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Charge and field distribution for spherical conductor with cavity

  1. Oct 20, 2013 #1
    say you have spherical metal conductor with a cavity with a positive charge inside, the field inside the cavity isn't zero and will induce an opposite charge/field on the surface of the cavity which will cancel the charge inside and lead to a zero Electric field inside the conductor. the negative charge on the wall/surface of the cavity (?) will in turn induce charges which move to the surface of the cylinder in uniform configuration. so this E field points normally outward though the E field inside the conductor is zero and information about the cavity is hidden from the outside because the E field is zero inside and the cavity charge induces an equal opposite charge on the cavity surface/walls
     
    Last edited: Oct 20, 2013
  2. jcsd
  3. Oct 20, 2013 #2

    Drakkith

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    Did you have a specific question?
     
  4. Oct 20, 2013 #3
    does the charge inside induce a charge on inner or outer cavity wall? and since the E field from the charge and induced negative charge cancel does the induced charge on the outer surface create an electric field that makes a contribution inside?
     
  5. Oct 20, 2013 #4

    clem

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    ...............
     
  6. Oct 20, 2013 #5

    Drakkith

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    Of course.

    No, it only affects the outside.
     
  7. Oct 28, 2013 #6
    why exactly is the charge inside a conductor equal to null and in the presence of an E field the electrons will rearrange themselves so as to cancel the outer field? do the electrons naturally move to some configuration that minimizes potential energy
     
  8. Oct 28, 2013 #7

    Drakkith

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    I don't know the terminology. All I can say is that the positive charge on the inside attracts the negatively charged electrons and pulls some towards the inside edge. No charges have left the conductor, so it's still neutral.
     
  9. Oct 28, 2013 #8

    WannabeNewton

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    The electric field in the material of the conductor is identically zero hence from Gauss's law we have ##\vec{\nabla} \cdot \vec{E} = 0 = 4\pi\rho##. In other words, all the charges move to the boundary of the conductor.

    We are talking about a system in electrostatic equilibrium so the charges will keep moving (which they can do since they are in a conductor) until the Coulomb force density vanishes.
     
  10. Oct 28, 2013 #9
    thank you that is really deep and beautiful! I only wish circumstances and health allowed me to become a physicist or mathematician by now

    so according to Gauss's law then E is 0 since the charge distribution is such that the net charge enclosed is 0 ( - I'm also studying these boundary discontinuities in Griffiths at surfaces). I really should study more of the Purcell Berkeley course
    so the Coulomb force is like a dense gas perhaps it is quantized into differential vectors at each point that may be uniform or not and are proportional to 1/r^2. but how do the random charge distribution always change so that force density or field vanishes, (except at the boundary I'm guessing that the E field points radially outward?)
     
    Last edited: Oct 28, 2013
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