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Electric Potential of a Sphere

  1. Feb 20, 2009 #1
    1. The problem statement, all variables and given/known data

    A conducting sphere 4.5 cm in radius carries 40 nC. It's surrounded by a concentric spherical conducting shell of radius 20 cm carrying -40 nC.

    Find the potential at the sphere's surface, taking the zero of potential at infinity.

    2. Relevant equations

    Inside Sphere: V = Q / (4*pi*ε0*R)
    Outside Sphere: V = Q / (4*pi*ε0*r)

    3. The attempt at a solution

    So I figured I would add these two potentials up, giving this:

    V = 9*10^9 [ ( 4*10^-5 ) / .045 + ( -4*10^-5) /.2 ] = 6200000V
     
  2. jcsd
  3. Feb 20, 2009 #2

    Delphi51

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    There is no place where the distance from the center of the charges is .045 and .2, so your calc can't be right. Also, what the heck is a nC? NanoCoulomb?

    I'm confused about this one - hope LowlyPion will help us out!
    I'm picturing the "sphere" inside a "shell" (hollow sphere). Are we interested in the potential at the surface of the inner sphere?

    Do you know about Gauss's Law and Gaussian surfaces?
    Using that, I think the E field outside the outer shell is zero, so if we were to find V by integrating E from infinity into the surface of the inner sphere, there would be no contribution from the outer charge at all. Only the inner one need be considered and only from 20 cm to 4.5 cm.
     
  4. Feb 20, 2009 #3
    Yes, nC is a nanoCoulomb, which is why my calculations were converted to Coulombs (4*10^-5). Also, I think what you're picture is correct.. and yes we are interested in the potential at the surface of the inner sphere.

    Also, I probably should have said this earlier: my calculations were based off this question:

    http://answers.yahoo.com/question/index?qid=20090120101033AAplj9b

    Except in that example the spherical shell is non-conducting..
     
  5. Feb 20, 2009 #4

    Delphi51

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    The site posted only the numerical answer to a different question - or did I miss something?
    Check that charge. 40 nC = 40 x 10^-9 C = 4 x 10^-8 C.

    No doubt there is an easier way, but using V = integral of E*dr I get
    V = kQ*integral (dr/r^2) from 20 cm to 4.5 cm and Q is 40 nC of the inner charge only.
    V = -kQ/r evaluated at the two radii
    V = -kQ(1/.045 - 1/.2)
     
  6. Feb 20, 2009 #5

    lanedance

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    the guassian surfaces and field description sound good

    so if the datum for the potential is zero at infinity, the potential will be zero at the boundary
    of the shell

    so how about you calculate the potential due to the sphere at the shell radius assuming there is no shell

    similarly calculate the potential at the sphere surface due to the sphere assuming no shell

    putting the shell in effectively shifts the potential at the shell to zero, so shift the potential at the sphere by the same amount
     
  7. Feb 20, 2009 #6
    Ha! Good catch! that ended up being the problem.. The correct answer was 6200V.. I've been doing problems with microCoulombs so I must've gotten screwed up between the two.

    Thanks a bunch

    And by the way, your last line of work is correct.. When I wrote it in my first post I just factored out the Q
     
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