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Homework Help: Electric field between two spheres

  1. Mar 17, 2013 #1
    1. The problem statement, all variables and given/known data

    If the inner sphere of radius a has charge +Q and the outer sphere of radius b has charge -Q/2. What's the electric field between them?

    3. The attempt at a solution

    If I use Gauss' law then I would have E*4*pi*a^2 = Q/ε then just solve for E. Is that correct? It seems like the outer sphere would affect the E-field on the inner sphere. By the way, they don't say that these are conducting spheres.

    Thanks for any help.
  2. jcsd
  3. Mar 17, 2013 #2

    rude man

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    You have the field at r = a correct, but what about a < r < b? No, the outer sphere does no affect the E field on the inner sphere. Believe in Dr. Gauss! And also no, it doesn't matter if the sphers are conducting or insulators in this case.
  4. Mar 17, 2013 #3
    Thanks for the reply. Is this the E-field between the two spheres:

    E = Q/(ε*4*pi*r^2) for a < r < b ?

    I'm still confused how the outer sphere doesn't affect the E-field between them..
  5. Mar 18, 2013 #4

    rude man

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    For the same reason that, if you go inside the Earth, the only part exerting gravity on you is the part below you.

    At a point r in your sphere, some of the charges outside r will set up a + field and others will set up a - field. Some will push a test charge at r one way, others the opposite way. The net result is complete cancellation of each others' fields. It's not an easy task to do that integration, so again - believe Dr. Gauss!
  6. Mar 18, 2013 #5


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    There are several additional pieces of information needed before this problem could possibly be solved.

    Are the spheres concentric? (Do they have a common center?)

    Is the outer sphere actually a spherical shell? -- That's was is implied, seemingly.

    Is the charge distributed uniformly? -- or at least in some sort of symmetrical manner

    Suppose you have a uniformly charged spherical shell. What is the electric field inside the shell?
    This situation is often covered even before introducing Gauss's Law. ​
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