# Gauss's law for a thick plane

ozone
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
$Q/ε = \oint E dA$

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

$Q = (ρ A (y))/ ε = E * 2A$

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 $ρy/ε.$

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

Thank you,
=

Staff Emeritus
Homework Helper
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
$Q/ε = \oint E dA$

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

$Q = (ρ A (y))/ ε = E * 2A$

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 $ρy/ε.$
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,
=

ozone
Well I guess I just don't understand why flux wouldn't be coming out of both the faces of our Gaussian pillbox.

Staff Emeritus
Homework Helper
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?

ozone
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.

Mentor
What's the field at the center of the slab?

ozone
E = 0 there. Is this why we ignore that face then?

Mentor
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?

ozone
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.

Mentor
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.

Staff Emeritus
Homework Helper
$$\oint \vec{E}\cdot d\vec{A} = E(\pi r^2) - E(\pi r^2) = 0,$$ 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.