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Electric induction

  1. Nov 23, 2013 #1
    Hi,

    I have been told that you take say a metal board and another board (both endless and paralell)
    leads to the following effect:

    The electrons are more or less freel in motion, they move to the surface until we reach an equilibrium. The electrons are assembled at the surface. Nice.

    Next lesson: Internally there is no electrical field. It holds:

    div(rho)=0 because rho is 0.

    Okay, rho is zero when the electrons are missing at the other side of the board...
    This counts only for the volume but not the surface!?

    Do I get something trivial wrong or is there a deeper reason?

    Thanks!
     
  2. jcsd
  3. Nov 23, 2013 #2

    WannabeNewton

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    Given an arbitrary conductor in electrostatic equilibrium, ##\rho = 0## in the material of the conductor because ##\vec{E} = 0## in the material of the conductor and from Gauss's law we then have ##\vec{\nabla}\cdot \vec{E} = 0 = 4\pi \rho##. But of course there is charge density on the surface of the conductor as you yourself noted; the surface charge density ##\sigma## is given by the jump condition in the electric field when passing from the interior of the conductor to the exterior: ##\hat{n}\cdot (\vec{E}_2 - \vec{E}_1) = 4\pi \sigma## or equivalently ##\hat{n}\cdot (\vec{\nabla}\varphi_2 - \vec{\nabla}\varphi_1) = -4\pi \sigma## where ##\vec{E}_2## is the exterior electric field, ##\vec{E}_1## is the interior electric field, and ##\hat{n}## is the unit normal to the surface of the conductor. This is assuming there are no dielectric materials present (the modification to the jump condition in the presence of dielectrics is very straightforward anyways). This is all in Gaussian units by the way.
     
  4. Nov 23, 2013 #3
    Thanks, but this is exactly what I read in every book. You say "surface charge density is given by the jump condition in the electric field when passing from the interior of the conductor to the exterior" but why?
    Take the not influenced object as electrically neutral. div(rho)=0 naturally holds. Apply a potential, you learn
    that the jump conditions hold. Jumps are something where you mathematically should be alarmed.
    But okay, it's fine if you accept that the distribution of the electrons is suddenly on the surface.

    My question is: Why?
    What about the other side?

    There is another reason which is not explained in detail. Could you give me a curve describing the electron density in this conductur?
     
  5. Nov 23, 2013 #4

    WannabeNewton

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    Why what? Why all the charges are on the surface of the conductor after reaching electrostatic equilibrium?
     
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