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Electric field & energy in a spherical distribution of charge

  1. Feb 16, 2017 #1
    1. The problem statement, all variables and given/known data
    upload_2017-2-16_22-20-39.png

    2. Relevant equations
    Gauss

    3. The attempt at a solution
    I am really confused with question a, I have an idea of how to answer b and c once I obtain an answer for part a... My best guess would be to use Gauss, but I am not sure. Would the field inside be 0? What will the bounds of integration be when I integrate to find the charge, right now I get a charge of 2pi(a^3)? I am working on my solution and will post a picture when I get something useful, but can someone explain? Thank you.
     
    Last edited: Feb 16, 2017
  2. jcsd
  3. Feb 17, 2017 #2

    BvU

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    Choose a suitable volume to write out your relevant equation in terms of the variables and post your working...
     
  4. Feb 17, 2017 #3

    haruspex

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    The question is a bit unclear. I assume that a/r is for the charge density at a distance r from the centre of the sphere radius a.
     
  5. Feb 18, 2017 #4

    BvU

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    I agree (as almost always :smile:). It is unclear in the sense that (most probably) the composer forgot to mention 'for ##r\le a##' in ##'\rho = ...' \ ##; and ## \rho = 0\ ## for ##r>a## ' so you make that assumption, mention it in your answer, and continue with the exercise.
     
  6. Feb 24, 2017 #5
    1. The problem statement, all variables and given/known data
    upload_2017-2-24_12-24-22.png

    2. Relevant equations


    3. The attempt at a solution
    I am at the first part of the question. For the inside field, I did the following (and the same for the outside, just using different bounds of integration)
    upload_2017-2-24_12-31-0.png
    In the given solution, they use electric displacement
    upload_2017-2-24_12-28-0.png
    Why not just use Gauss' law like usual, like I did?
     
  7. Feb 24, 2017 #6

    BvU

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    They do. ##\vec D=\varepsilon \vec E##.

    I will ask a mentor to move this to the identical thread you started a week ago ...
     
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