I Thought I Understood Gauss' Law (Sheets)

[flyusx]

Homework Statement

Is the electric field near a conducting sheet $\textbf{E}=\frac{\sigma}{\varepsilon_{0}}$?

Relevant Equations

$\displaystyle\oint \textbf{E}\cdot d^{2}\textbf{r}=\frac{Q_{enc}}{\varepsilon_{0}}$

We know Gauss's law for an infinite sheet yields $\textbf{E}=\frac{\sigma}{2\varepsilon_{0}}$. This is relatively elementary and I completely understand the derivation. This is also valid when looking at a parallel plate capacitor (the electric field is additive between the plates yielding $\textbf{E}_{cap}=\frac{\sigma}{\varepsilon_{0}}$, which can indeed be confirmed using the capacitance-charge-voltage and V=Ed relationship).

Here's an example problem:
A square plate of copper with 0.5m sides has no net charge and is placed in a region of uniform electric field of 80000$\frac{V}{m}$ directed perpendicularly to the plate. Find the charge density and total charge on each face.
The total charge can be found using Gauss's law by imagining a box passing through both sides of the sheet. $$2EA=2E(0.5m)^{2}=\frac{Q_{total}}{\varepsilon_{0}}\rightarrow Q_{total}=354nC\rightarrow Q_{side}=177nC$$
And of course, the charge density on each face is just the charge on each face divided by the area, yielding 708$\frac{nC}{m^{2}}$.

Great, this all makes sense. But here's the point of confusion. The electric field near the surface of a conductor is $\frac{\sigma}{\varepsilon_{0}}$. As I've read online, this is because the charge on a conductor resides on the surface (which I completely agree), and thus only one surface contains all the charge. In other words, $EA=\frac{Q_{enc}}{\varepsilon_{0}}$ which yields the previous equation. Simply draw a Gaussian surface that goes from somewhere inside the conducting sheet to the outside to get the EA. See here and here.

However, this directly contradicts the prior intuition I've developed since $\frac{\sigma}{2\varepsilon_{0}}$ should hold valid near a conducting sheet. I think the problem is where the charge resides on the conducting sheet. It must reside on both faces since the faces are the surfaces. Indeed, my solution to the example problem is correct and makes this assumption. However, the first link that includes a video dictates that the charge on the other face is zero because it is not a surface...but it clearly is a surface?
Here's another example of a discrepancy. When applied to a parallel plate capacitor that is made of metal, we generally take charges to reside on both faces of each capacitor and from there, we can derive the electric field in between the plates. However, if the plates of a conductor was made of a conductor, the charge density for each plate would instead be $\frac{\sigma}{\varepsilon_{0}}$ which wouldn't match up with reality.

Interestingly enough, for this example problem, the solution manual uses $\frac{\sigma}{\varepsilon_{0}}$. $$E=80000\frac{V}{m}=\frac{\sigma}{\varepsilon_{0}}\rightarrow\sigma=708\frac{nC}{m^{2}}$$ and the charge on each plate can be solved quite easily from there.

I'd appreciate some guidance.
Edit: I think packages that allow a double closed integral command oiint aren't automatically loaded, but that's what I meant in the relevant equations section.

 

[haruspex]

total charge can be found using Gauss's law

The total charge on the sheet?

A square plate of copper with 0.5m sides has no net charge

 

[flyusx]

It's the total charge on each face. Doesn't charge redistribute when placed in an electric field?

 

[kuruman]

You solved your example problem incorrectly. If it has no net charge, how can it have a total charge of 354 nC with 177 nC on each side?

Consider the conducting sheet in an external electric field. The electric field inside the conductor is zero. Consider the picture below. The left face has charge distribution $\sigma_1$ (magenta) and the right face $\sigma_2$ (red). There are three Gaussian pillboxes, red, magenta and blue. Write Gauss's law for each one of them and see what you can say about the charge distributions.

 

[Delta2]

Ye there is conservation of charge in the system, if initially its has no net charge, it just cant gain net charge just because it was placed in the external E-Field, unless you tell us where and how the net charge came from or went to.
Maybe you wanted to say that it has negative charge on one side and equal positive charge in the other side due to the effect of the external electric field? And so that the net charge keeps remaining zero (positive and negative charge cancel out).

 

[Delta2]

I found the problem not very well stated btw, it had to state that the plate had some thickness, even infinitesimal as @kuruman figure at post #4 give us.

 

[haruspex]

it had to state that the plate had some thickness

All plates do. It is only necessary to state idealisations.

 

[Orodruin]

What Gauss’ law really tells you is that the discontinuity in the orthogonal component of the electric field is $\sigma/\epsilon_0$. For a conductor, the field inside must be zero so the field outside must be $\sigma/\epsilon_0$.

 

[flyusx]

Thanks for your replies.
My example problem comes from Serway/Jewett Physics for Scientists and Engineers with Modern Physics, 9th edition.
The exact wording of the question is as follows:
A square plate of copper with 50.0-cm sides has no net charge and is placed in a region of uniform electric field of 80.0 kN/C directed perpendicularly to the plate. Find (a) the charge density of each face of the plate and (b) the total charge on each face.
The answer key gives the answers I got above. I assume in part (b), the answer key means +177nC on one side and -177nC on the other (though the answer key just says 177nC and leaves it at that). I also thought the problem/answer might have been poorly worded so I just took 177nC to be the magnitude on each side.

Write Gauss's law for each one of them and see what you can say about the charge distributions.

View attachment 339648

For the red: $$\oint_{S}\textbf{E}\cdot d^{2}\textbf{r}=EA=\frac{q_{2}}{\varepsilon_{0}}$$
For the magenta: $$\oint_{S}\textbf{E}\cdot d^{2}\textbf{r}=EA=\frac{q_{1}}{\varepsilon_{0}}$$
For the blue: $$\oint_{S}\textbf{E}\cdot d^{2}\textbf{r}=E2A=\frac{q_{1}+q_{2}}{\varepsilon_{0}}$$
Which yields $E_{red}=\frac{\sigma_{1}}{\varepsilon_{0}}$, $E_{magenta}=\frac{\sigma_{2}}{\varepsilon_{0}}$ and $E_{blue}=\frac{q_{1}+q_{2}}{2A\varepsilon_{0}}=\frac{0.5\sigma_{1}+0.5\sigma_{2}}{\varepsilon_{0}}$ respectively.
So if the charge distributions are identical on each side, $\sigma$ is identical? I don't really see how I could get $\frac{\sigma}{\varepsilon_{0}}$ versus $\frac{\sigma}{2\varepsilon_{0}}$.

 

[haruspex]

It's the total charge on each face. Doesn't charge redistribute when placed in an electric field?

But your method was completely invalid. Your Gaussian box passed

through both sides of the sheet

so should have contained zero net charge, but somehow you arrived at 354nC, a charge which does not exist anywhere in the setup.
Note that you would have arrived at the same result for an insulator.

A correct approach is to have the box only half into the sheet and use the fact that there is no field inside a conductor. Then you would have arrived validly at a magnitude of 177nC

 

[flyusx]

Note that you would have arrived at the same result for an insulator.

A correct approach is to have the box only half into the sheet and use the fact that there is no field inside a conductor. Then you would have arrived validly at a magnitude of 177nC

If I understand your message the right way, for an uncharged situation like my example, one should draw the Gaussian box to not pass through such that a net charge will be enclosed on one surface or the other. In the situation of a charged sheet, the Gaussian box can pass through because it's a charged sheet and will enclose charge either way.
If a sheet had a net charge, say 1μC total, and an area of 1$m^{2}$, then it would be applicable to use a Gaussian box that passed through all the way: $$\oint\textbf{E}\cdot d^{2}\textbf{r}=2EA=\frac{1μC}{\varepsilon_{0}}\rightarrow E=\frac{1μC}{2A\varepsilon_{0}}=\frac{\sigma}{2\varepsilon_{0}}$$
Or, one could draw a Gaussian box that passed through one side and ended in the middle of the conducting sheet and used the fact that the charge was distributed half and half on each surface: $$\oint\textbf{E}\cdot d^{2}\textbf{r}=EA=\frac{0.5μC}{\varepsilon_{0}}\rightarrow E=\frac{0.5μC}{A\varepsilon_{0}}=\frac{\sigma}{2\varepsilon_{0}}$$
Where $\sigma=\frac{1μC}{2A}$ in both scenarios.
Is my logic right? And what about a charged sheet of metal? When people write $E=\frac{\sigma}{2\varepsilon_{0}}$, they label it as an infinite sheet but don't discriminate between a conducting and non-conducting sheet. My textbook gives this derivation, but also says that the electric field of a sheet very close to a conducting (metal) surface is $E=\frac{\sigma}{\varepsilon_{0}}$. Isn't this contradictory, or is the definition of $\sigma$ different?

 

[nasu]

The flux through the blue box is zero and not 2EA. There is nothing wrong with using such a box, just not useful for this case.

 

[Delta2]

Guys the total net E-field in the space outside conductor wouldn't be altered by the charge redistribution on the two sides of the plates and would no longer be exactly homogeneous and with the value given, so the whole calculation is approximation under some assumption (that should also be stated in the problem IMO).

 

[Vanadium 50]

Guys the total net E-field in the space outside conductor wouldn't be altered by the charge redistribution on the two sides of the plates and would no longer be exactly homogeneous and with the value given, so the whole calculation is approximation under some

Incorrect and unhelpful.

The problem states what the field is. Not how it got there. Further, the important thing is to get the OP to understand the factor of two - not to quibble about irrelevancies.

 

[Delta2]

Incorrect and unhelpful.

The problem states what the field is. Not how it got there.

According to my interpretation it states what the field is BEFORE we insert the plate in the field.

Further, the important thing is to get the OP to understand the factor of two - not to quibble about irrelevancies.

I don't think its irrelevant, it adds extra insight. What's that insight? It is the insight that the charge density due to redistribution of charge in the surface of the conductor causes its own E-field which alters the total E-field.

But it is something more important than that: The thing we quite often say about physics that the art and science of doing physics is the art and science of doing the correct approximations. Even when we do high school physics -kinematics and we take that g to be constant, that is an approximation and strictly speaking the suvat equations don't hold near the surface of the earth but they approximately hold.

 

[Vanadium 50]

Once again, the problem states what the field is. Not how it got there.

 

[Delta2]

Once again, the problem states what the field is. Not how it got there.

So you claim that the problem states that the field is homogeneous and has a value of 80000V/m AFTER we have inserted the plate? Maybe it lets that to be implied somehow but I don't see it explicitly stated.

 

[Delta2]

You can blame me that many times what I do is to "refine the mosquito, but swallow the camel" (direct translation to English from an ancient Greek saying, especially that thing I said about g being not constant is like refining the mosquito indeed. Yes it is not a good thing and I am battling my self to fix it.

 

[haruspex]

If I understand your message the right way, for an uncharged situation like my example, one should draw the Gaussian box to not pass through such that a net charge will be enclosed on one surface or the other.

Then you have not understood it.

Gauss' law refers to the net flux. If the sheet is uncharged, the flux in one side equals the flux out of the other. The net flux is zero. If your box spans the sheet the net flux leaving that is zero, and Gauss' law says, correctly, the net contained charge is zero.
This is true whether the sheet is a conductor or an insulator.

To find the charge on each surface you need to use the fact that it is a conductor. If it were not, the surfaces would not be charged. You use that fact by specifying a box that only goes half way into the sheet. Since there is no field in a conductor, there is now only flux on the one side, and you can now use Gauss' law to find the charge on the enclosed surface.

 

[flyusx]

Gauss' law refers to the net flux. If the sheet is uncharged, the flux in one side equals the flux out of the other. The net flux is zero. If your box spans the sheet the net flux leaving that is zero, and Gauss' law says, correctly, the net contained charge is zero.

So to enclose charge, if a sheet has no net charge, then you'd use a box that starts from inside the conductor to out one end, whereas if it was charged, the box can span through the entire sheet.

 

[Orodruin]

So to enclose charge, if a sheet has no net charge, then you'd use a box that starts from inside the conductor to out one end, whereas if it was charged, the box can span through the entire sheet.

The box can be whatever you want for Gauss’ law to hold. The question is which choice gives you the relevant information. To pick a suitable volume takes afterthought and carefullness.

 

[Vanadium 50]

The box can be whatever you want

Thia is the key.

However, there are wise and unwise ways to draw the box. If you have a six-sided box where 5 of the sides have no field lines crossing (5 because they are parallel and one because it is zero) your calculation is simpler than if your box were, say, a dodecahedron.

Usually your box shape is related in some way to the symmetry of the problem.

 

[flyusx]

Well, I see now that it would make sense if a sheet had no net charge but had a charge distribution (say +Q on one end and -Q on the other) to only have a Gaussian box that extends from the middle of the metal to an exterior region on one end.
Would this still hold if the sheet had a net charge, say +Q on both ends in a region without external electric fields? Then one could draw a box to extend through the entire sheet to get $$E=\frac{\sigma}{2\varepsilon_{0}}$$ or through just one end to get $$E=\frac{\sigma}{\varepsilon_{0}}$$. Is the difference due to the differing definitions of $\sigma$ in this case since the charge enclosed is 2Q for the first and Q for the second?

 

[haruspex]

Is the difference due to the differing definitions of σ in this case since the charge enclosed is 2Q for the first and Q for the second?

Yes.

 

[nasu]

Well, I see now that it would make sense if a sheet had no net charge but had a charge distribution (say +Q on one end and -Q on the other) to only have a Gaussian box that extends from the middle of the metal to an exterior region on one end.
Would this still hold if the sheet had a net charge, say +Q on both ends in a region without external electric fields? Then one could draw a box to extend through the entire sheet to get $$E=\frac{\sigma}{2\varepsilon_{0}}$$ or through just one end to get $$E=\frac{\sigma}{\varepsilon_{0}}$$. Is the difference due to the differing definitions of $\sigma$ in this case since the charge enclosed is 2Q for the first and Q for the second?

The field at a given location should be the same, no matter how you pick the Gaussian surface. The flux depends on the choice of surface. So does the charge enclosed. But not the result you get for the field. As long as you do it correctly.

 

[Orodruin]

The field at a given location should be the same, no matter how you pick the Gaussian surface. The flux depends on the choice of surface. So does the charge enclosed. But not the result you get for the field. As long as you do it correctly.

This is an important point. Gauss’ law is very often misapplied or misused. I wrote an Insight on that (using different charge configurations as examples) a while back. OP might find it interesting: https://www.physicsforums.com/insights/a-physics-misconception-with-gauss-law/

 

[flyusx]

This is an important point. Gauss’ law is very often misapplied or misused. I wrote an Insight on that (using different charge configurations as examples) a while back. OP might find it interesting: https://www.physicsforums.com/insights/a-physics-misconception-with-gauss-law/

Just had a read of that. It was always under my impression that Gauss's law is rarely useful since cylindrical/spherical/planar sheet symmetries associated with uniform electric fields are very rare in actuality.
Sidenote: it's nice to see $\rho_{l}$ instead of $\lambda$ for linear charge density; the former always made more sense to me.