Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Staff Emeritus
    Science Advisor
    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

    User Avatar
    Science Advisor

    You can still use a spherical surface though.

  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

    User Avatar
    Science Advisor

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

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook