# Homework Help: Flux due to infinite long wires

1. Feb 15, 2013

### Saitama

1. The problem statement, all variables and given/known data
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. (see attachment)

2. Relevant equations

3. The attempt at a solution
I have actually solved the problem but I think there's a more simpler way.

Okay, I figured out that electric field due only four out of twelve wires will contribute to the electric flux through any face. Lets consider the bottom face. The flux passing through this face is due to only four wires of the top face.

Calculating flux due to a single wire. (see attachment 2)
In the figure $x=l \tan \theta$. Or, $dx=l \sec^2 \theta d\theta$.
Electric field due to this wire, $$\vec{E}=\frac{\lambda \cos \theta}{2 \pi \epsilon_o l}$$
Value of the differential area is
$$\vec{dA}=(l)(dx)=l^2 \sec^2 \theta d\theta$$
Flux is the dot product of the electric field vector and area vector.
$$d\phi=\vec{E} \cdot \vec{dA}$$
$$d\phi=\frac{\lambda \cos \theta}{2 \pi \epsilon_o l} \cdot l^2 \sec^2 \theta d\theta \cdot \cos (180-\theta)$$
$$\phi=\int_{0}^{\pi/4} \frac{\lambda l}{2 \pi \epsilon_o} d\theta$$
Solving this, the flux due to a single wire is
$$\phi=\frac{\lambda l}{8 \epsilon_o}$$
Last step is to multiply it by 4.

I think that there is a much easier way instead of doing the math. Most of the questions I have done based on finding flux through cube or its faces have some kind of symmetry and I guess that this question too involves some trick which I can't figure out.

Any help is appreciated. Thanks!

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2. Feb 15, 2013

### ehild

What abou Gauss' Law?

ehild

3. Feb 15, 2013

### Saitama

How?

What should be the gaussian surface?

4. Feb 15, 2013

### BruceW

I can think of one way. Think of the most obvious Gaussian surface. I get a different answer to your calculation. I don't know why you have assumed that only electric field due to the wires on the opposite face contributes..

Edit: a hint: don't be surprised if the answer depends on the limit in which you take the Gaussian surface.

5. Feb 15, 2013

### Saitama

I sketched the field lines due to each wire and found that only four wires contribute to electric flux through any face.

6. Feb 15, 2013

### BruceW

hmmm... what about the four wires which extend perpendicular to the face?

Also, the symmetry argument I am thinking of only works for the flux through the face which is defined either just inside or just outside the cube... I now realise that you are trying to find the flux through an area which is exactly on the plane of one of the faces, and contained within the boundary of the wire. So the symmetry argument I was thinking of doesn't work in that case.

7. Feb 15, 2013

### ehild

The surface of the cube? :tongue2:

ehild

8. Feb 15, 2013

### BruceW

I think the problem is that if it is exactly the surface of the cube, then what is the enclosed charge? There is no problem if we are talking about flux just inside or outside the cube, but flux at exactly the cube faces is more difficult.

edit: for sure, the electric field is not continuous at the face of the actual cube. (unless we make the wires have finite width, but since we are not told this, I think he is meant to assume the wires are effectively dirac-delta 'functions').

9. Feb 15, 2013

### Saitama

Draw the field lines due to one of those wires. The field lines are always normal to the direction of area vector.

I am still at loss. We have got a line charge here. If it was a point charge, it would have been easy to calculate the flux (given that it is at one of points where flux can be calculated using symmetry) but I don't get how we can calculate flux of a line charge using symmetry arguments.

10. Feb 15, 2013

### BruceW

No. It would be, if the wire was infinitely long, but these wires are not infinitely long.

The idea is that you can calculate the flux through the face due to all the wires. (but we can only do this for the face just inside and just outside the cube). Use a Gaussian cube just inside the actual cube. What is the charge enclosed? And by symmetry, what does this tell you about the flux through each face? And also, do this for a Gaussian cube just outside the actual cube.

Edit: darnit. I'm so sorry, I thought the wires were supposed to be not infintely long. Sorry about that. I should have read your post more carefully.

11. Feb 15, 2013

### BruceW

I'm pretty sure the symmetry argument will still work even though the wires are infinitely long, because there is still a symmetry in the system. But as I said, this only works for the flux just inside the actual cube or just outside the actual cube. And then your method gives the flux which is exactly on the face of the actual cube. I can't think of any way to get the answer for exactly on the face of the cube, using symmetry..

12. Feb 15, 2013

### ehild

You do not need to calculate the electric field. But you can use symmetry. Is the flux different at the different faces of the cube?

ehild

13. Feb 15, 2013

### Saitama

Due to a single wire, the flux is zero through four faces and equal for the other two faces.

14. Feb 15, 2013

### ehild

I mean due to all wires.

ehild

15. Feb 15, 2013

### Saitama

Then its same for all the faces.

16. Feb 15, 2013

### BruceW

yep :)

17. Feb 15, 2013

### ehild

And what is the whole flux for a surface just enclosing the cube?

18. Feb 15, 2013

### Saitama

It would be 6 times the flux through each face but for that I will have to find the flux through each face.

19. Feb 15, 2013

### BruceW

yes, it is 6 times the flux through each face. And the key is to use Gauss' law now. I'm still stressing that this is for flux just outside the face.

Edit: I'm sure it will seem obvious, once you realise it.

20. Feb 15, 2013

### ehild

What does Gauss' Law state?

ehild

21. Feb 15, 2013

### Saitama

$$\int \vec{E} \cdot \vec{dA}=\frac{q_{enclosed}}{\epsilon_o}$$

22. Feb 16, 2013

### ehild

What is the enclosed charge if you cover the cube with a closed cubical surface just outside the faces?

ehild

23. Feb 16, 2013

### Saitama

$12λl$?

24. Feb 16, 2013

### ehild

yes.

25. Feb 16, 2013

### Saitama

The flux through the whole cube is $12\lambda l/ \epsilon_o$. Then flux through a single face is $\frac{2\lambda l}{\epsilon_o}$ but this is not the right answer.