# Homework Help: Electrostatics, conductor infinite sheets

1. Apr 13, 2012

### fluidistic

1. The problem statement, all variables and given/known data
Two infinite plane lattices (conductor) of uniform thickness t1 and t2 respectively, are placed parallely to each other with their adjacent faces separated by a distance L. The first lattice has a total charge by unit of area (the sum of surface densities of each sides of the lattice) worth q1 and the second lattice, q2.
Demonstrate that:
1)The surface charge densities of the adjacent sides are equal in magnitude and opposite in sign.
2)The charge densities of the 2 external sides are equal.

2. Relevant equations
Electric field due to an infinite sheet in the system of units where c=1 (I believe), is $E=4\pi \sigma$.
$q1=\sigma _1 + \sigma _2$
$q2=\sigma _3 + \sigma _4$.
$\int _{\partial \Omega } \vec E \cdot \hat n dS = \int _ \Omega 4\pi \rho dV$.
3. The attempt at a solution
I've made several attempts, demonstrated part 1) with Gauss theorem but couldn't solve part 2). But anyway I want to redo everything.
What I really don't understand is, since the E field due to an infinite charged sheet is a constant, so is the total E field in my problem, because due to the linearity of Poisson equation, I can sum the 4 E fields up to get the total E field. The sum of 4 constants is a constant. However I know that inside the conductors the E field is worth 0 and outside it, it's different from 0. How is that possible? I don't understand this at all.
Therefore I don't know how to solve even part 1).

2. Apr 13, 2012

### darkxponent

When the charges are free to move, as this is a conductor, the charges will allign thensemselves so that the total potential energy of system is minimum. Try soving it now

3. Apr 13, 2012

### fluidistic

Ok I agree but I must not use the potential at all for this exercise. I omitted a part in the problem statement which stated to use a previous result (namely that $E=4\pi \sigma$ ) to demonstrate what they ask me. I include the equation in the relevant equations though. So I should not use the potential function at all.

4. Apr 14, 2012

### darkxponent

how do you get E = 4*pie*sigma? It is not correct

5. Apr 14, 2012

### fluidistic

I used Gauss law as follows: I am over the sheet, I take a sphere of radius r as Gaussian surface. Gauss law tells me that $\int _{\partial \Omega } \vec E \cdot \hat n dS = \int _ \Omega 4\pi \rho dV$. In my case this implies that $2\pi r^2 E=4\pi \cdot \pi r^2 \sigma \Rightarrow E=2\pi \sigma$. Ok I see I made a mistake, I had forgotten the flux of the other side of the sheet. Hmm this result still looks wrong. I would have made r tends to infinity but anyway my result doesn't depend on r so this wouldn't change. The problem of the result is that it tells me that the E field of a charged disk doesn't depend on the distance along its axis of symmetry...

6. Apr 14, 2012

### darkxponent

You can use the potential as potential is calculate using the electric field only. So you are using the above result.

7. Apr 14, 2012

### darkxponent

yes that is why it doesnt matter of you use the value of E correct or incorrect. But we should know how to calculate Electric field for infinte charge. Take a gaussian surfacd which is symmetrical to sheet i.e. A cuboid or a cyllinder

8. Apr 14, 2012

### fluidistic

I took a cylinder in my last post, not a sphere as I wrote.

9. Apr 14, 2012

### darkxponent

I dont think the statement for gauss law written by you is correct. I have never seen such a equation for gauss law in my physics books or any where else. Gauss law is

closed integral,E.dA= q/(epsilanot).

There is no epsilanot in your equation . Use this and you will get E=
sigma/epsilanot.

Now you can take any surface. Take cuboid for simplicity.

One more thing. You have to take the flux on one side only. On the other side it willbe zero as electric field inside a conductoris zero so is flux.

