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Infinite Wire/Surface charge question

  1. Oct 30, 2016 #1
    1. The problem statement, all variables and given/known data
    A thin infinite wire with linear charge density λ is located parallel to an infinite conducting surface, which is coincident with the x-y plane (i.e., z = 0). The wire is parallel to the ˆx direction and is located a distance z = d from the conducting surface. The figure on the left shows a section of the infinite wire and infinite surface in 3D, while the figure on the right shows the configuration projected in the y-z plane, where the wire is going into/out of the page.

    (a) Sketch the location and charge configuration of the image charge.
    (b) Sketch the total electric field E~ for region above the plane (z > 0).
    (c) At a point located on the conducting surface ~r = (x, y, 0), sketch the electric field vector E~wire due to the charged wire. At the same point, sketch the electric field vector E~ image due to the image charge. Argue that the total electric field E~ = E~wire + E~ image points in the −zˆ direction.
    (d) How are z-components of E~wire and E~ image related to each other?
    (e) Determine the total electric field E~ at an arbitrary point ~r = (x, y, 0) on the surface. Your answer should depend on y and other constants given above. Hint: You just need to compute the z-component of E~wire, using your arguments in parts (c) and (d).
    (f) What is the surface charge density σ on the conducting surface? (Note: The surface density is not constant; your answer should be a function of position y.) Sketch your answer for σ(y) as a function of y. (3 points)
    (g) The amount of charge on the conducting surface must be the same as that of the image charge. Verify that the linear charge density of the surface charge along the x-direction, which is given by ∫ ∞ −∞ dy σ(y), (1) is the same as the linear charge density −λ of your image charge. (3 points)
    2. Relevant equations
    Field due to infinite wire: E=λ/2πε0r*cosθ

    E = σ / ε0

    Et = - (Eimage+ Ewiire)
    Et = - λ/πε0r*cosθ

    r=√(d2+y2)
    cosθ = d/√(d2+y2)

    σ = λ/πr3/2

    3. The attempt at a solution
    https://drive.google.com/file/d/0B76TBRMyBuffYzdZYUpVUk8zdkU/view?usp=sharing [Broken]


    I only really need help with part g) I'm obviously doing something wrong as I either get 0 or an indefinite... Not quite sure what I"m doing wrong.
     
    Last edited by a moderator: May 8, 2017
  2. jcsd
  3. Oct 30, 2016 #2

    TSny

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    upload_2016-10-30_12-54-8.png

    The power of 3 on the right side is incorrect.
     

    Attached Files:

  4. Oct 30, 2016 #3
    It would be that easy....

    Thanks a lot.
     
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