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Electric field and Gauss' law

  1. Feb 4, 2016 #1
    1. The problem statement, all variables and given/known data

    Assume that the infinite wall −2 < y < 2 is filled with a uniform charge of density ρv = 3 C/m3. Obtain an expression for Ey at a general point (0,y,0) and plot Ey as a function of y over −5 < y < 5.

    2. Relevant equations

    E = ρv / ( 4 π ε0 ρ)

    ∫∫∫D⋅ds = Qenc

    3. The attempt at a solution

    I will be honest. I am not really sure where to begin. I am not concerned about the plot. I need to know how to set this up. I am thinking gauss law or columns law. If I use coloumns law I set up a triple integral that will go to nfinity due to the limits of x and z. Gauss law seems like the way to go. If this is the case, do I wrap my gausian surface around a chunk of the wall, or do I take some segment? I need lost of help on this. I honestly do not know where to begin
     
  2. jcsd
  3. Feb 4, 2016 #2
    Please, someone help me out here. If only with the set up
     
  4. Feb 4, 2016 #3

    TSny

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    Gauss' law is the way to go.

    Before trying to set up any equations, use the symmetry of the problem to determine the direction of E for any point inside or outside the slab (wall).

    Also, determine all points where E = 0.
     
  5. Feb 4, 2016 #4
    because the charge is positive it points outward at all points. the only place I think it might be 0 is at the center. I think that here it is equal and opposite and therefor 0
     
  6. Feb 4, 2016 #5
    Is this correct?
     
  7. Feb 4, 2016 #6
    Can anybody help me on this, I am running out of time fast
     
  8. Feb 4, 2016 #7
    Note that outside of the wall (since the electric field lines cannot diverge because the wall is infinite) the field
    must be constant everywhere outside of the wall. This should provide some hints at what is happening inside
    of the wall.
     
  9. Feb 4, 2016 #8
    inside I suspect it is 0 at the middle. as I move outward in either direction it would begin to rise.

    My main concern is getting the expression. From that I can plot it.

    Can you help me. where do I start?
     
  10. Feb 4, 2016 #9

    TSny

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    Yes, the field is zero everywhere on the plane y = 0.
    Try a Gaussian surface in the shape of a rectangular box with one face at y = 0.
     
  11. Feb 4, 2016 #10
    so my Qenclosed = ρv L H where L and H are edges of the box
    and ∫∫∫D⋅dv = DLH
    can I say D= ρv
    and then
    E = ρv / ε0
     
  12. Feb 4, 2016 #11
    I'm really lost with this. We are just covering gauss law this is my first problem
     
  13. Feb 4, 2016 #12

    TSny

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    ρ is a volume charge density. So, to get the charge enclosed, you need to multiply ρ by a volume. Is L H a volume?
    In Gauss' law, the integral of the field is a surface integral representing flux through the surface. Which side or sides of the box will have nonzero flux?

    No.
     
  14. Feb 4, 2016 #13
    ok if 1 face is at y =0 then that face has no flux through. The only face we care about is the one poking out the side (that's where we are asked to find E) but the top and bottom would have flux too.
    so Qenc =ρv L W H
    and then I'm not sure what to do with the integral if I do ∫d⋅dv then I get D L W H and I get the same expression E = ρv / ε0
    I don't think this is right
     
  15. Feb 4, 2016 #14

    TSny

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    I'm not sure what direction your y-axis is pointing. I'm guessing that your y-axis is running horizontally so that y = 0 is a vertical plane in the middle of the vertical wall of charge. So, we have a Gaussian box placed such that one face of the box is at y = 0. As you say, there will not be any flux through this face because the field is zero on the this face. There are 5 other faces. Which of these faces has nonzero flux?

    Your expression Qenc =ρv L W H will be correct if you interpret L W and H properly. You will need to deal separately with 2 cases: (1) finding the field at a point inside the wall, and (2) finding the field at a point outside the wall.
     
  16. Feb 4, 2016 #15
    lets deal with case 1 first where the box is completely inside. at this point the outside face will be the only one with flux thru. the flux will be parallel to all other faces. Am I right?
     
  17. Feb 4, 2016 #16

    TSny

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    Yes. Good.
     
  18. Feb 4, 2016 #17
    ok then I need to set up ∫D⋅ds
    isn't this just D * surface area or D L H
     
  19. Feb 4, 2016 #18

    TSny

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    Yes.
     
  20. Feb 4, 2016 #19
    alrighty then so if I set eqns equal to eachother D= ρv W / ε0

    wont the expression be the same for the outside of wall case?

    I know I have to get rid of the w this is not given how do I do this?
     
  21. Feb 4, 2016 #20

    TSny

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    How is W related to the value of y for the face which had nonzero flux?

    For points outside the wall, your box will only be partially filled with charge.
     
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