Gauss's Law: Cond. versus Non-cond. Sheets: Conflicting Results

[Ackbach]

Homework Statement

Compare and contrast the electric field due to a uniform sheet of charge with charge density ## \sigma## for the conducting and non-conducting cases.

Homework Equations

Gauss's Law: $$\oint \mathbf{E}\cdot d\mathbf{A}=\frac{q_{\text{en}}}{\varepsilon_0}.$$

The Attempt at a Solution

I'm very familiar with the standard solution here - at least I think I am.

Let's examine the conducting case first. The issue at hand, is that some books (such as Halliday and Resnick, 9th Ed.) put one end of the Gaussian surface (either cylinder or rectangular prism) inside the conductor, and other authors (Griffiths, 4th Ed.) put the Gaussian surface entirely straddling the infinite sheet, one endcap on either side of the sheet. If you take the Halliday and Resnick approach, you get ##E=\sigma/\varepsilon_0##, because there is no field inside the conductor, so the endcap inside the conductor does not contribute to the flux of the Gaussian surface. On the other hand, if you take the Griffiths approach, there appears to be flux on both endcaps, so you get ##E=\sigma/(2\varepsilon_0).##

So, I have several questions here:

1. Are these two authors solving the same problem? I wonder if Halliday and Resnick isn't assuming the charge is only on one surface, effectively, while Griffiths has charge on both sides? And if Griffiths does have charge on both sides, why wouldn't the enclosed charge be ##2\sigma A## instead of only ##\sigma A##? In Griffiths, I'm talking about page 74, Example 2.5, which reads thus: "An infinite plane carries a uniform surface charge ##\sigma##. Find its electric field."
2. Wouldn't a realistic conducting sheet, if it had net surface charge density, have to have that surface charge density on both sides? The charges are going to want to get as far away from each other as possible. If so, should I interpret most authors as, essentially, putting half of that surface charge density on one side, and have on the other?
3. Related to the previous point: how does the thickness of the sheet play into the question?
4. In the non-conducting case, one resource I have used says that the field only goes one way, making the flux come out of only one endcap, and hence you get ##E=\sigma/\varepsilon_0##. But it's not entirely obvious to me why the field should be zero inside, or on the other side of, an infinite non-conducting sheet. What's going on here?

Thanks very much for your time!

 

[SammyS]

Homework Statement

Compare and contrast the electric field due to a uniform sheet of charge with charge density ## \sigma## for the conducting and non-conducting cases.

Homework Equations

Gauss's Law: $$\oint \mathbf{E}\cdot d\mathbf{A}=\frac{q_{\text{en}}}{\varepsilon_0}.$$

The Attempt at a Solution

I'm very familiar with the standard solution here - at least I think I am.

Let's examine the conducting case first. The issue at hand, is that some books (such as Halliday and Resnick, 9th Ed.) put one end of the Gaussian surface (either cylinder or rectangular prism) inside the conductor, and other authors (Griffiths, 4th Ed.) put the Gaussian surface entirely straddling the infinite sheet, one endcap on either side of the sheet. If you take the Halliday and Resnick approach, you get ##E=\sigma/\varepsilon_0##, because there is no field inside the conductor, so the endcap inside the conductor does not contribute to the flux of the Gaussian surface. On the other hand, if you take the Griffiths approach, there appears to be flux on both endcaps, so you get ##E=\sigma/(2\varepsilon_0).##

So, I have several questions here:

1. Are these two authors solving the same problem? I wonder if Halliday and Resnick isn't assuming the charge is only on one surface, effectively, while Griffiths has charge on both sides? And if Griffiths does have charge on both sides, why wouldn't the enclosed charge be ##2\sigma A## instead of only ##\sigma A##? In Griffiths, I'm talking about page 74, Example 2.5, which reads thus: "An infinite plane carries a uniform surface charge ##\sigma##. Find its electric field."
2. Wouldn't a realistic conducting sheet, if it had net surface charge density, have to have that surface charge density on both sides? The charges are going to want to get as far away from each other as possible. If so, should I interpret most authors as, essentially, putting half of that surface charge density on one side, and have on the other?
3. Related to the previous point: how does the thickness of the sheet play into the question?
4. In the non-conducting case, one resource I have used says that the field only goes one way, making the flux come out of only one endcap, and hence you get ##E=\sigma/\varepsilon_0##. But it's not entirely obvious to me why the field should be zero inside, or on the other side of, an infinite non-conducting sheet. What's going on here?

Thanks very much for your time!

I suspect that the Halliday and Resnick approach has a charge density of σ on a single surface of the conducting sheet, while Griffiths has a total of σ which includes both surfaces, thus σ/2 on each surface.

 

[Ackbach]

I suspect that the Halliday and Resnick approach has a charge density of σ on a single surface of the conducting sheet, while Griffiths has a total of σ which includes both surfaces, thus σ/2 on each surface.

So, in setting up Gauss's Law in the Griffiths approach, the charge enclosed on one side of the sheet would contribute towards the electric field at the endcap on the other side of the sheet? Odd that Griffiths nowhere mentions this. It does make sense, though, and if you're thinking of the sheet as thin, then saying the surface charge density is ##\sigma## works ok, I suppose. Am I on the right track here?

 

[SammyS]

So, in setting up Gauss's Law in the Griffiths approach, the charge enclosed on one side of the sheet would contribute towards the electric field at the endcap on the other side of the sheet? Odd that Griffiths nowhere mentions this. It does make sense, though, and if you're thinking of the sheet as thin, then saying the surface charge density is ##\sigma## works ok, I suppose. Am I on the right track here?

Yes. It all works out, either way you look at it.

 

[Ackbach]

Or I suppose another way of looking at the Griffiths setup is to imagine the sheet to be one particle thick.