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Electric field, hollow sphere

  1. Apr 17, 2012 #1
    Hey im currently studying for my final and im stuck on a question i have the solution but im not sure what he did, could someone explain

    A hollow sphere of the inner radius R1 and outer radius R2 is uniformly charges with total charge Q. Calculate the electruc fiels in the three regions shown in the cross-section view below 1) r<R1, 2) R1 < r <R2, and 3) r >R2

    pPKRZ.png

    What i dont get is part two, i know he is using gauess law Q inclosed/e = E(r)*A
    but i dont know how he got Q and reduced it
     
  2. jcsd
  3. Apr 17, 2012 #2

    tiny-tim

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    welcome to pf!

    hi underground! welcome to pf! :smile:
    the total charge is Q, so the charge density is Q/volume = Q/{4π/3(R23 - R13)},

    so the charge inside radius r is the density times the volume within radius r, ie 4π/3(r3 - R13) :wink:
     
  4. Apr 17, 2012 #3
    Thanks for the respond im getting there but can you explain one thing

    I get what your saying but i dont see where the Q overall charge disappeard

    P7rGZ.png

    This is my logic of the question and the Q which is overall charge remained in the equation
     
  5. Apr 17, 2012 #4

    tiny-tim

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    suppose R1 was 0 (ie, a solid sphere) …

    then the charge within radius r would be Q(r/R2)3, wouldn't it?

    this is similar … the charge is Q times (volume/total-volume) :wink:
     
  6. Apr 17, 2012 #5
    Thanks so much for your help :)
     
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