I am trying to figure out why in a conductor, lets for example say a uniform sheet of metal (infinitely large to neglect edge effects) that has a net positive charge, at all points inside the conductor the electric field is zero. I understand the Gauss's law explanation that there is no net free charge enclosed by a Gaussian surface inside the conductor because all the free charge is on the surface. This makes sense, but then I get confused if I think further. If you look at a point inside the slab that is closer to one side of the sheet of charged metal, shouldn't there be a stronger field as you approach that side due to closer proximity to the charges on that surface? I appreciate the help


1 Answer 1


This example illustrates the answer for spherical geometry but the argument is the same in the planar geometry of your question.

enter image description here

On the near side of the sphere, closer to the point inside the sphere, the amount of charge captured in the angular opening is small, but the charges are closer. On the far side, the amount of charge captured by the angular opening is large but the charges are farther away. Because the area of the angular opening grows like $r^2$ while the electric field $\vert E\vert\sim 1/r^2$, the effect of the larger distance is exactly balanced by the effect of the larger area.

This other example shows how this works for an arbitrary surface.

enter image description here

By inscribing a sphere centred at any point inside, one shows that the same must hold for the opposing areas of the surface, so that the contribution from the larger number of farther charges exactly cancels the contribution from the nearer number of fewer charges.

To be a little more explicit, the right dashed lines define a cone (in 3d) that would intercept the sphere and so contain on the intercepted surface a certain amount of charge $dq_R\sim \sigma \ell_R^2$, where $\ell_R$ is the distance from the point to this patch of surface. The contribution to the field from this patch of charge is $dE_R\sim \sigma \ell_R^2/\ell_R^2\sim\sigma$. The same goes for the dashed lines on the left, where $dq_L\sim \sigma \ell_L^2$ and $dE_L\sim \sigma \ell_L^2/\ell_L^2\sim\sigma $. The magnitude of $dE_L$ is thus the same as that of $dE_R$ but their directions are opposite, so they exactly cancel at the point $P$ inside your surface.

Obviously in the case of planar geometry, this can only hold if the surfaces are infinite in extent.

I should add that strictly speaking this argument only works for infinitesimal openings, so that all the points "on the right" are at the same distance from the point inside, and all the points "on the left" are also at the same distance from the point inside.

  • $\begingroup$ I am not sure I understand the setup for the explanation using the angular openings? Could you explain this logic further for me? Thanks $\endgroup$
    – Joe
    Jul 17, 2017 at 23:11
  • $\begingroup$ @Joe I added a paragraph which hopefully helps. $\endgroup$ Jul 17, 2017 at 23:24
  • $\begingroup$ @ZeroTheHero Does the demonstration for the arbitrarily shaped conductor assume a uniform surface charge density $\sigma$ ? $\endgroup$
    – Hilbert
    Aug 14, 2019 at 22:40
  • 1
    $\begingroup$ @Hilbert Good question. I think yes inasmuch as charges would distribute themselves uniformly on a sphere or on an infinite slab, as the example of the OP. $\endgroup$ Aug 15, 2019 at 0:23

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