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Calculating Charge at the Center of a Spherical Shell has me stumped

  1. Dec 4, 2015 #1
    Edit: Forgot to type "stumped" at the end of the title

    1. The problem statement, all variables and given/known data

    Instead of typing it out, a link to a scanned document of the problem is here: http://imgur.com/Be3jSLp.

    2. Relevant equations
    The equations to use are stated in the problem here: http://imgur.com/Be3jSLp

    3. The attempt at a solution
    For 3.a: Wouldn't you have to draw a gaussian sphere around the outer edge of the sphere, which would yield an enclosed charge of Q_o? Then after that, wouldn't you draw another one inside the conducting shell, resulting in an enclosed charge of zero? Therefore, wouldn't the charge at the center be -Q_o?

    For 3.b: Using the equation given, I don't know exactly how to integrate the equation after you substitute in the E value given in the beginning of the problem.

    Any advice?
     
  2. jcsd
  3. Dec 4, 2015 #2

    SammyS

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    Here's the image:
    upload_2015-12-4_16-5-30.png

    Have you tried applying Gauss's Law ?
     
  4. Dec 4, 2015 #3
    Hi Physicsman69:

    I suggest you keep the following hints in mind:
    1. The total field is radially symmetric.
    2. The field out side of the sphere at radius r is exactly the same as if the total charge inside
    Q0 + q0
    was at the center.

    Hope this helps.

    Regards,
    Buzz
     
  5. Dec 4, 2015 #4
    Yes, you can draw a gaussian surface outside of the sphere which would yield $$E4\pi r^2 = Q_0/\epsilon$$

    And, then a gaussian surface drawn inside the conducting shell would yield $$E4\pi r^2 = 0/\epsilon$$

    Wouldn't this mean the charge in the center equals $$-Q_0$$ To offset the positive charge on the surface of the shell?

    Being that it is radially symmetric, the integral for Gauss' law would yield: $$E4\pi r^2$$ That is what you are implying right?

    And for #2 are you saying that the unknown charge at the center, q0, would have to be opposite and equal of Q0?
     
  6. Dec 4, 2015 #5
    Wait, for:

    For #2, this would mean that q0 would be actually be zero, correct? A charge uniformly distributed over the surface, can act like a charge concentrated at the center, therefore Q0 + q0 = Q0, so q0 = 0?
     
  7. Dec 4, 2015 #6
    Just make a spherical net around that spherical shell and shrink it to just a point at the centre carrying unaltered total of charge.

    Edit
    Maybe it's right to call it Gauss's net?
     
  8. Dec 4, 2015 #7
    A gaussian surface just around the point at the center wouldn't yield anything. The charge at the center is unknown.
     
  9. Dec 5, 2015 #8
    Gauss's Law
    dQ=D.dA
    For spherical surface,
    Q=DA, where D=εE.
    Where Q is net charge at the center.
     
  10. Dec 5, 2015 #9

    SammyS

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    The total charge enclosed is Q0 + q0, is it not?

     
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