A slab? Infinite area? Electric field? Help please

[askcr9]

Homework Statement

Consider a slab centered at the origin with uniform volume charge density, ρ. The slab is infinitely long in the y-z plane, but has thickness d along the x axis. Calculate the magnitude of the electric field due to the slab
i. outside (x > d/2,) and
ii. inside (0 < x < d/2).

[see attachment for figure]

Relevant Equations

Well, it's electric field, so I'm assuming Gauss' Law is the best way to solve this thing… and… I think that's the only formula you need.

So, yeah, Gauss' Law is the integral of E dot dA equals charge enclosed divided by permittivity of free space… yeah, you know the whole deal… the equation is in the first line of my work which is attached.

The first time I saw this question I had no idea how to do it (as you can see in the figure, I lost a lot of points :s) because I was confused on how to even approach it with area of the slab from all sides being infinity. Right? That's problematic, no?

Today, I just tried the problem again for the first time in a while and I got an answer but I don't know if it's correct. (Well, I got an answer for part i; I haven't done part ii yet, but once I find out how to do part i, I know doing part ii will be real easy. Mathematically, it'll be, like, the same exact thing as part i with just a different value of charge enclosed. But anyway, yeah, the point is, I've only done part i so far.) See attachment for my work; the answer I got was ρ/xϵ0. I'm guessing this isn't correct, cause area canceled out when I was solving for it so it's independent of area… and how could the electric field generated by this thing be independent of its area? That doesn't make any sense… right?
Like, the electric field generated by this infinite slab would have to be different from the electric field generated by a finite slab, right? My answer goes against that notion.

Any help would be appreciated. Thank you!

 

[Orodruin]

You have not specified what Gaussian surface you are using.

 

[askcr9]

Yeah, sorry, I'm using the Gaussian surface which is exactly the same as the surface in the problem just with a thickness of 2x rather than d.

 

[askcr9]

(and I would use the same exact Gaussian surface for part ii as well; x would have a different domain, though, as the problem itself says)

 

[Orodruin]

Then your result for the enclosed charge is not correct.

Also, the area is infinite. You should choose a smaller volume.

 

[askcr9]

My enclosed charge is incorrect? Ok… I don't know why…

Oh wait, I think I know the problem. It's because ρ as given in the problem is the charge density for the surface given in the problem. It isn't the charge density of my Gaussian surface.

So Qen = ρAd, right?

So my E would be ρd/ϵ0.

But this is wrong too? It's still independent of area…
Regarding the second thing you said, if I chose a smaller volume, then the A's wouldn't cancel out when solving for E and I would have two different area values… that would be problematic too. I would need to simplify the answer further but I wouldn't be able to with two different area values. Most problematically, I wouldn't be able to express the Gaussian surface area value in terms of what's given in the problem.

I hope my writing is comprehensible!

 

[Orodruin]

But this is wrong too? It's still independent of area…

Again, what area? The slab is infinite.

Most problematically, I wouldn't be able to express the Gaussian surface area value in terms of what's given in the problem.

This is irrelevant. You can choose whichever Gaussian surface you want, but you need to do so intelligently for it to help you.

I wouldn't be able to with two different area values.

What two different area values?

You also need to think about what the flux through the different parts of your surface is.

 

[askcr9]

Ok, I get it, it's dumb for to think it'd be dependent on area. Every other Gauss' law problem I've ever done the answer has been dependent on area… but this is just an exception and the area is infinity so the answer in reality make sense, at least to me.

When I said two different area values I meant an area value for the actual surface (called A here) and an area value for my Gaussian surface which if I made my Gaussian surface have less volume then I wouldn't be able to express this area in terms of what's given in the problem.

The flux is infinity cause the area is infinity. Flux is BAcostheta. So the area being infinity makes the flux infinity as well.

So is ρd/ϵ0 correct? You never said it was wrong so I'm just wondering

 

[gneill]

Have you studied (or at least been shown) the derivation for the electric field of an infinite sheet of charge?

 

[askcr9]

Oh yeah, it's like charge density / 2ϵ0. I had heard the result before, but I had never seen the derivation for it until I looked it up just now. I wasn't thinking about that anyway, I guess it's similar to this question.

In the derivation I saw, they took into account both sides of the infinite sheet of charge, so the total area ended up being 2A and there ends up being a 2 in the denominator. Is that something I would have to do here?

 

[gneill]

In the derivation I saw, they took into account both sides of the infinite sheet of charge, so the total area ended up being 2A and there ends up being a 2 in the denominator. Is that something I would have to do here?

Yup.

You'll want to come up with a strategy (using Gaussian surfaces) for the region inside the slab.

 

[askcr9]

The region inside the slab? Um, I'm sorry, I don't think I quite understand. I'm solving for the area outside the slab (in part i, at least). Why would I need to worry about what's inside the slab?

All I would need to do is just sum up the areas of all of the faces of my Gaussian slab, right? The face areas are infinite, so that's kinda weird… but whatever, stuff will cancel out in the fraction, it'll work out. And so my answer would gain a 2 in the denominator, it would simply be ρd/2ϵ0.

So what are you saying about the inside? Are you saying I would also have to account for the areas of the inside faces as well? I've never seen anything like that done before; the inside faces are just duplicates of the outside ones anyway, that seems weird.

Have I misunderstood you?

 

[Orodruin]

All I would need to do is just sum up the areas of all of the faces of my Gaussian slab, right? The face areas are infinite, so that's kinda weird…

It is not weird, it is wrong. You should choose a different Gaussian surface.

