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Solid Insulator Sphere Inside Hollow Sphere Conductor

  1. Feb 27, 2015 #1
    1. The problem statement, all variables and given/known data
    I was looking for some practice problems in my textbook and found this problem that I was just a little stuck on. I drew the diagram from my textbook with the givens of the problem.

    Screen shot 2015-02-27 at 10.14.00 AM.png
    2. Relevant equations
    ∲E*dA = Q (inside) / ɛ0

    3. The attempt at a solution

    For r less than/equal to a:

    E(4pir^2) = 3Q/ɛ ---> E = 3Q / 4*pi*ɛ(r^2)

    For a<r<b:

    E= -5Q/ (4*pi*r^2)

    Can somebody verify with me that I've done the first two above correctly?

    Also the part that I got stuck on was the r is greater than/equal to c...
    I see that if r was equal to c there would be no charge, hence e-field will be zero inside the conductor. However, when r is greater than c, then there would be a charge of -5Q. How would I usually deal with the case of "greater than/equal to"?

    Thank you Physics community!
     
    Last edited: Feb 27, 2015
  2. jcsd
  3. Feb 27, 2015 #2

    mfb

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    Staff: Mentor

    The 3Q are distributed over the whole sphere, they are not all at the center ("inside").
    I don't see why you use -5Q here.

    Not only the -5Q.
    Ignore the case of r exactly equal to c, that does not have a reasonable answer in this simplified model.
     
  4. Feb 27, 2015 #3

    Screen shot 2015-02-27 at 6.25.50 PM.png

    Sorry this was the actually image from the textbook I was working on. There was a Q in the middle, but the statement of the problem still applies.

    So for r is less than/equal to a, I would draw a gaussian surface at r=a and one for r<a. I notice that at r=a, the only charge enclose in the middle of the insulator is the whole +3Q and +Q. But for the r<a, the only charge inside is the +Q. Is that correct?
     
  5. Feb 27, 2015 #4
    Hi welcome to PF:smile:

    No,for r<a the charge inside is not just +Q.As it is a insulating solid sphere the charge +3Q is distributed uniformly over its total volume.So for exampleif you are considering a Gaussian surface of radius a/2 then you also have to consider the portion of +3Q distributed in the volume inside the Gaussian surface!
     
  6. Feb 27, 2015 #5
    ah I see...so in that case to find Q inside , it would be Q= (4/3*pi*r^2 / 4/3*pi*a^2) * Q => Qinside = (r^2/a^2)*Q?
     
  7. Feb 28, 2015 #6
     
  8. Feb 28, 2015 #7

    mfb

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    Staff: Mentor

    r and a should be to the third power. You have volumes here, not surfaces.
     
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