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Charge outside and inside a hollow ball

  1. Feb 3, 2013 #1
    1. The problem statement, all variables and given/known data
    http://home.phys.ntnu.no/brukdef/undervisning/tfy4155/ovinger/Ov03.pdf

    Hi. Take a look at the picture under assignment 2 ("oppgave 2"). Let's say there is a charge +q inside the cavity of the conducting metal ball.

    The problem is:

    "How will (free) charge be distributed when the system is in an electrostatic equilibrium?
    Draw the field-lines for the electrostatic field E. Find an expression for E outside the sphere".

    2. Relevant equations
    Gauss' law

    3. The attempt at a solution
    I know from Gauss' law the electric field is 0 inside the solid part of the ball. I also know -q charge will accumulate at the edge of the cavity close to the q charge, and +q charge will again accumulate on the edges of the ball.

    So my questions are:

    1) Will the electric field be at its strongest in the lower part of the figure?

    2) Will the electric field be like that around a point charge +q, inside the cavity? Ie, will the field lines go radially outwards from q inside the cavity?
     
    Last edited: Feb 3, 2013
  2. jcsd
  3. Feb 3, 2013 #2

    mfb

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    The link looks broken.

    What do you know about field lines ending on a conductor (here: the outer ball)? Does your idea satisfy that condition?
     
  4. Feb 3, 2013 #3
    Apologies. It should work now.

    Yeah, true, the field lines should go towards the small negative charges inducted by the main positive charge, right? So the field lines bend.

    However, will the electric field outside the hollow ball be at their strongest on bottom, or will they spread radially outwards from the ball? Check the link (oppgave 2) to know what I talk about http://home.phys.ntnu.no/brukdef/undervisning/tfy4155/ovinger/Ov03.pdf
     
    Last edited: Feb 3, 2013
  5. Feb 3, 2013 #4

    mfb

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    Right

    What do you know about the electric potential on the outer surface? ;)
     
  6. Feb 3, 2013 #5
    I think i get it. Gauss law somehow says the E-field is constant there, but I am a bit sketchy on this. Could you give a thorough explaination?
     
  7. Feb 4, 2013 #6

    mfb

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    The potential has to be the same everywhere on the outer surface, this gives a spherical symmetry for the field outside.
     
  8. Feb 4, 2013 #7
    Why does it have to be the same?
     
  9. Feb 4, 2013 #8

    mfb

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    The surface is conducting, a potential difference would lead to a current.
     
  10. Feb 4, 2013 #9
    We havent started on potensials yet, so is there another explaination? Mayve 1 using gauss law?
     
  11. Feb 4, 2013 #10

    mfb

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    Just argue with spherical symmetry then.
    For a proper explanation, you have to use that the surface is conducting (otherwise the field can be different), and therefore the electric field along the surface vanishes. But this is just another way to say "the surface has the same potential everywhere".
     
  12. Feb 4, 2013 #11

    vela

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    As you noted, the electric field inside the conductor is zero, so the charge distribution in the cavity can't affect the charge distribution on the outside surface.
     
  13. Feb 4, 2013 #12
    ∫E*dA = E*4pi*r^2 = q/ε. => E = q*k/r^2, with r> radii of the ball.

    But here E= the total field passing through a shell with area 4pi*r^2. How can I show the E-field is constant by using spherical symmetry?

    You see, we haven't gone through electric potential, so we're supposed to solve this using knowledge of Coulomb's law, electric fields and Gauss' law.

    Why not?
     
  14. Feb 4, 2013 #13

    haruspex

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    The induced charge on the inside surface of the cavity is exactly that needed to neutralise the field created by the charges within the cavity. The neutralisation is effective throughout the conductor and beyond (outside). It also changes the field in the cavity but does not neutralise it.
     
  15. Feb 4, 2013 #14

    vela

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    F = qE. Zero electric field means no force.
     
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