# I don't understand gauss's law for sheet surfaces

1. Jul 10, 2013

### iScience

hi all, gauss's law states that for any surface, the total electric flux coming through an open surface is always q/ε0. okay.. i understand this...

so for a sphere.. the simplest case... http://imgur.com/bwJzf7t,2KFO8Xw,O9Mgafg

$\Phi$=$\int$$\vec{E}$$\bullet$$\hat{n}$dA=EA= $\frac{q}{4(pi)ε}$ (4$\pi$r2)=$\frac{q}{ε}$.

okay... so this says... that ALL of the electric field produced by the charge q0 contributes to the value of $\Phi$. However, let's now look at the case of two charged particles...

http://imgur.com/bwJzf7t,2KFO8Xw,O9Mgafg#1

forgive the ambiguousness if the image is not very clear at first.. i just have two charges and the lines/curves are the electric fields, more specifically.. Ʃ(E1+E2).. in otherwords superposition.

now, at every point that does NOT lie on the plane where the two electrons sit, there will exist a y-component of $\vec{E}$. However at every point that lies on the plane where the two electrons sit, there will exist only an x-component $\vec{E}$.

okay.. so then let's consider a plane/sheet of charge carriers.

http://imgur.com/bwJzf7t,2KFO8Xw,O9Mgafg#2

everywhere i look i see that the two gaussian surfaces are supposed to be JUST the two top and bottom circular areas and that the flux produced/going through these two areas is supposed to be equal to $\frac{q}{ε}$. i don't agree with this (obviously i am wrong so please show me where my reasoning is wrong) because..

if we go back to the point when i earlier said.. that ALL of the electric field lines went into contributing to the value of $\Phi$. however if we now look at just the top and bottom two circular surfaces, i don't see how they incorporate ALL the field lines produced by the charges enclosed in the cylindrical volume, ie.. the 2-D sheet. the two top and bottom circular areas should ONLY account for those electric field lines that have an initial y-component of $\vec{E}$ (speaking with respect to the plane of the sheet). what about all the rest of the x-component $\vec{E}$? their superpositioned $\vec{E}$ doesn't get pushed up into the y-direction at all. which means that these x-component $\vec{E}$ reside on the plane of the sheet. so why then do we only include the two top and bottom circles for gauss's law?

EVEN IF the x-component $\vec{E}$ were to all cancel out.. the y-component $\vec{E}$ alone should not contribute to the entire value of "$\frac{q}{ε}$"... but it apparently does... why?

Last edited: Jul 10, 2013
2. Jul 10, 2013

### iScience

sorry guys im still trying to fix the format for latex...

3. Jul 10, 2013

### Mmm_Pasta

First, it's a closed surface, not an open one. I'm guessing that the plane you describe is infinite and has a uniform charge density. In that case the x-components cancel everywhere and there exists only a y-component. In fact, you can use Gauss's law to show that the electric field is constant everywhere.

4. Jul 12, 2013

### iScience

regardless of whether or not the x-components cancel, the each charge carrier produces a radial field and a fraction of that field radiates into the x-direction. so why are we still treating the flux going through both circular areas as if all of the E-field lines being produced by the charge/s gets put through the two circular areas? because that's essentially what's happening when we set the flux equal to $\frac{q}{\epsilon}$.

so im still at a loss

5. Jul 13, 2013

### Zubeen

There are several mistakes...
The E due to the point charges is in all directions in the 3D space, actually they are spherically symmetric and spread in all directions. Hence the fields are not only in x or y component, but also in z component.
Considering the plane including the charges, the E have x component or z component or both and in the second image, the cylinders including the charges are to be shown as having E in just one direction, but as I stated, its in all directions.
Also Gauss theorem is ALWAYS valid. The point is if you can use it to find E fields easily or not.
Using symmetry in certain cases, we can logically reason that the E is constant over the Gaussian surface and then we can take out the E out of integral sign in the formula of gauss theorem, and then we are left with the integral of area which can easily be calculated.

6. Jul 13, 2013

### BruceW

yes, the x-component all does cancel. This is in the limit of a very large sheet, then the x-component will tend to zero. And why don't you think the y-component contributes to the entire value?

