# Help with gauss, Sheet vs conducting surface

1. Sep 27, 2010

### Markel

I'm having trouble using Guass' law to find electric fields.

Where is the magnitude of the Electic field that you end up calculating? I had assumed it would be anywhere on the Gaussian surface, but it doesn't make sense now that I think of it.

for instance, calculating the field of a charged cunducting surface.
The gausian surface is a cylinder, and the left half is within the surface, and the right half is outside the surface.

The calculations show the field is E= (surface charge density)/(permitivity of vacuum)

But where is this field? It can't be along the gausian surface because part of it is within the conductor, where the field is 0, and part is outside where it is definitely not 0.

Also, I don't understand the difference between finding the field due to a conducting surface:

E= (surface charge density)/(permitivity of vacuum)

and the Field from a sheet of charge:

E= (surface charge density)/( 2 x (permitivity of vacuum))

What is the difference? Both are simply just a wall of charge. In my book it says that the expression for a conducting surface is only valid very close to the surface ( I assume this is so that we can approximate the surface as a straight plane), but even that makes no difference. The only difference is for the conducting surface, the Gaussian cylinder is protruding into the sheet, and for the second, it pierces the sheet completely and comes out the other side. But why does the electric field change depending on how you draw your gaussian surface? Shouldn't it be for any closed surface? Of any size or shape?

I thought I understood this, but I'm fairly confused. Any help would be appreciated.

2. Sep 28, 2010

### cragar

We choose our Gauss surface to be a pill box so it goes above and below the plane .
And since this is an infinite plane the field lines point directly away from the plane .
so the field lines only go through our end pieces .
$$E(2A)=\frac{(\sigma)(A)}{\epsilon_0}$$
sigma is the charge per area
And then we can do the same for a uniformly charged hollow sphere
and we can treat this like a point charge.
So we have
$$E(4)(\pi)(r^2)=\frac{Q}{\epsilon_0}$$
The angle between the vectors is zero so we can forget about the dot product
and we know the surface area of a sphere, so we don't haft to do surface integral .
But your right about the flux, the flux through any closed surface is the same but we choose our Gauss surface in a clever way to make the problem easy .
I hope this helps

Last edited: Sep 28, 2010
3. Sep 28, 2010

### Markel

Yes, I understand these calculations- but my question is where is the field?

For the sphere it seems to be obvious that the value E is on the gausian surface.

But for the sheet. the value of E changes from
$$E(2A)=\frac{(\sigma)(A)}{\epsilon_0}$$
if the Gausian surface goes all the way through the conductor

to
$$E(A)=\frac{(\sigma)(A)}{\epsilon_0}$$
If the Gaussian surface only penetrates one side

If the gausian surface only goes through the outer surface of the conductor. Where is this field that you are calculating?? (there is no dependance on X, so it is essentially everywhere, but how can this be if the inside of a conductor has 0 field?)

My confusion is that the field depends on how we draw our gaussian surface.

4. Sep 28, 2010

### cragar

for the infinite plane if the E field did not point directly outwards it would not be so easy.
We have to make sure we enclose all the charge. We can't make a box that just touches the plane it hast to go through it .

5. Sep 28, 2010

### Markel

Yes but how far through it?
How far we draw the gaussian surface changes the value of E at distance X from the surface. (as in, we keep one end of the gausian surface at X, then change how far the other end protrudes into the conducting surface).
This to me seems wrong, and I thought that determining the field did not depend on the properties of the gaussian surface (as long as it was enclosed).

If we draw the gaussian surface partway through the conductor we get one answer. If we draw it all the way through we get another answer. So we can get two answers, differing by a factor of 2 for the the value of a field at the same distance x from the surface.

this means that the value for the field at a distance is basically arbitrary.

I still haven't found a suitable answer. I went wandering around the physics department today looking for a prof. Only found a grad student and we thought about it together for a while, agreed something was wrong (something probably simple) but couldn't really figure out what was wrong.

6. Sep 28, 2010

### Staff: Mentor

Realize that a charged conducting sheet has charge on both sides. You can find the field at its surface in two ways:

(1) Taking advantage of the fact that the E field inside the conductor is zero and thus having your gaussian surface intersect the interior of the conductor. The field would be given by

$$E=\frac{\sigma}{\epsilon_0}$$

(2) Treating it as two identical sheets of charge, each sheet of charge giving a field contribution of:

$$E=\frac{\sigma}{2\epsilon_0}$$

Either way, you get the same value for the field at the conducting surface.

7. Sep 29, 2010

### Markel

I don't see how this value is for each seperate sheet. You enclose both in the gaussian surface, so naturally I would assume that the value for E is dependant on both of these sheets, not just one.

I think I'm close, but I still don't see it quite yet....

8. Sep 29, 2010

### Staff: Mentor

What's the E field from a sheet of charge? Convince yourself it's what I gave.
Yes, the E field from a charged conductor depends on both sheets, which is why you can add them for the total.

The point is: Applying Gauss's law (with a surface inside the conductor) already includes the effect of the other sheet of charge. Nonetheless, you can always just treat the two sheets of charge separately and add their fields to get the total. (The principle of superposition.) Two different ways of solving the problem, which give the same answer of course.

Perhaps the stumbling block is this: If instead of putting your gaussian surface within the conducting material you extend it beyond the other surface. In that case, you must be sure to include all the charge enclosed, which is now twice as much. (Rereading your posts, I'm sure this is it!)

9. Sep 29, 2010

### Markel

Perhaps the stumbling block is this: If instead of putting your gaussian surface within the conducting material you extend it beyond the other surface. In that case, you must be sure to include all the charge enclosed, which is now twice as much. (Rereading your posts, I'm sure this is it!)

yes but the final expression doesn't depend on charge, only charge density. Which is the same for both sides. so I only see a difference of a factor of 2... if I cuold just make a sketch I could better explain where I'm having a problem

10. Sep 29, 2010

### Staff: Mentor

Two cases:
(1) Gaussian surface inside conductor: Total charge enclosed = σA; total flux = EA.
(2) Gaussian surface encloses both sides: Total charge enclosed = 2σA; total flux = 2EA.

Works out the same either way! (Let me know if something here needs additional explanation.)

11. Sep 29, 2010

### Markel

Ok, I understand this aspect now.

But for my next problem- where is this field with value $$E=\frac{\sigma}{2\epsilon_0}$$ ?

Is it along the gaussian surface?

because part of the gaussian surface is inside the conductor, so there can't be a field there. so then the field strengths on different parts of the gaussian surface are different?

12. Sep 29, 2010

### Staff: Mentor

That's the field from a single sheet of charge. You'd solve for that field as a separate problem--nothing to do with conductors. Once you have that solution, you can use it to model the conducting surface problem. Since the field outside a conducting surface is the total field from all charges--both sides--you can just add up the field from each surface, if you want.

Not in this problem.

Not sure what you mean. The field outside the conducting sheet is the same on both sides. And the field inside the conductor is zero.

13. Sep 29, 2010

### Markel

Ah, sorry. I coppied the latex code for the wrong one.

I mean ;

$$E=\frac{\sigma}{\epsilon_0}$$

ok, there we go.

So where is E in this problem? Maybe more genrally: where is E for Gauss's law of electric fields in General?

e_0 times the integral of the Electric field (at what point??) dot product the differential of area
= Qenclosed.

I had just assumed that the Integral was along the gaussian surface (and we chose a surface where it was constant, and thus we could have a nicer integral)

But here E is obviously not constant over the gaussian surface which is half in, and half out of this conductor.

14. Sep 29, 2010

### Staff: Mentor

OK, that field is the field outside the surface of a charged conducting sheet, where σ is the surface charge.

On either side of the charged conducting sheet.

On the gaussian surface.

Everywhere on the gaussian surface.

You choose a surface where you can get a nice integral. And nothing's nicer than a surface where the field is zero!

True. So what? All that matters is: Can we get a simple expression that allows us to solve for the E field? (Where it's non-zero of course! We already know the field inside the conductor.)

Generally you want a surface where the field is constant, or zero, or where the flux is zero.

15. Sep 29, 2010

### Markel

Ok, in the post above, you say that E is indeed everywhere on the gaussian surface.

So we do our calculations and find $$E=\frac{\sigma}{\epsilon_0}$$

then we look at one part of the Gaussian surface, where E=0 (inside the conductor)

this could only be true if $$0=\frac{\sigma}{\epsilon_0}$$

which is obviously not true for a charged conductor.

16. Sep 29, 2010

### Staff: Mentor

Perhaps I wasn't clear. When applying Gauss's law, you must integrate the field everywhere along the gaussian surface.

That's the value of the field outside of the conducting surface. (Not inside the conductor! Inside the conductor the field is zero.) That's the answer we get after we apply Gauss's law to the problem.

No. Inside the conductor the field is zero. (Gauss's law must be applied to the entire gaussian surface.)

Note: I'm not saying that the field is everywhere equal to $E=\frac{\sigma}{\epsilon_0}$; that's only true outside the conductor.

17. Sep 29, 2010

### Markel

Ok, I think I found my problem.

I was confused because E is not constant.
Yet I thought we took it out of the integral because it was constant.

But really we break it into integrals along it's 3 faces. on the face inside E is constant (0). On the cylindrical face E is not constant, but that expression falls away due to the dot product. So we are left with just the circle that is outside the conductor, where the field is constant. And has the value of:

$$E=\frac{\sigma}{\epsilon_0}$$

Does this sound correct? Or am I making some more mistakes I can't see?
If not, I think I finally get it all and am much less confused.

Thanks so much for taking the time to go through that with me! I apreciate that. I did all these computations before, but I never really got the intuition of what was really going on. But now I think I understand much better.

18. Sep 29, 2010

### Staff: Mentor

Sounds like you have it exactly right.

(Just to belabor the obvious, when you're first solving this problem you don't know the value of the field outside the surface, all you know is that it is constant and non-zero in that region. Once you apply Gauss's law to the entire surface you will solve for the field and find that it equals $E=\frac{\sigma}{\epsilon_0}$.)

19. Sep 29, 2010

### Markel

Awesome, thanks so much!