1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Flux due to infinite long wires

  1. Feb 15, 2013 #1
    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, [tex]\vec{E}=\frac{\lambda \cos \theta}{2 \pi \epsilon_o l}[/tex]
    Value of the differential area is
    [tex]\vec{dA}=(l)(dx)=l^2 \sec^2 \theta d\theta[/tex]
    Flux is the dot product of the electric field vector and area vector.
    [tex]d\phi=\vec{E} \cdot \vec{dA}[/tex]
    [tex]d\phi=\frac{\lambda \cos \theta}{2 \pi \epsilon_o l} \cdot l^2 \sec^2 \theta d\theta \cdot \cos (180-\theta)[/tex]
    [tex]\phi=\int_{0}^{\pi/4} \frac{\lambda l}{2 \pi \epsilon_o} d\theta[/tex]
    Solving this, the flux due to a single wire is
    [tex]\phi=\frac{\lambda l}{8 \epsilon_o}[/tex]
    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!
     

    Attached Files:

  2. jcsd
  3. Feb 15, 2013 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    What abou Gauss' Law?

    ehild
     
  4. Feb 15, 2013 #3
    How? :confused:

    What should be the gaussian surface?
     
  5. Feb 15, 2013 #4

    BruceW

    User Avatar
    Homework Helper

    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.
     
  6. Feb 15, 2013 #5
    I sketched the field lines due to each wire and found that only four wires contribute to electric flux through any face.
     
  7. Feb 15, 2013 #6

    BruceW

    User Avatar
    Homework Helper

    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.
     
  8. Feb 15, 2013 #7

    ehild

    User Avatar
    Homework Helper
    Gold Member

    The surface of the cube? :tongue2:

    ehild
     
  9. Feb 15, 2013 #8

    BruceW

    User Avatar
    Homework Helper

    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').
     
  10. Feb 15, 2013 #9
    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.
     
  11. Feb 15, 2013 #10

    BruceW

    User Avatar
    Homework Helper

    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.
     
  12. Feb 15, 2013 #11

    BruceW

    User Avatar
    Homework Helper

    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..
     
  13. Feb 15, 2013 #12

    ehild

    User Avatar
    Homework Helper
    Gold Member

    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
     
  14. Feb 15, 2013 #13
    Due to a single wire, the flux is zero through four faces and equal for the other two faces.
     
  15. Feb 15, 2013 #14

    ehild

    User Avatar
    Homework Helper
    Gold Member

    I mean due to all wires.


    ehild
     
  16. Feb 15, 2013 #15
    Then its same for all the faces.
     
  17. Feb 15, 2013 #16

    BruceW

    User Avatar
    Homework Helper

    yep :)
     
  18. Feb 15, 2013 #17

    ehild

    User Avatar
    Homework Helper
    Gold Member

    And what is the whole flux for a surface just enclosing the cube?
     
  19. Feb 15, 2013 #18
    :confused:

    It would be 6 times the flux through each face but for that I will have to find the flux through each face.
     
  20. Feb 15, 2013 #19

    BruceW

    User Avatar
    Homework Helper

    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.
     
  21. Feb 15, 2013 #20

    ehild

    User Avatar
    Homework Helper
    Gold Member

    What does Gauss' Law state?

    ehild
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Flux due to infinite long wires
Loading...