7. Jul 13, 2013

### iScience

it does not contribute to the entire thing because look at this image again

http://imgur.com/bwJzf7t,2KFO8Xw,O9Mgafg

see how ALL of the field lines radiating from the charge hits the surface?
so you can say that the surface(or surface area) from which the field lines are radiating from that are getting caught in the gaussian surface is 4pir^2. that is..4pir^2 is the surface area of the charge that is responsible for the field lines being radiated that are getting caught in the gaussian surface
and it is THIS total flux that gives rise to q/epsilon

but now lets look at the case of a single charge on sheet

http://imgur.com/5Dm3VHj

it doesn't matter whether or not the x-components cancel out. you see how the x-component field lines just lie on the sheet and stay there? the area of the charge now responsible for producing the field lines that are now getting caught in the gaussian surfaces (4pir^2)-2pir.
"2pir" because of the 2-D circular symmetry around the charge that gets stuck on the plane of the charge.

8. Jul 14, 2013

### omega_minus

I'm not sure I follow your argument fully iScience, but remember that Gauss' Law uses a limit as the walls of the cylinder (or pill box as my profs call it) tend to zero. The y components are the only thing left that don't cancel because your cylinder becomes, in the limit, arbitrarily flat and there is no surface area left on the sides for field lines to penetrate. Not sure if this is what you're getting at with your question, just my two cents...

9. Jul 14, 2013

### BruceW

you have misunderstood what I meant by "the x-component cancel out'. If you look at the case of the single charge on the sheet, and look at one of the field lines that is 'diagonal'. What I meant is that the field from another charge on the sheet will 'cancel' any x-component that is in this diagonal field line, so the field line becomes vertical.

In other words, when you add up two diagonal field lines coming from different charges on the sheet, then you get a vertical field line going away from the sheet (in the y-direction).

yet another way to think about it, let's say the sheet goes from x=-a to x=a. Then at a point above the sheet at x=0, the field must be purely in the y direction (due to symmetry). And we can also say that at any point close to x=0, the field must be almost completely in the y direction. For example, at x=a/1000, the field will be almost completely in the y direction. Now here's the interesting part: if we let 'a' tend to infinity, then a/1000 is still infinity, so we can say that the field is almost completely in the y direction everywhere.

10. Jul 16, 2013

### iScience

could you clearify this part? when you say diagnal, do you mean field lines NOT parallel to the sheet? i am not concerned with field lines that are not parallel to the sheet. i'm concerned about the field lines that are parallel to the sheet because they indeed do remain on the sheet.

Last edited: Jul 16, 2013
11. Jul 16, 2013

### BruceW

Oh right. I see what you mean now. I though you were concerned with diagonal field lines. But you are actually concerned with field lines parallel to the sheet. OK, well in the middle of the sheet, there is no parallel field line (due to symmetry). And now if we look at points on the sheet which are close to the centre of the sheet, then also the parallel field line will be negligibly small. (There is no discontinuous jump of the electric field inside continuous matter). So if we make the sheet infinitely large, then we are essentially always going to be 'close to the middle of the sheet' in a certain sense. Therefore, the parallel field line tends to zero when the sheet tends to infinitely wide.

Now I've explained this fairly qualitatively. But rest assured that if you go through the calculation, using limits (i.e. the proper way), then the answer does come out right.

12. Jul 16, 2013

### iScience

can you specify what you mean when you say that the field lines parallel to the sheet will negligibly small?
(d=distance from sheet (from the direction normal to the sheet))

yes, if we use limits as the limit of d tends towards 0, i understand that the vector field(Efield) will be constant all throughout. however @ the point d=0, i something else happens; as i've stated, on the surface d=0, there are NO z-component E-fields, there are only x and y so no z-components can come out of this plane we are working on (the plane of d=0). so why would there not be a discontinuity between d=0 and all other planes d$\neq$0? @ d=0 there would be just, the E-field components that remain on the plane, and at all other points in space (d$\neq$0), the E-field is pointing away from the plate. there must be a discontinuity here. but if there isn't may you show me the proof?

13. Jul 16, 2013

### BruceW

hmm. Alright, let's say we are on the plane d=0. Then in the middle (x=0), the electric field parallel to the x direction will be zero (due to symmetry), and if we move in the x direction, we are not crossing any discontinuity, so we can expect that close to the middle, the electric field parallel to the x direction will be negligibly small.

And to show this the nice way, you can start off with a rectangle of charged material, then take the limit that one side is much longer than the other (i.e. it tends to a sheet). Then you will find that there is a very similar situation inside the rectangle as there is just outside the rectangle. The only difference will be that the electric field in the y direction will vary inside the rectangle.