I am studying introductory electricity and magnetism, and this is a conceptual question.

For an infinite conducting sheet, any excess charge is localized onto one side. There is no electric field in the interior of the surface or the other side. The reasoning, as I understand it, is that any such field that exists would cause movement of charge within the surface, and then the surface would not be at equilibrium. Thus, when applying Gauss' Law, the flux through the surface is taken to be 0. See this for what I mean: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gausur.html#c3

However, if we were to treat the charges present on the one side of the surface as point charges (as is so often done when deriving an electric field for a surface using integration), Coulomb's Law means that those charges will exert an electrical force on any charged particle, even on the other side of the surface. Coulomb's Law for the attraction between two charges applies regardless of whether there is material between them, right? Won't this mean than if we held a positive point charge close to the other side of the infinite conducting sheet, it would feel a repulsion? This, then, would mean that the positive charges located on one side of the sheet DO generate an electric field that goes through the sheet to the other side, and that there IS an electric flux within the surface. What's wrong with this way of thinking?

  • $\begingroup$ I assume by "the flux through the surface is taken to be 0", you meant the flux within the conductor is 0. $\endgroup$ May 1, 2017 at 18:04

1 Answer 1


Charges could be distributed along only one side of the infinite conducting plate, with one of two Cases:

  1. The conductor has no thickness, you just have a theoretical distribution of charges on a plane, or
  2. The conductor has thickness, and the charges have to be perfectly straight for electrostatic equilibrium.

Since you said the E field could "go through the sheet", I'm gonna assume you meant Case 2.

In Case 2, you could try to keep all charges in only one plane. However, any charge not perfectly in line with the other charges will cause other charges to move, breaking the electrostatic equilibrium. So we call the configuration of charges along only one side of the conductor an unstable equilibrium.

Generally we don't consider unstable electrostatic distribution of charges as being under electrostatic equilibrium. For example, the excess charges on a conducting sphere at electrostatic equilibrium is always on the outer surface of the sphere, even though having all charges in the exact center of the sphere is also a valid (but unstable) electrostatic equilibrium.

My above reasoning sounds sketchy, and could be wrong. But I was just trying to get you a feel of the Uniqueness Theorem. If you can understand the math, you will know why there is only one possible distribution of charges for a conductor in electrostatic equilibrium.

So by the Uniqueness Theorem, the excess charges are distributed on both sides of the infinite conducting plate.

A positive test charge on the other side feels a repulsion away from the conducting plate only because of E field generated by excess charges on the side closer to the positive test charge.

The E field generated by excess charges on the further side do not affect the positive test charge. This is because within the conductor, the E field generated by excess charges on the further side is cancelled out by the E field generated by excess charges on the closer side.

Thus there is no net E field within the conductor. You can also prove there is no net E field in the conductor by drawing a Gaussian surface inside the conductor (excluding the surface charges). Since there is no net charge enclosed, electric flux, and thus electric field, is zero.

P.S. you don't need to put the positive test charge near the infinite conducting plate for repulsion. The E field is uniform, thus does not vary with distance from the plate. A positive test charge very far away will experience the same repulsion.

  • $\begingroup$ As I understand it, for an infinite sheet of nonzero thickness (yes, #2), all the excess charge is localized to the surface of one side of the sheet. $\endgroup$
    – Yunfei Ma
    May 1, 2017 at 22:30
  • $\begingroup$ I said that was wrong in my answer. Do you think that's correct because of the figure you linked to? The Gaussian surface in the figure only intersects one side of the conductor, making you think the charge is only on one side? By the Uniqueness Theorem, there is only one unique distribution of charges for a conductor at electrostatic equilibrium. Case 2 (where charges are distributed on one side) is unstable, so we won't see that in nature. Case 2 will transform to a more stable configuration, where charges are distributed on both sides. $\endgroup$ May 1, 2017 at 23:22
  • $\begingroup$ Try this similar question as a reference. physics.stackexchange.com/q/61810/131827 $\endgroup$ May 1, 2017 at 23:33

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