Is it true that $\vec{E}\neq 0$ inside a 1- or 2-dimensional conductor?

Question

It is known that when a conductor is placed in an electric field, the charges redistribute themselves such that $E=0$ inside the conductor. I was also told that the same is NOT true for the 2D and 1D analogues. Why is this true? And what happens in these cases?

Thank you.

Answer 1

"Physically" speaking you're right. In any dimensions the charges are always at the conductor boundary. In particular this is true for 3D, 2D and 1D cases.

However one may think of a 2D/1D object in a 3D space. For instance, you may pick a disk (with zero depth). The charges will not be just on the disk boundary (circle), but also on the disk surface.

Anyway there's no contradiction here. For a 2D object in a 3D space every point is actually a boundary point. That is, a point in which the charge movement is constrained to the 2D object.

Answer 2

In any dimension, if you have a conductor, the electric field will be zero on the interior. The correct lower dimensional analogs are found by reducing the dimension of space, not by reducing the dimension of the conductor only. The one dimensional analog is a space which is a one dimensional line, just the x-axis, with a certain conducting interval in the middle stretching from $x=a$ to $x=b$, and an electric field coming in from infinity, so that the electric field is $E$ at $x=\pm\infty$.

The three dimensional version of this is found by extending everything into two perpendicular dimensions, so you get an infinite block conductor which makes an infinite plane, sandwiched in the x-direction by the two ends of the interval. The electric field does not extend into the conducting block. The dimensional reduction of this is a field which does not penetrate into the interval [a,b].

Similarly, when you consider a 2d conducting disk, you are equivalently looking at a cylinder in three dimensions, and then the field is zero on the interior of the cylinder.

If you just make a nearly flat disk conductor, the charge density is in general nonzero on all the surfaces of the boundary, including the top and bottom planes. This means that the disk limit will potentially not have all the charge on the outer circle, but it will have charge density on the interior

But there is a simple heuristic which suggests otherwise--- in the limit of a thin plate, shouldn't the charge on the top and bottom surface of the plate be exactly cancelling? This limit is what I will discuss here.

When you ahve a thin plate, with the electric field coming in directly perpendicular to the plate, the induced surface charge densities on the top and bottom are equal to the external perpendicular field, and they cancel in the thin-plate limit. This means that the charge density on the plate does have a certain sense in which it is like a dimensionally reduced problem--- the field in the direction perpendicular to the plate has nothing to do with the charge density on the plate, when you sum over the top and bottom surface.

But when there is a field in the plane of the disk, it will induce charges on the ring, the field of the induced charges on the ring will be along the direction of the disk, and in order to cancel this field in the interior, both the top and bottom charge densities need an additional component which has the same sign. The field from the induced charge on the ring will not be zero on the interior of the disk, so it needs to be cancelled by field from induced charges on the disk.

This is a two dimensional problem, but not the dimensionally reduced two dimensional problem, a different one. This problem gives an integral equation (an infinite dimensional linear matrix equation) for the charge distribution in 2d or 1d.

Answer 3

This is discussed in my question on the MathOverflow forum:
http://mathoverflow.net/questions/80731/minimize-energy-for-charge-distributions

And you will find references therein for the explicit counterexamples of 1D+2D.

[Edit]:
IF the 2D objects are embedded in $\mathbb{R}^3$, then the charge still moves to the boundary, because the whole object is a boundary! So there is no contradiction here.
BUT IF you consider the 2D objects as resting in $\mathbb{R}^2$, then no the charge does not all go to the boundary (the 'tips' of the needle, or 'outer ring' of the disk). This is because the Coulomb potential is not harmonic in dimensions 1 or 2.
*The above assumes we use a potential that scales as $\frac{1}{r}$. But we could alternatively use a potential that scales as $\ln\frac{1}{r}$ in 2D, in which case again the charges will lie on the boundary of the objects.