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E-field vanishing *at* the surface of a conductor

  1. Sep 5, 2014 #1
    Suppose we apply a uniform field to an infinite conducting slab (i.e. like an infinite parallel plate capacitor, but the interior is included as part of the conductor). What is the resulting field?

    The simple answer is that a surface charge develops on the boundary planes of the conductor so as to cancel the field inside. Outside, the field is unaffected by the surface charges. Since the field inside the conductor is zero, the charge configuration is static.

    But, it seems that the field on the boundary of the conductor is not zero (and hence, a small charge [itex]d\sigma[/itex] on the boundary would move). To see this, let a small charge [itex]d\sigma[/itex] be located at a point p on the left plate. Then the field acting on [itex]d\sigma[/itex] due to the rest of the left plate is zero, so the field acting on it is due only to the external field and the right plate. But the right plate alone is insufficient to cancel the external field. Hence, there is a non-zero field acting on [itex]d\sigma[/itex].

    I am not looking for a physical resolution of this problem. I do not care what actually physically happens (i.e. that there are no true surface charges, that the surface charge dies off exponentially, etc.). I am, rather, interested in whether there is a consistent solution within the idealized model. So, I can restate my question more precisely as follows:

    Suppose we have a conducting slab, C, where the slab fills the closed region [itex]R_C = \{(x,y,z)|-1 \le x \le 1\}[/itex]. My question is whether there exists a static charge distribution (along with whatever boundary conditions you need) such that:

    (1) E(x,y,z) is constant at all points outside [itex]R_C[/itex] and is perpendicular to the conductor surface
    (2) E(x,y,z) is well-defined at every point in space
    (3) If [itex](x,y,z) \in R_C[/itex], then E(x,y,z) = 0.
    Last edited: Sep 5, 2014
  2. jcsd
  3. Sep 5, 2014 #2

    Simon Bridge

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    That is not correct.
    1st of all, Nature does not break down the physics like that - the entire combination of fields has to result in a net zero field inside the conductor. If you removed that bit of surface charge, the net field would not be zero inside the conductor.

    Handling surfaces is a little tricky conceptually. In an ideal conductor - charge is considered to be "on" the surface and not part of the conductor. Much like a hockey-puck slide "on" the surface of ice. But the bottom line is that you are overthinking the model.

    2nd. IRL there are no perfect conductors - the better description is that the charges orient to a minimum potential energy configuration.

    3rd - the continuous description is only an approximation that does not withstand microscopic examination.
    This is because charge is quantized.
    Last edited: Sep 5, 2014
  4. Sep 6, 2014 #3


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    I don't think it can be done (although I'd love to see a counterexample). There's nothing wrong with the ##\vec{E}## field, but the infinite charge density at the surface is more problematic.

    However, I also don't think this is a serious problem. Consider that Coulomb's law fails at ##r=0## but we couldn't live without it everywhere else. Likewise, we don't need a charge density that works at the idealized surface as long as we can get useful ##\vec{E}## values arbitrarily close to that surface with some physically reasonable (not the zero-thickness infinite-density charge distribution) even closer.
    Last edited: Sep 6, 2014
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