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Homework Help: Electric field:neutral sphere

  1. Jan 27, 2008 #1
    It is known that a neutral metal ball will have 0 electric field inside the sphere, will there be a non zero electric field inside the neutral metal ball if this neutral ball is placed near a charge? I know that it will induce charges of the opposite kind ont he outside, what about the inside.

  2. jcsd
  3. Jan 27, 2008 #2
    Based on Gauss's law, there should not be a charge enclosed, because all of the charge is on the surface. So at some region r<R (r is a radius less than the Radius of the sphere) of the sphere would have zero charge. But the surface of the sphere would have a charge.

    This applies for a hollow sphere.
  4. Jan 27, 2008 #3

    Ben Niehoff

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    If the sphere is a conductor, then the electric field will be zero everywhere inside the sphere. Think about it: in a conductor, charges are free to move. So if there is a net electric field, then the charges will move...and they will continue moving until there is no net electric field. That is, the charges will arrange themselves in a way that precisely cancels the field inside the sphere.

    There will, however, be an induced charge distribution on the surface of the sphere, and that means that the electric field outside the sphere will get complicated.
  5. Jan 27, 2008 #4


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    Not too very complicated. You can solve the problem using the method of images if the exterior charge distribution is simple enough.
  6. Jan 29, 2008 #5
    Does anyone have the mathematical proof that the E-field is 0?

    Ie that in a hollow sphere made of insulating material, if the charge is evenly distributed on the surface (so the charges cannot move) then the E-field is zero?
  7. Feb 1, 2008 #6
    I would say that for a conductor (hollow or solid) the field is zero *IF* the charge on the surface (whether induced or not) is static -- if there was a field then the charges would not be static (as was mentioned before). So the assumption is that you waited to look for the field (a millionth of a second or so) until the charges reached their equiilbrium positions and if they are staying put then the field must be zero. Note that the electric field inside a current-carrying conductor is not zero -- that's what makes the charges move!
  8. Feb 1, 2008 #7

    Ben Niehoff

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    Yeah, I know, I just worked through 11 problems from Chapter 2 of Jackson. :P

    I meant, complicated relative to the OP's (assumed) understanding of electrostatics.
  9. Mar 31, 2008 #8
    can anyone prove it mathematically using probably gauss law or something?
  10. Apr 3, 2008 #9
    Gauss law is only really tractable analytically for charged cases about three cases (spherical symmetry, infinite line of charge, infinite plane of charge). Nevertheless, it's always true. if there's no charge inside your region, the surface integral over your gaussian surface is zero.
  11. Apr 3, 2008 #10
    You can't use Gauss law in general, but you may be able to do it with Coulomb's law and a good maths background in integration (analytical integration here is not trivial, see the Shell Theorem of Newton for an example - similar maths can be used for a hemispherical shell too).
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