Flux due to infinite long wires

[Saitama]

you have the correct answer for the flux just outside the face. To summarise:
flux just outside = 2 \lambda l / \epsilon_0
flux just inside = zero
flux exactly on the face = \lambda l /2 \epsilon_0
the electric field (and flux) is not continuous, as I said. Which is why there is the discontinuous 'jumps' between these three answers. And I don't think it is easy to get the answer for flux exactly on the face, by using symmetry. But you have calculated it by using the method of adding electric field contributions from each wire. So all is good.

Edit: to make this clear, all three answers are correct. It just depends on whether you want to talk about the flux just inside, just outside, and exactly on the face.

Thanks BruceW for the explanation. Appreciated!

The Gaussian surface has to wrap up the whole cube, including the edges. The flux across the cylindrical surfaces add to the flux of the faces. If the radius of the wrapping cylinders is δ the flux of a single wire across its own 3/4 cylindrical wrapping surface is

ψ=λ/(2πε₀ δ) L 3/4 (2πδ)= 3/4 λL/ε₀.
As δ tends to zero, the flux of the other edges through the cylindrical surfaces tend to zero.

The flux across a single face is Φ. So the total flux is 12λL=6Φ+12[3/4 λL/ε₀---> Φ=λL/(2ε₀)

ehild

How did these cylindrical surfaces come into the scene?

 

[ehild]

Thanks BruceW for the explanation. Appreciated!
How did these cylindrical surfaces come into the scene?

I am not appreciated? Feeling hurt

You need to make a closed surface around that cube. The surface has to surround the wires. But the electric field tends to infinity very near to a wire. The edge of the cube can be approximated with a curved surface and there is flux through it. The electric field is inversely proportional to the distance from the wire, but the area is proportional to the same distance. So the edges contribute with some finite amount to the flux.

ehild

 

[Saitama]

I am not appreciated? Feeling hurt

You are always appreciated. :)

You need to make a closed surface around that cube. The surface has to surround the wires. But the electric field tends to infinity very near to a wire. The edge of the cube can be approximated with a curved surface and there is flux through it. The electric field is inversely proportional to the distance from the wire, but the area is proportional to the same distance. So the edges contribute with some finite amount to the flux.

ehild

I still don't get it completely. I have never seen such examples while studying Gauss law, do you have any links where I can look at more examples of this kind?

 

[ehild]

No, I have no links, I figured it out by myself. Now it looks quite obvious, and Bruce has right: "Flux though a face" can mean the face itself, without the wires, or it can include also the wire edges. And a wire has a cylindrical surface, and the electric field is very intense close to a wire. You determined the flux through a face-without the edges. The method based on Gauss' Law gave the flux where the wires were inside the Gaussian surface. ehild

 

[BruceW]

The Gaussian surface has to wrap up the whole cube, including the edges. The flux across the cylindrical surfaces add to the flux of the faces. If the radius of the wrapping cylinders is δ the flux of a single wire across its own 3/4 cylindrical wrapping surface is

ψ=λ/(2πε₀ δ) L 3/4 (2πδ)= 3/4 λL/ε₀.
As δ tends to zero, the flux of the other edges through the cylindrical surfaces tend to zero.

The flux across a single face is Φ. So the total flux is 12λL=6Φ+12[3/4 λL/ε₀---> Φ=λL/(2ε₀)

ehild

Yes! very nice. Maths is a creative subject :) It takes imagination for problems like this. There is also the issue of how the cylinders join up at the vertices of the cube, but when we take the limit of very small cylinders, the length of cylinder which is effected by the edge effect tends to zero, so there is not a problem.

Also, I am glad to see that this way of doing the problem gives the same answer as Pranav's original post. It is always nice in physics when two different methods agree. Yeah, so this method gives the 'flux exactly on the face' answer.

I'll see if I can explain it. The Gaussian surface is almost a cube, where the faces coincide with the faces of the actual cube, except near the edges, where the Gaussian surface is 3/4 of a cylinder going around the wire. We know the total flux, because we know enclosed charge. Some of the flux will go out of the faces, and some of the flux will go out of the little cylinders at the edges. Now we take the limit that the little cylinders get very little. (i.e. the distance from the cylinder to the wire tends to zero). You might think that the flux through them will tend to zero because they are very little. But this does not happen because as the cylinder gets smaller, it gets closer to the wire, so the electric field tends to infinity. So in the limit of very small cylinders, we get a finite answer for the flux coming out of them.

As ehild said, as we make the cylinders very small, the flux through each cylinder is due to only the electric field coming from the wire it is very close to. This is because the electric field from the other wires is finite, and the area of the cylinder tends to zero, so the flux due to the other wires tends to zero. This makes it fairly easy to calculate the flux through each wire.

So now you have the total flux and the flux through each wire, so you can calculate how much flux goes through the faces. (It is just what is left over). Also, as the cylinders get very small, the area of the face tends to the area of the face of the actual cube. so what we have really done is an improper integral.

Hope that explanation helped. I think the more times you practice electrostatics problems, using Gaussian surfaces, the better you get at 'guessing' what kind of Gaussian surface would be most useful. The book 'Introduction to electrodynamics' by Griffiths might be useful. I don't know if there are practice questions in the book though.

One last thing: you could use a slightly different Gaussian surface which uses 1/4 cylinders, which are inside the wires. In this case, the enclosed charge is zero, so the total flux is zero. And so the flux through the cylinders is equal and opposite to the flux through the faces, so you can also work out the answer this way, which gives the same answer.