10. Apr 14, 2012

### darkxponent

In your eqn for gauss law, on the RHS (1/4pi*epsilanot) is missing. Multiply RHS by it and you will get right vale

11. Apr 14, 2012

### fluidistic

I'm confused somehow, I redo this.
First of all, my professor uses a unit system where k=1 (not c=1 as I wrote in my first post, sorry for this error). So my "4 pi" is worth your 1/epsilon_0. I think they are called Gaussian units but I am not sure, my professor never mentioned them.
$k=\frac{1}{4\pi \varepsilon _0 }=1 \Rightarrow \underbrace {4 \pi }_{\text {mine} } = \underbrace { \frac{1}{\varepsilon _0} }_{\text {yours}}$.
I calculate the electric field due tue a single infinitely thin sheet, assuming that the E field leaves the sheet from both sides. As a Gaussian surface I take a cylinder of radius r. In this case $\int _{\partial \Omega } \vec E \cdot \hat n dS = 2 \pi r^2 E$. The enclosed charge in this cylinder is worth $\int _ \Omega 4\pi \rho dV =4 \pi\cdot \sigma \pi r^2$.
Equating both results and solving for E, I get $E=2\pi \sigma$. In your units this means $E=\frac{ \sigma }{2 \varepsilon _0}$. What bothers me about this result is that I didn't have to make r tends to infinity (the radius of the cylinder, in order to take into consideration ALL the sheet). As if the E field due to a charged disk of radius r was independent of the distance if I had assumed that it varies only on an axis orthogonal to it.
By the way my result is half of yours. Did I do something wrong?

12. Apr 15, 2012

### darkxponent

This true only for a non conducting sheet. in a conducting sheet E leaves from one side only as Electric field inside a conductor is zero.

13. Apr 15, 2012

### fluidistic

Even for an infinitely thin sheet as I said in my post...? There's not really an "inside" the infinitely thin sheet or once again I'm missing something?
Edit: I calculate the E field due to an infinitely thin sheet because my idea is to add 4 of them to represent the whole problem. That's the reason why I didn't consider a thick sheet first.

Last edited: Apr 15, 2012
14. Apr 15, 2012

### darkxponent

Electric field for inside a conductor is always zero for a perfect conductor. So we have to take it zero only

15. Apr 15, 2012

### fluidistic

And what would be the "inside the conductor" for a sheet of no thickness? You mean right over it?
I feel like you're considering a conductor of thickness t1 or t2 like in the OP while I'm talking about a conductor sheet of thickness 0. My strategy to solve the problem is in post #13.

16. Apr 16, 2012

### darkxponent

well of they have not stated the thickness to ne zero then we must not assume it to be zero. They have given thickness t1 and t2. So that is a reasonable thick.
And what is the point of saying a metal sheet in question if it is of thickness zero. It is non comductiong sheet only.

One more thing. There are two layers of charge disrribution in the metal sheet given. So there must be some space between the surface charge distribution. That space is nothing but conductor.

17. Apr 16, 2012

### fluidistic

I agree.
Wrong, we can assume this I believe and this will be my strategy to solve the problem.
Yes, true.
That's the point of my strategy. I consider the whole problem as 4 infinitely THIN sheets. The space between the first and second sheet is a conductor indeed, I translate this as saying that the E field must be 0 in this region. The same apply for the space between the 3rd and 4th infinitely THIN sheets. I think that that should be ok for the problem?

I agree 100% with you here.

18. Apr 17, 2012

### darkxponent

That is really good thinking.

19. Apr 17, 2012

### Steely Dan

You can think about it that way if you want, but that biases you towards memorizing formulas related to infinitely thin conducting sheets rather than being able to use Gauss's law to derive any results you need. Of course, if you can do both that's even better.

