1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Electric potential of a negatively charged sphere

  1. Jan 17, 2005 #1
    A negatively charged sphere is brought near an uncharged metal object. Positive charges appear on the side of the uncharged metal object nearest to the charged sphere, negative charges on the opposite side.
    [​IMG]
    I don't understand how the potential could be the same everywhere... There's a clear separation of charges which shoud produce a field imo.
     
  2. jcsd
  3. Jan 17, 2005 #2
    first, let's make it clear.. the potential is not the same "EVERYWHERE", however, it has the same value INSIDE the metal. you can prove it even without mathematics.
    In an zero resistant objecct, charge flows from high potential to low potential since nothing prevent it to do so.. so, if the potential is not same everywhere inside the metal, the charge will flow around, the high potential area will drop its potential because the charge is leaving.. the opposite is true for the low potential area.. until the system reaches an equilenvence point.. the charge will stop flowing around and the system has equipotential..
    I think you confused the concept about electric field and potential.
     
  4. Jan 17, 2005 #3
    it must be that :(
    I see there some induced charge on the right side and the equivalent opposite charge on the other side. My idea is that there should be an electric field created due to that separation. And if you are an electron and you travel through that field, you are passing through a potential difference.
     
  5. Jan 17, 2005 #4

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Consider this rectangular slab of metal under the influence of an externally applied field E0:

    |-_______+|
    |-_______+|
    |-_______+|
    |-_______+|
    |-_______+|
    |- <----E1+|
    |-_______+|
    |-_______+|
    |-_______+|
    |-_______+|

    ----> E0

    Fig. 2.42

    It makes sense that the externally applied field will induce movement in the charges within the conductor, creating a separation of charge. That separation of charge results in an electric field of its own, but the movement of charges continues only until this new field E1 exactly balances the applied field E0. Here's a more detailed explanation:

    I think that addresses exactly the question you were asking.
     
  6. Jan 18, 2005 #5
    thank you!
    so, from which I deduce that if the object was a dielectric it would have a different potential on the sides because the charges are not free to move. All they can do is shift a bit creating an electric field inside which does not cancel the field outside.
     
  7. Jan 19, 2005 #6

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Exactly...it partially cancels the field outside, acting like a poor conductor. In my first year physics book (not Griffiths), it says that dielectrics can become true conductors under *very* strong applied fields, and that this is called "dielectric breakdown" if I remember right.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?