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Gauss's law for a thick plane

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  • #1
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Alright so I already checked the answer to this problem in my solution manual, but I can't figure out why our results are differing

say we have an infinite plane slab of thickness 2d, with uniform charge density ρ. Find the function of the electric field with regards to y where y=0 is at the center of the slab.

Then
[itex] Q/ε = \oint E dA [/itex]

so I draw my cylinder inside of the plane. I imagine 1 face at y=0 and 1 face at y < d. then,


[itex] Q = (ρ A (y))/ ε = E * 2A [/itex]

where A is the area of the face of our cylinder, and we can ignore the sides of the cylinder since the flux is parallel there.

However in my book they draw the second face coming out of the cylinder, and the final answer is just [itex] ρy/ε.[/itex]

I for the life of me don't understand why we differ by a factor of two.

Thank you,
=
 

Answers and Replies

  • #2
vela
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Alright so I already checked the answer to this problem in my solution manual, but I can't figure out why our results are differing

say we have an infinite plane slab of thickness 2d, with uniform charge density ρ. Find the function of the electric field with regards to y where y=0 is at the center of the slab.

Then
[itex] Q/ε = \oint E dA [/itex]

so I draw my cylinder inside of the plane. I imagine 1 face at y=0 and 1 face at y < d. then,


[itex] Q = (ρ A (y))/ ε = E * 2A [/itex]

where A is the area of the face of our cylinder, and we can ignore the sides of the cylinder since the flux is parallel there.

However in my book they draw the second face coming out of the cylinder, and the final answer is just [itex] ρy/ε.[/itex]
I'm not sure what you mean by "second face coming out of the cylinder."

Your mistake is in thinking that
$$\int \vec{E}\cdot d\vec{A} \to 2EA$$ for the Gaussian surface you've chosen.

I for the life of me don't understand why we differ by a factor of two.

Thank you,
=
 
  • #3
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Well I guess I just don't understand why flux wouldn't be coming out of both the faces of our Gaussian pillbox.
 
  • #4
vela
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That could be the case, but you seem to be assuming that the flux through one face is the same as the flux through the other. Why do you think that?
 
  • #5
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Well the box has two faces, and charge is enclosed entirely within said box. Since the electric field radiates in all directions I guess I don't see why there wouldn't be flux coming through both faces of the box we've drawn. Although I do see your converse argument of the fact that the flux could be different between the two faces. It is definitely different in sign.
 
  • #6
Doc Al
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What's the field at the center of the slab?
 
  • #7
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E = 0 there. Is this why we ignore that face then?
 
  • #8
Doc Al
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E = 0 there. Is this why we ignore that face then?
You're not ignoring it. If the field is 0, then what's the flux through that face?
 
  • #9
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Alright fair enough, but when were drawing our Gaussian surface it is arbitrary isn't it? so what if I decided to draw a very thin box inbetween y = 0 and y < d. Then I would get the result I came up with earlier wouldn't I?

I'm sorry if I am being a bit obnoxious with these questions but Gauss's law wasn't taught to me very well at my previous uni, and now I am afraid I will run into troubles applying in a upper level physics course.
 
  • #10
Doc Al
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Alright fair enough, but when were drawing our Gaussian surface it is arbitrary isn't it? so what if I decided to draw a very thin box inbetween y = 0 and y < d. Then I would get the result I came up with earlier wouldn't I?
What counts is drawing a surface that allows you to apply symmetry or to have the flux through surfaces equal zero. That allows you to calculate the field where the flux is non-zero.

If I understand what you mean by a thin box, that won't change anything: The flux through the sides will still equal zero and the field at y = 0 would still equal zero.
 
  • #11
vela
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Well the box has two faces, and charge is enclosed entirely within said box. Since the electric field radiates in all directions I guess I don't see why there wouldn't be flux coming through both faces of the box we've drawn. Although I do see your converse argument of the fact that the flux could be different between the two faces. It is definitely different in sign.
Suppose we had your cylindrical surface in empty space. There's no charge anywhere, so the electric field is 0 everywhere. The total flux through the surface is obviously 0, and Gauss's law tells you the charge enclosed is 0. So far, so good.

Now suppose there's an infinite sheet of charge with charge density σ, and your surface lies completely to one side of the sheet with its ends parallel to the sheet. The electric field everywhere (except right on the sheet) has a magnitude of σ/2ε0 and is perpendicular to the sheet. The charge enclosed in your surface is still 0, but this time, there is a non-vanishing electric field at the ends of the cylinder. The field enters at one end and leaves at the other end, so if you calculate the flux, you'd get
$$\oint \vec{E}\cdot d\vec{A} = E(\pi r^2) - E(\pi r^2) = 0,$$ where r is the radius of the cylinder. You can't deduce E=0 from that result, which is a good thing because E certainly isn't 0.

In the second case, the charge outside of the surface affects the value of E at points on the surface, but it doesn't change the total flux through the entire closed surface, which is still 0. Note that the E that appears in the integral is the electric field due to all charges, inside and outside the surface. The charge outside the surface, however, won't contribute to the total flux through the surface. The same thing is happening in your problem.
 
  • #12
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Alright thanks to both of you. Admittedly I'm still not 100% sure on the concept but I think I'm pretty close and I have some other classes I need to worry about so this is as good as it will get for the moment.
 

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