20. Apr 17, 2012

### fluidistic

Not really. As I not clearly I have to agree, stated/implied in post #3, I had first to derive the electric field due to an infinite sheet and then use that result to solve this problem. I had to use Gauss's law for that.
Furthermore, even using $E=2\pi \sigma$, I still have to use Gauss's law to solve the problem. Unless of course that you tell me I can do it without applying Gauss's law.
So let's get back to the problem. I divide the system into 5 regions.
I've made a sketch with the 4 parallel sheets, I've made the vertical. My first region is the one extending from - infinity to the first sheet whose superficial charge density is $\sigma 1$. Region 2 is the space between the 2 first sheets, namely inside the conductor. Region 3 is the space between sheet 2 and 3 (outside the conductors). Region 4 is the region inside the 2nd conductor, namely between sheet number 3 and 4. Region 5 is the space from sheet #4 and infinity.
I have to show first that $\sigma_3 =- \sigma _2$. I apply Gauss's law. As a Gaussian surface I consider a cylinder going through region 2 to 4. Net flux through it is 0 because the E field at the left and right side (or top and bottom of the cylinder) is 0 due to the conductor region. Hence the net charge enclosed by the cylinder is 0. Mathematically this means $\vec E \cdot \hat n =0 \Rightarrow \int _{\partial \Omega } \vec E \cdot \hat n dS=0 \Rightarrow \int _{\Omega } 4\pi \rho dV= \int _{\Omega } 4\pi (\sigma_2 + \sigma _3 )dV \Rightarrow \sigma _3 = -\sigma _2$.
I'll think about how to solve the rest. If I made any mistake please let me know and feel free to help me for the next part(s).
Thanks for the help so far.

21. Apr 17, 2012

### fluidistic

I don't know how to solve part 2).
Basically I don't know how to set up the directions of the E fields in each region.

22. Apr 17, 2012

### Steely Dan

Evidently you misinterpreted my comment. What I was trying to say is that I expect that the motivation for your approach, of pretending this problem is four infinite sheets with regions of no electric field in regions 2 and 4, comes from a desire to avoid using Gauss's law in favor of memorizing the result for infinite conducting sheets. What I am recommending is to not memorize those results, and just use Gauss's law whenever you need it, to solve a problem.

This part is correct.

What you are saying is correct for the endcaps, but there's also the question of flux through the actual cylindrical part of the surface. This is zero as well, but for a different reason. Once you have that the net flux is zero, then it's pretty simple to prove that the two sheets have equal and opposite charge, as you found.

By symmetry concerns, the direction of the electric field from an infinite sheet is always perpendicular to the surface. Try taking that same cylindrical surface through all the regions, with endcaps on either side of the four sheets. What would the net charge enclosed be inside this cylinder?

23. Apr 17, 2012

### fluidistic

Ok thanks a lot.
Yes, I assumed that the E fields were orthogonal to each sheet for symmetric reasons. This is why the net fluxes through my Gaussian surfaces only pass through the top and bottom of the cylinders.
With the data of the problem, q1+q2. If I use part 1), this reduces to $\sigma 1 + \sigma _4$. I cannot really apply Gauss's law if I don't know the direction of the E fields. Particularly for the "$\vec E \cdot \hat n$" part. The direction of the E field is extremely important in my case.

24. Apr 17, 2012

### Steely Dan

That's alright. Two thoughts. First, in a situation like this, it is often good to make some assumption about the charge of $\sigma_1$ and $\sigma_4$. Say you assume the first is positive and the second is negative. Well, then you would know that the electric field outside the left sheet is positive and the field outside the right sheet is negative (by our conventions for what positive and negative charge mean). If your assumption is backward, then when you solve these problems you will just get a negative number for each charge density. If it was correct, you'll get positive numbers.

But second, the field orientations don't much matter for this problem. If you just use that equation, you have too many unknowns as it is. You want to combine Gauss' law applied to the whole system with Gauss' law applied to just sheets 1 and 4. This will allow you to eliminate the unknown fields $E_1$ and $E_4$.

25. Apr 17, 2012

### fluidistic

Hmm but I'm asked to prove that $\sigma _1 = \sigma _4$. The assumption that the sheets 1 and 4 have opposite sign charge densities looks strange or I'm missing something?