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Electric fields of a spherical shell

  1. Jan 22, 2014 #1
    My class hasn't delved into Gauss's Law much besides describing conductors at electrostatic equilibrium to have no net electric field or force within itself.

    For the picture, the question is:

    What are the magnitudes of the electric fields at:

    1) r = a
    2) r = 3/2 a
    3) r = b

    Firstly, if r = 0, E = 0, correct?

    At r = a, is it correct to assume that the charge on the smaller spherical shell is also 0 since it has no electric fields from it? Is this the correct explanation?

    For r = 3/2a, I was having a bit of trouble and was wondering if we are to find

    ##Q = ρ\frac {4}{3}π [(1.5a)^3 - a^3] ##

    Then:

    ##E = kQ/(1.5a)^2## ?

    Would the same steps for r = 3/2a be followed for r = b? Any help regarding how to answer this question, with an explanation (please!), and possibly links to links that you find very informative with tho topic would be greatly appreciated!

    Also, does anyone know where I could find a list of electric field formulas for various objects (infinitely thin and long sheets, spheres, quadrupoles, etc.) and accompanying proofs?

    And finally, is it incorrect to denote electric field as ε? I've seen both around but E more commonly and was uncertain if it's common convention or not.
     

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    Last edited: Jan 22, 2014
  2. jcsd
  3. Jan 22, 2014 #2

    haruspex

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    Yes
    What "smaller spherical shell"? There is only one shell. Do you mean the inner surface of the shell?
    The charge is uniformly distributed through a volume. That means the charge on a surface is effectively zero.
    Yes
    Yes. In fact, you could save yourself a bit of work by solving for a radius which is an arbitrary multiple of a, r = ca, then plug in c = 1.5, c = 2.
    Only by doing what you can do - use a search engine.
    Pass.
     
  4. Jan 22, 2014 #3
    Do you mind expanding on this point? If charge is uniformly distributed, won't it be the same everywhere, including on the surface of the inner sphere?
     
  5. Jan 22, 2014 #4

    haruspex

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    Since it is a finite charge uniformly distributed over a volume, the charge in a given portion is proportional to the volume of that portion. An area has no volume, so contains no charge.
     
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