# Why the electric field $\vec{E}$ is constant (=position independent) for an infinite 2D sheet of constant charge?

So I'm reading a text on electricity and it talks about using the integral to compute the total charge of a collection of points, which I mostly understand. But then we get to finding the electric field due to a charged collection of points and I find things that don't make sense to me. For instance, for an infinite sheet of constant charge, the text says that the electric field is constant on any one side of the sheet. But that seems intuitively wrong to me, since I would think the field should be stronger the closer a point is to the sheet. I mean, if I'm standing 10 meters from a sheet, holding a charged particle, and walk closer to the sheet, I'd think the particle would react more strongly. I follow the mathematical derivation fairly well, which leads me to think I must not be thinking about the physics correctly. Can anyone help make this make sense?

-
You can get this result without Gauss's law, just on dimensional grounds. There's no way to put together any equation in terms of the charge density and the distance $r$ from the sheet that will have the right units and vary with $r$. –  Ben Crowell Nov 17 '14 at 16:02

for an infinite sheet of constant charge, the text says that the electric field is constant on any one side of the sheet. But that seems intuitively wrong to me, since I would think the field should be stronger the closer a point is to the sheet.

There's a geometric scaling argument at hand, and you probably need to appreciate Gauss' Law to get a real sense of it. It follows the same thinking that the $1/r^2$ law does.

In this argument for $1/r^2$ field strength from a point particle, it is seen that the solid angle that "A" occupies decreases at larger radii. If you consider a charged ball, then think about squishing it from a 3D object into a 2D piece of paper. That represents the area-based charge density.

An unsaid assumption for this line of thinking are that the field strength is determined by:

• The solid angle occupied by charged material times
• The 2D charge density presented by that material

So extend this to an infinite sheet. No matter how close or far away it is, the sheet occupies exactly half of your field of vision. Furthermore, the surface charge density and angles of the sheet are also irrelevant of your normal distance.

Illustratively, you could apply this via the image above. Just remove the "A", and consider that this is an infinite sheet. As you expand the distance, you expand the area you sweep, but the angle times the charge density is invariant. Thus, the field is constant with distance.

-
I see, I'll go re-read Gauss's Law. But as I see it right now, the explanation seems to be: As you get close, sure, the field strength from any one point is stronger, but the component of that strength which is directing you toward/away from the sheet is diminishing because of the angle at which it "hits" you. The rest of the components just cancel with the forces coming from a symmetrically located other point on the sheet. Hrrrrrrn, very cool! –  Addem Jul 11 '14 at 18:45

It's definitely strains one's intuition. But think about this: if you are in the vicinity of an infinite sheet of charge, how can you determine whether you are close to it or you are far away? I can tell if I'm close to or far from a sphere by comparing my distance to the sphere with the radius. If I'm close the sphere looks big. If I'm far away it will look small, even like a point. The size of the sphere "sets the scale". Not so for an infinite sheet. No matter where you are, the sheet looks exactly the same. There is no scale. From this point of view one might say "what else could it be but constant" (and pointing perpendicular to the sheet).

-
But what about a point charge? The scaling argument doesn't work here, does it? –  Christian 8 hours ago

Interestingly, as AlanSE mentioned above, this can be derived from Gauss's Law very intuitively if you consider that an infinitely large sheet (a plane) is identically the same as a sphere of infinite radius. Therefore, the flux of your test particle through this sphere (and vice versa) is the same regardless of its position within this sphere, which can be said to extend to all points in the space.

-
With a point in space and a plane running through the origin, there is a (shortest) distance to the plane. I don't see how we can similarly regard this as a sphere of infinite radius unless you mean some kind of analogue to higher dimensions. –  Addem Jul 11 '14 at 21:16