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Why is the magnitude of the electric field equal in all cases?

  1. Jul 12, 2005 #1
    Theres a series of circles inside eachother. A central metal ball, two spherical metal shells, and three spherical Gaussian surfaces of radii R, 2R, 3R, all with the same center. The uniform charges on the three objects are: ball, Q; ssmaller shell, 3Q; Larger shell, 5Q. Rank the Gaussian surfaces according to the magnitude of the electric field at any point on the surface, greatest first. The image is attached.I don't get it because The electric field outside a spherical shell of charge with raidus R and total charge q is directed radially and has a mag of: E = 1/[4piEo] (q/r^2);

    So I ignored the 1/[4piEo] because they all will have that, the only difference is the charge and radius. So....
    I ended up getting Q/R^2; 3Q/4R^2; 5Q/9R^2; How can these be equal? :bugeye:
     

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  2. jcsd
  3. Jul 12, 2005 #2

    quasar987

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    you're forgetting to add the charges. the charge enclosed by the first surface is Q, that'd right. But the charge enclosed by the second surface is Q (ball) + 3Q (first shell) = 4Q. The charges enclosed by the third one is Q + 3Q + 5Q = 9Q. So you see, after cancelation of numerator and denominator, they are equal.
     
  4. Jul 12, 2005 #3
    Can you please tell us the magnitude of the radius of the respective metal ball and the two shells. Your image is yet to be approved.

    That is not correct. The charges are resting 'on' the surface of the respective shells and the metal ball . The charge enclosed by the first shell is simply Q , while 4Q is being enclosed by the outermost shell. While the metal ball which is a solid sphere has charge Q uinformly distributed along its volume.

    BJ
     
  5. Jul 12, 2005 #4

    quasar987

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    I'm not talking about the surfaces enclosed by the ball and shells, but the charge enclosed in the gaussian surfaces, which I imagined to be placed in such ways that the problem had the simple solution I wrote.
     
  6. Jul 12, 2005 #5
    You should have referred to the Gaussian Surface in your first post, anyway I apologise :approve: . Lets first see the image (yet to be approved) , what it says , and then we can discuss it better.

    BJ
     
  7. Jul 12, 2005 #6
    Last edited: Jul 12, 2005
  8. Jul 12, 2005 #7

    quasar987

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    yes the image showed up and I had guessed it right.
     
  9. Jul 12, 2005 #8
    awesome thanks for the help! :biggrin:
     
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