Edit: also, I have attached a picture of ehild's method. The picture is a slice through the cube. i.e. the wires are coming out of the page, and are enclosed by the cylinders.

 

[ehild]

That is a very good explanation, Bruce. And the Gaussian surface with the 1/4 cylinders is devilishly clever with the inward flux .

ehild

 

[Saitama]

Yes! very nice. Maths is a creative subject :) It takes imagination for problems like this. There is also the issue of how the cylinders join up at the vertices of the cube, but when we take the limit of very small cylinders, the length of cylinder which is effected by the edge effect tends to zero, so there is not a problem.

Also, I am glad to see that this way of doing the problem gives the same answer as Pranav's original post. It is always nice in physics when two different methods agree. Yeah, so this method gives the 'flux exactly on the face' answer.

I'll see if I can explain it. The Gaussian surface is almost a cube, where the faces coincide with the faces of the actual cube, except near the edges, where the Gaussian surface is 3/4 of a cylinder going around the wire. We know the total flux, because we know enclosed charge. Some of the flux will go out of the faces, and some of the flux will go out of the little cylinders at the edges. Now we take the limit that the little cylinders get very little. (i.e. the distance from the cylinder to the wire tends to zero). You might think that the flux through them will tend to zero because they are very little. But this does not happen because as the cylinder gets smaller, it gets closer to the wire, so the electric field tends to infinity. So in the limit of very small cylinders, we get a finite answer for the flux coming out of them.

As ehild said, as we make the cylinders very small, the flux through each cylinder is due to only the electric field coming from the wire it is very close to. This is because the electric field from the other wires is finite, and the area of the cylinder tends to zero, so the flux due to the other wires tends to zero. This makes it fairly easy to calculate the flux through each wire.

So now you have the total flux and the flux through each wire, so you can calculate how much flux goes through the faces. (It is just what is left over). Also, as the cylinders get very small, the area of the face tends to the area of the face of the actual cube. so what we have really done is an improper integral.

Hope that explanation helped. I think the more times you practice electrostatics problems, using Gaussian surfaces, the better you get at 'guessing' what kind of Gaussian surface would be most useful. The book 'Introduction to electrodynamics' by Griffiths might be useful. I don't know if there are practice questions in the book though.

One last thing: you could use a slightly different Gaussian surface which uses 1/4 cylinders, which are inside the wires. In this case, the enclosed charge is zero, so the total flux is zero. And so the flux through the cylinders is equal and opposite to the flux through the faces, so you can also work out the answer this way, which gives the same answer.

Edit: also, I have attached a picture of ehild's method. The picture is a slice through the cube. i.e. the wires are coming out of the page, and are enclosed by the cylinders.

Thanks both of you!

I never thought that there is a very simple solution to this. Though I haven't understood it properly but I will try to. I could never think of those cylindrical surfaces if I did not discuss this question here.

I don't have time to study an Intro book on electrodynamics because I have already completed the electrodynamics section in my course so I will just keep practicing question based on Gauss law and selecting Gaussian surfaces. Any links for that?

Thank you once again! :)

 

[BruceW]

I don't know of any links. I'm sure there probably will be some webpages if you search around on google for a bit. physicsforums is pretty good. This is how I stop myself from forgetting a lot of physics stuff :)

 

[ehild]

Google the Web, you find interesting places. A few example:

http://physicspages.com/2011/10/04/gausss-law-examples/
http://www.jfinternational.com/ph/gauss-law-exercises.html
http://www.iit.edu/arc/workshops/pdfs/Gauss_Law.pdf
http://www.phys.utk.edu/courses/Spring%202007/Physics231/chapter22.pdf

ehild

 

[TSny]

Here's another possible choice for the Gaussian surface that can also get the answer fairly easily. Suppose you want the flux through face A due to the blue wire in the top figure shown. Then introduce 3 more cubes as shown and make a Gaussian surface with the 8 gray faces and the front and back of the big box.

 

[Saitama]

Here's another possible choice for the Gaussian surface that can also get the answer fairly easily. Suppose you want the flux through face A due to the blue wire in the top figure shown. Then introduce 3 more cubes as shown and make a Gaussian surface with the 8 gray faces and the front and back of the big box.

Excellent! Thanks TSny!

This is the same method I use when a point charge is kept at the center of one of the edge. But here in this case, flux is zero through eight faces. :

 

[ehild]

Reading the question again

Twelve infinite long wires of uniform linear charge density (λ) are passing along the twelve edges of a cube. Find electric flux through any face of cube.

I would say that the correct answer is zero. The wires do not belong to the cube. The picture shows a cross section of the cube. At the top face, the wires along the edges of that face produce inward flux while the wires along the edges of the opposite face produce outward flux across the top face. The cube itself has zero charge. The Gaussian surface that is made from the faces of the block does not enclose any charge. So the net flux is zero. But the flux is the same across each face, because of symmetry. So the flux across a face is zero.

ehild

 

[BruceW]

Well, for sure "find flux through any face of cube" is too vague. If we take it to mean the improper integral of the flux through any face of the cube, then the answer is \lambda L / 2 \epsilon_0 This is the flux through the face (not including the wire), but in the limit that the area integral gets very close to the wire. I think this is most likely to be what they meant.