Gaussian cylinder enclosing cylinder of charge

[annamal]

Homework Statement

A very long non-conducting cylindrical shell of radius R has a uniform surface charge density $\sigma$. Find the electric field at a point outside the shell.

Relevant Equations

$\frac{R\sigma}{\epsilon_0*r}$

I am confused why we don't take into account the lids of the cylinder since the Gaussian cylinder is of finite height L as shown in the image

 

[kuruman]

Homework Statement:: A very long non-conducting cylindrical shell of radius R has a uniform surface charge density $\sigma$. Find the electric field at a point outside the shell.
Relevant Equations:: $\frac{R\sigma}{\epsilon_0*r}$

I am confused why we don't take into account the lids of the cylinder since the Gaussian cylinder is of finite height L as shown in the image
View attachment 300627

Of course we take into account the lids. What is the electric flux through each lid?

 

[annamal]

Of course we take into account the lids. What is the electric flux through each lid?

The actual solution did not take into account the lids.

 

[kuruman]

Can you post the actual solution as was given to you? Maybe it explains what to do about the lids and you missed it. Alternatively, calculate the flux through the lids and see what you get. Maybe you don't need to worry about the lids.

 

[annamal]

Can you post the actual solution as was given to you? Alternatively, calculate the flux through the lids and see what you get. Maybe you don't need to worry about the lids.

It is alternatively here: https://opentextbc.ca/universityphysicsv2openstax/chapter/applying-gausss-law/

 

[kuruman]

OK, they don't mention the lids. I hd to make sure. So what do you think the electric flux is through the lids? What do you have to know to calculate it?

 

[annamal]

OK, they don't mention the lids. I hd to make sure. So what do you think the electric flux is through the lids? What do you have to know to calculate it?

Are you saying the problem is wrong not calculating the electric field at the lids?

I am guessing including the lids, $\lambda_{enc} = 2*\pi*R*\sigma_0$
$q_{tot} = 2\pi*R*\sigma_0*L$
$E = \frac{2*\pi*R*\sigma_0*L}{2*\pi*r^2 + 2\pi*r*L}$

 

[kuruman]

Try not to guess. What does Gauss's law say? The solution suggests "apply Gauss's law strategy given earlier." Maybe the answer is there.

 

[Orodruin]

Are you saying the problem is wrong not calculating the electric field at the lids?

I am guessing including the lids, $\lambda_{enc} = 2*\pi*R*\sigma_0$
$q_{tot} = 2\pi*R*\sigma_0*L$
$E = \frac{2*\pi*R*\sigma_0*L}{2*\pi*r^2 + 2\pi*r*L}$

Apart from what has already been said, it is easy to conclude that what you have done here is wrong as your result makes the field dependent on what L you pick. The actual field cannot depend on how you choose to draw your Gaussian surface.

Also, please don’t write out the * in LaTeX. It makes things more unreadable.

 

[Delta2]

I see that no one here wants to mention the self implied piece of information which is critical for your question.

This is that due to symmetry reasons the electric field is in the radial direction and only that direction. There is simply no component in the vertical (z) direction. So can you answer now what is the flux through the lids?

 

[Orodruin]

I see that no one here wants to mention the self implied piece of information which is critical for your question.

That would be because we are in the homework forums and I think it was pretty clear that @kuruman was trying to guide the OP towards that realization. That the field is radial is shown in the figure accompanying the problem.

 

[Orodruin]

But in general, I agree that this kind of omission by textbooks are way too common and often lead to exactly the type of misunderstandings displayed here.

 

[kuruman]

But in general, I agree that this kind of omission by textbooks are way too common and often lead to exactly the type of misunderstandings displayed here.

I wanted to give this textbook the benefit of the doubt. The solution under "Strategy" mentions "apply the Gauss's law strategy given earlier". The contribution from lids may have been discussed there and that is why I directed OP to (re)read that part. Understanding why the lids are ignored is important in understanding Gauss's law and is better achieved by discovery on one's own rather than being told.

 

[Orodruin]

I wanted to give this textbook the benefit of the doubt. The solution under "Strategy" mentions "apply the Gauss's law strategy given earlier". The contribution from lids may have been discussed there and that is why I directed OP to (re)read that part. Understanding why the lids are ignored is important in understanding Gauss's law and is better achieved by discovery on one's own rather than being told.

I typically doubt this since it is kind of the simplest non-spherical geometry. I agree that things are better if students can figure things out for themselves but the prevalance of students misinterpreting Gauss law as EA = Q_enc/epsilon without further qualification is way too high for me to be convenient with not giving students an example where the flux through some side is not EA. Sure, something like a pillbox around a surface charge may have been covered earlier in this particular book, but I find that it is often not the case and students are left lost and confused.

