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Still more problems with Gauss Law

  1. Oct 6, 2004 #1
    Ok i'm realllly unsure about this!

    If a spherical cavity of radius 3.66cm in a piece of metal (kind of cube, but not a perfect cube) has a charge of +Q at it's centre and there is a point P1 located half way between the spherical cavity and it's surface and a point P2 located in the metal piece , use gauss Law to find the electric field at
    a) Point P1
    b) point P2

    a) This is fine for me

    Let e0 = permittivity of free space epsilon 0

    e0 SurfaceIntegral (E dA) = q
    e0 E (4 pi r^2 ) = q
    E = kq /r^2
    which is coulomb's law derived! and just plug and away i go!

    b) But for point P2

    err can i actually use a spherical surface here??

    as in do the same thing as before but use a bigger value for r i.e. radius of sphere + distance of sphere to point P2

    appropriate? or not?
     
  2. jcsd
  3. Oct 6, 2004 #2

    Gokul43201

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    Gold Member

    Umm...the answer to that would be : NOT YET. Mostly because you don't yet know what charges are enclosed by this Gaussian surface.

    Read what your text has to say about induced charges and electric fields inside a piece of metal.
     
  4. Oct 7, 2004 #3

    Claude Bile

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    You can still use a spherical surface though.

    Claude.
     
  5. Oct 7, 2004 #4

    i'm not quite sure how i would use it , so the shere would have a radius of 3.66cm + A ??

    e0 E 4 pi (3.66+A)^2 = 126 x 10^-9

    something liek this??
     
  6. Oct 10, 2004 #5

    Claude Bile

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    You need to take into account the induced charges on the inner surface of the metal.

    Claude.
     
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