charge Q rests at the center of an uncharged thin spherical condcting shell. with is the the electric field E, as a function of r.....then it asks for r less than the inner radius of the shell... then it asks for inside the shell, and finally beyond the shell.

    now how im invisioning the figure is almost like a washer, with a thickness, like the rings of saturn....

    so for the first part i need to find E as a function of r....so i thinks to myself... E0 (constant) = 1/ (4(k)(pi)) and gauss's law states
    flux = EA flux also = q/E0 so EA = q/ Eo after a little algebra
    i get E= Qk/ R^2 using SA of a sphere for area...... ok so im pretty sure thats right for the first question....if not please tell me where i went wrong......next part asking for r less than inner radius of shell????
    i have no idea what to put............one thing i do understand is that when there is a charge in a shell, the electric field continues as if there was no shell present, by this i mean if Q is positive , negative charges accumulate on inner shell circumference, and positive accumlate on SA of total spherical shell........... this is all kind of new to me, and i dont think i have a great grasp on all of this,,,,,,also a coule off topic questions....what is the diff,,,between q (test charge) and Q////// as always thanks a million for any help/ information.
     
  2. jcsd
  3. Jul 13, 2005 #2
    I assume that the second part is asking you what is the electric field less than the outer surface of the shell but more than the inner surface of the shell, i.e. what is the electric charge inside the conducting shell.

    Your right, outside of the whole shell, you will not see the shell because it will not insulate the charge, so it will be like the shell didn't exist.

    And also inside of the shell, it will be like the shell doesn't exist because the charge doesn't need to move through the shell to get from the charge to the point where you are testing.

    Hope this helps,

    ~Lyuokdea
     
  4. Jul 13, 2005 #3
    yes, it does,,, so i guess i would just describe what the electric field does,,, i think im getting confused because of the lack of numbers, i guess i just explain whats happening. thanks
     
  5. Jul 13, 2005 #4

    Gokul43201

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    You are asked to calculate E(r), when :

    (i) r < R(in)

    (ii) R(in) < r < R(out)

    (iii) r > R(out)

    In the first and third cases, the total enclosed charge is Q. So, you can find the fields from Gauss' Law, and they will have the same form. But this is not so in the second case. The charge enclosed here will be different because of the induced charges on the inner surface of the shell. So, find the new value of the enclosed charge, and again calculate the field from GL.

    Is the result you get for (ii) surprising, or expected ?
     
  6. Jul 14, 2005 #5
    so would Q=0 inside the shell since all charges are found on the surface of conductors? thanks everyone for the help
     
  7. Jul 14, 2005 #6

    Doc Al

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    I think you are talking about the total charge Q within a spherical Gaussian surface that is inside the material of the shell. If so, since the electric field is zero within the conducting shell, the total charge Q contained in that Gaussian sphere is zero.
     
  8. Jul 15, 2005 #7

    Gokul43201

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    In this case, though, I think you want to reason the other way round. You are asked to find the field on this Gaussian surface, so you must determine the charge enclosed. First there's +Q at the origin, and then there's -Q distributed on the inside of the shell. So the total enclosed charge is zero, and hence, the field within the material of the shell is zero (as expected, since the field inside the body of a conductor must be zero).
     
  9. Jul 15, 2005 #8

    Doc Al

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    OK, but how do you know that there's -Q distributed on the inside of the shell? (You need to start with some knowledge; I usually make use of the fact that the field must be zero within the conductor.)
     
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