# Electric field: infinite sheet vs infinite plane

Griffiths' Electrodynamics says that the electric field of a uniformly charged infinite plane, surface charge density sigma, is sigma/2e0. The field of an infinite sheet of charge is said to be sigma/e0, twice that of the plane.

What is the supposed difference between the sheet and the plane?

thanks.

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PS--My confusion increases the more I look at Griffiths (Electrodynamics 3rd edition). On p. 89 note 6, while discussing the field of a sheet of charge, Griffiths writes,

if you're only interested in the field due to the essentially flat local patch of surface charge itself, the answer is (sigma/2e0)n immediately above the surface, and -(sigma/2e0)n immediately below... If you are close enough to the patch it "looks" like an infinite plane. Evidently the entire discontinuity in E is attributable to the local patch of charge.

Perhaps someone can explain this? Thanks.

jtbell
Mentor
The field of an infinite sheet of charge is said to be sigma/e0, twice that of the plane.
Where does Griffiths (or whoever, if it's from some other source) say this? (I have Griffiths 3rd edition, so if you have some other edition, give the section number and name and I can probably figure it out.)

##E = \sigma / \epsilon_0## is the magnitude of the field outside the (plane) surface of a conductor, where ##\sigma## is the charge density on that surface. I don't remember seeing this situation referred to as a "sheet of charge".

I also have Griffiths 3rd. On p. 74 he gives the electric field of an infinite plane with surface charge density sigma. This is (sigma/2e0)n.

On p. 89 he gives the difference in the fields between the upper and lower surface of charge sigma. This is (sigma/e0)n.

I see now that I misquoted him: he talks about a surface, not a sheet. I don't know whether that misquote is important. In any case, I believe that Griffiths' point is that the difference between the sheet's upper and lower fields is just twice the magntitude of the field on either side. That makes sense the upper and lower fields point in opposite directions. Griffiths' Fig. 2.36 shows these fields going the same direction. Perhaps the figure is wrong?

jtbell
Mentor
[...] the upper and lower fields point in opposite directions. Griffiths' Fig. 2.36 shows these fields going the same direction. Perhaps the figure is wrong?
I can see how that diagram would be confusing. It probably confused me a bit too when I first saw it, years ago! However, I can think of two things that might have influenced that diagram:

1. Further down the page, he says:
For consistency, we let "upward" be the positive direction for both [Eabove and Ebelow].
This makes it easier to "see" their difference, Eabove - Ebelow.

2. There are situations in which both fields do point in the same direction. Suppose the surface by itself produces an E with magnitude 10 V/m, upwards above the surface and downwards below it. Place that surface in a room which already has an E = 100 V/m upwards inside it. Now we have E = 110 V/m upwards above the surface, and E = 90 V/m upwards below it. In either case, the difference Eabove - Ebelow = 20 V/m upwards.

ehild
Homework Helper
Surface charge density is used in two different meanings.
If you have a planar arrangement of charge with surface charge density σ it means that unit area of the plane has σ charge. Half of the field lines emerging from the charges appear at one side of the plane, the other half of the lines appear at the other side, so the electric field intensity is σ/(2ε0) on both sides and point away from the plane if σ>0.
A metal sheet has two surfaces and the charge appears on the surfaces. If the charge density on the surfaces is σ, all the field lines emerging from these charges appear at one side of the sheet. So the electric field intensity is σ/(ε0) at both sides. Thanks for your replies, jtbell and ehild.

Looking at ehild's diagrams, I wonder if my understanding is correct. Imagine a Gaussian pillbox of unit area on the right side of the charged sheet, the box enclosing just the surface charges on that side and extending only halfway into the sheet. The enclosed charge is sigma/2, the field is sigma/(2e0). If the sheet were somehow made progressively thinner, it would approach a plane of charge. If the sheet were to become a plane, the pillbox would enclose a charge of sigma and the field would be sigma/e0. Yes?

Griffiths pointed out that the change in electric field on crossing a surface charge sigma is sigma/e0, and that the change is discontinuous. Changing a sheet of charge to a plane amounts to the same thing, no?

ehild
Homework Helper
The surface charge σ on both sides of the sheet means 2σ charge on unit area of the plane.