 

[gneill]

The region inside the slab? Um, I'm sorry, I don't think I quite understand. I'm solving for the area outside the slab (in part i, at least). Why would I need to worry about what's inside the slab?

The question has two parts: Outside and inside the slab. I've offered a suggestion for the "inside the slab" part.

Outside should be fairly straightforward if you've followed the derivation of the electric field due to a sheet of charge. Inside, you'll want to consider employing a couple of Gaussian surfaces to find the net electric field at the given position.

 

[askcr9]

@gneill

Ok

So then, back to part i, when I sum up all the faces on my Gaussian slab, I end up getting a 2 in the denominator of my fraction, so I get ρd/2ϵ0. Is that right or am I still wrong?

(I mean, I'm guessing it's wrong, I think that's what Orodruin was saying in the post above yours. My Gaussian surface is wrong? Am I suppose to be instead using a finite Gaussian surface…?)

 

[gneill]

@gneill

Ok

So then, back to part i, when I sum up all the faces on my Gaussian slab, I end up getting a 2 in the denominator of my fraction, so I get ρd/2ϵ0. Is that right or am I still wrong?

(I mean, I'm guessing it's wrong, I think that's what Orodruin was saying in the post above yours. My Gaussian surface is wrong? Am I suppose to be instead using a finite Gaussian surface…?)

The 2 in the denominator is okay. Your result looks good.

I believe that what @Orodruin was referring to (and I stand to be corrected) was that you were thinking of Gaussian surfaces that were infinite in extent, but that you should consider ones that are finite instead.

The example of the electric field due to the infinite sheet of charge that was mentioned earlier should give you that hint. Don't try to encompass the entire slab with your Gaussian surface. Instead use a finite region as in the aforementioned example.

 

[Orodruin]

I mean, I'm guessing it's wrong, I think that's what Orodruin was saying in the post above yours. My Gaussian surface is wrong? Am I suppose to be instead using a finite Gaussian surface…?

The result is not wrong, but your way of arriving at it is as formally you have two infinities on either side of the equation.

As @gneill said, you should choose your Gaussian surface carefully and be careful in comsidering the contributions to the flux through it.

Edit: Unfortunately, the case of a spherically symmetric charge distribution is often the only example shown to students, or at least explained thoroughly. This gives a false impression of the need to encompass the entire charge distribution and places less emphasis on the need to carefully choose a Gaussian surface from the symmetries of the problem and from there obtain a simple expression for the surface integral.

 

[askcr9]

Alright alright. Things are making a little bit more sense now.

A little bit.

I have just a couple of questions relating to all of this.

1. So a Gaussian surface can never be infinite? I guess that'd make sense, right, on a purely common sensical level.

But then again, to me, it makes zero sense how a problem can give me a surface with dimensions that are literally infinity. So at this point I'm not sure how important it is that things make sense to me lol.

Like, hmm, is this question supposed to be taking place in a fantasy land…? How is a length measurement of infinity even a real thing

2. When I make my Gaussian surface finite, how come the charge enclosed by my Gaussian surface is the same as the charge enclosed in the infinite slab? The Gaussian surface is an infinitesimally small portion of the infinite slab… so wouldn't the charge enclosed by the Gaussian surface be an infinitesimally small portion of the charge inside the infinite slab?

 

[Orodruin]

1. So a Gaussian surface can never be infinite? I guess that'd make sense, right, on a purely common sensical level.

It can, but you need to be extra careful in those cases and make sure you are not missing contributions at infinity and that your integrals are finite.

Like, hmm, is this question supposed to be taking place in a fantasy land…? How is a length measurement of infinity even a real thing

Usually, it is a theoretical construct and/or an approximation of a physical situation.

When I make my Gaussian surface finite, how come the charge enclosed by my Gaussian surface is the same as the charge enclosed in the infinite slab?

It is not. The charge in the infinite slab is infinite, the charge in the finite Gaussian surface is not.

The Gaussian surface is an infinitesimally small portion of the infinite slab… so wouldn't the charge enclosed by the Gaussian surface be an infinitesimally small portion of the charge inside the infinite slab?

Not infinitesimal. You should choose a finite Gaussian surface. And again, the charge inside it will not be infinite.

 

[gneill]

Gaussian surfaces can be of any extent and shape. However, their usefulness in terms of analysis of a given situation depends upon careful consideration.

Pay attention to the derivation of the electric field due to the infinite sheet of charge! Was the Gaussian surface employed of infinite extent? Nope. A carefully defined segment of the sheet was selected and the charge encompassed by that segment was considered. The charge encompassed by the surface was related to the overall charge density.

Clearly the total charge of an infinite sheet of charge would be infinite, right? Infinities are awkward to deal with, so a Gaussian surface that enclosed an infinite charge is not the right approach. Instead, a finite "sample" of the whole is considered with an appropriate Gaussian surface. In the case of the sheet of charge, a finite cylindrical "slice" of the sheet is chosen. That's a good hint as to how to approach this problem. The charge density gives you the net charge within the portion of the surface selected.

The idea can be readily extended to the case of the slab. Choose your Gaussian surface appropriately and use the given charge density to determine the charge within a given Gaussian surface (or volume).

When you look to find the field within the slab, you will need to employ more than one Gaussian surface, as there will be charges "inside" and "outside" your position with respect to the center of the slab.

 

[askcr9]

My understanding has improved tremendously on this subject! I know the answer to part i and understand how to get it correctly and I'm pretty sure I can do part ii by myself now. It has been a privilege to receive help from such knowledgeable and experienced users on this website. Thanks for your help, gneill and Orodruin!