 

[annamal]

Try not to guess. What does Gauss's law say? The solution suggests "apply Gauss's law strategy given earlier." Maybe the answer is there.

Gauss's law says $q_{enc}/\epsilon_0 = EA$
Therefore this follows.
$\lambda_{enc} = 2\pi R\sigma_0$
$q_{enc} = 2\pi R\sigma_0L$
$E = \frac{2\pi R\sigma_0L}{2\pi r^2 + 2\pi r L}$

 

[annamal]

I wanted to give this textbook the benefit of the doubt. The solution under "Strategy" mentions "apply the Gauss's law strategy given earlier". The contribution from lids may have been discussed there and that is why I directed OP to (re)read that part. Understanding why the lids are ignored is important in understanding Gauss's law and is better achieved by discovery on one's own rather than being told.

The contribution of the lids are none if the length of the cylindrical charge is infinite. Here we have a finite lengthed Gaussian surface so there should be contribution from the lids.

 

[annamal]

That would be because we are in the homework forums and I think it was pretty clear that @kuruman was trying to guide the OP towards that realization. That the field is radial is shown in the figure accompanying the problem.

How can the electric field just be radial when the Gaussian cylinder is of finite length L. Because the Gaussian surface has finite length, there should be electric field from the lids.

 

[Orodruin]

Gauss's law says $q_{enc}/\epsilon_0 = EA$
Therefore this follows.
$\lambda_{enc} = 2\pi R\sigma_0$
$q_{enc} = 2\pi R\sigma_0L$
$E = \frac{2\pi R\sigma_0L}{2\pi r^2 + 2\pi r L}$

No. This is not what Gauss’ law says. Gauss’ law states that the flux of the electric field through a closed surface is equal to the enclosed charge divided by epsilon_0. This is equal to EA only of the electric field is perpendicular to the surface and has the same magnitude on the entire surface.

 

[Orodruin]

How can the electric field just be radial when the Gaussian cylinder is of finite length L. Because the Gaussian surface has finite length, there should be electric field from the lids.

The field’s value does not depend on whatever Gaussian surface you choose to draw. In the case of the infinite line charge, the reflection symmetry around any plane perpendicular to the line means the field can have no component in the direction of the line. The reflection symmetry through a plane containing the line and the point where you want to know the field implies there can be no tangential component of the field.

 

[annamal]

The field’s value does not depend on whatever Gaussian surface you choose to draw. In the case of the infinite line charge, the reflection symmetry around any plane perpendicular to the line means the field can have no component in the direction of the line. The reflection symmetry through a plane containing the line and the point where you want to know the field implies there can be no tangential component of the field.

I am not saying that at the lids the electric field is parallel to the lid. The black arrows are what I am suggesting.

 

[Orodruin]

I am not saying that at the lids the electric field is parallel to the lid. The black arrows are what I am suggesting.
View attachment 300672

Please reread my post. By the symmetries of the line charge, the electric field can have no component in any direction other than the radial direction. This means that the field is parallel to the lid, which is why it is a problem that you are not saying that it is.

 

[SammyS]

I typically doubt this since it is kind of the simplest non-spherical geometry. I agree that things are better if students can figure things out for themselves but the prevalence of students misinterpreting Gauss law as EA = Q_enc/epsilon without further qualification is way too high for me to be convenient with not giving students an example where the flux through some side is not EA. Sure, something like a pillbox around a surface charge may have been covered earlier in this particular book, but I find that it is often not the case and students are left lost and confused.

You were right on target with this post !

In the very next Post, which is in response to @kuruman asking OP what Gauss's Law says, the very first thing stated by OP is:

Gauss's law says $q_{enc}/\epsilon_0 = EA$
. . .

 

[Orodruin]

How can the electric field just be radial when the Gaussian cylinder is of finite length L. Because the Gaussian surface has finite length, there should be electric field from the lids.

Rereading this thread you seem to be under a very serious and fatal misconception that the Gaussian surface you draw somehow affects the field. It does not. The field is generated by the infinitely long line charge and the Gaussian surface has nothing to do with it. The Gaussian surface is a tool to use to help you figure out what the field generated by the line charge is by cleverly selecting what surface to use, but it carries no charge in itself and does not carry any charge. It encloses part of the charge that is carried by the line, but that is a different issue and just relates to the flux of the field out of the surface.

You were right on target with this post !

In the very next Post, which is in response to @kuruman asking OP what Gauss's Law says, the very first thing stated by OP is:

I’d say it surprised me, but that would be a lie … ;)
I have seen this misconception way too many times not to strongly suspect its presence in this case.