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Electric potential for flat sheet

  1. Jul 31, 2012 #1

    WannabeNewton

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    1. The problem statement, all variables and given/known data
    A flat non conducting sheet lies in the x - y plane. The only charges in the system are on this sheet. In the half space above the sheet, z > 0, the potential is [itex]\varphi = \varphi _{0}e^{-kz}coskx[/itex]. Describe the charge distribution on the sheet.


    2. Relevant equations



    3. The attempt at a solution
    We have that [itex]E = -\triangledown \varphi = (\varphi _{0}ke^{-kz}sinkx)\hat{i} + (\varphi _{0}ke^{-kz}coskx)\hat{k} [/itex] for z > 0 and [itex]\int_{S}E \cdot dA = 4\pi\int_{S} \sigma dA [/itex] but I'm not sure how to use it here; all I can see is that if I enclose a part of the sheet with a box then the only parts of the box with a net flux through them will be the top part of the box and the bottom part and for that case the total flux through the two together will be double the flux through the top part of the box and the field lines are not perpendicular to that top part. Not sure what to do with that.
     
    Last edited: Jul 31, 2012
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  3. Aug 1, 2012 #2

    TSny

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    I think you're close to the answer. Assume the Gaussian box is small enough so that the electric field is essentially constant over the top and bottom surfaces of the box and the charge density within the box is essentially constant. Let the top of the box be essentially right at the top surface of the sheet and the bottom of the box at the bottom surface of the sheet. You won't need to worry about the fact that E is not perpendicular to the sheet because you can assume that the "height" of the box is essentially zero so that you can neglect any flux through the "sides" of the box.

    Find an expression for the total flux through the box using your expression for E. Then use Gauss' law to relate the flux to the charge density.
     
  4. Aug 1, 2012 #3

    WannabeNewton

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    So would I just have [itex]E_{z} = 2\pi \sigma [/itex] since [itex]2\int_{S}E\cdot dA = 2\int_{S}E\cdot dxdy\hat{k} = 2E_{z}\int_{S}dA[/itex] under the limiting case you described for the box where the field would essentially be constant over the top and bottom surfaces of the box, the flux would all be essentially upwards and downwards, and the charge density would be constant in the section of the sheet we are covering? Do I then take [itex]\lim_{z \rightarrow 0}E_{z}[/itex]?
     
    Last edited: Aug 1, 2012
  5. Aug 1, 2012 #4

    TSny

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    Yes, I think that should get it.
     
  6. Aug 1, 2012 #5

    WannabeNewton

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    Thank you! I end up with [itex]\sigma = \frac{1}{2\pi }k\varphi _{0}coskx[/itex]. Does this technique of taking the limiting case of the box for sheets come up often (and maybe even other geometries)?
     
    Last edited: Aug 1, 2012
  7. Aug 1, 2012 #6

    TSny

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    Your expression for σ looks correct to me :smile:. A limiting Gaussian box is often used to determine the discontinuity in the normal component of the electric or magnetic field at a surface separating two different materials when there is a charge or current on the surface. Most standard texts will derive these "boundary conditions".
     
  8. Aug 1, 2012 #7

    WannabeNewton

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    Thanks! And just one final question, my apologies: when we made the box really small that included taking the limit as height, width, depth -> 0 for the box? So as long as the E - field isn't dis-continuous at that sheet and the charge distribution doesn't fluctuate wildly we should be able to get a constant E - field and charge distribution within that really small box?
     
  9. Aug 1, 2012 #8

    TSny

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    I think of it as sort of an odd limit. We take the top and bottom of the box to have "very small" but finite areas ΔA (we don't think of taking the limit if these areas to zero). Thus, think of these areas as finite, but small enough that any variation of the field over the top (or bottom) is negligible and any surface charge density is essentially uniform inside the box. However, for the height of the box, we do in fact think of taking the limit to zero. That way, any flux through the sides of the box will go to zero, and the net flux is due just to the flux through the top and bottom. In this limit, the charge enclosed in the box would be σ[itex]\cdot[/itex]ΔA. Gauss' law would then allow us to derive the connection between the normal component of the field on each side of the surface and the surface charge density.
     
  10. Aug 1, 2012 #9

    WannabeNewton

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    Ah ok so the [itex]\delta A << 1[/itex] for each side of the box allows us to consider E and [itex]\sigma [/itex] as constants over the sides and within the box, respectively, and letting z -> 0 is what allows us to say that only the top and bottom will be considered so that we can just consider [itex]\delta A\hat{k}[/itex] and [itex]\delta A(-\hat{k})[/itex]?
     
  11. Aug 1, 2012 #10

    TSny

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    Yes, I think that's the right way to think about it.
     
  12. Aug 1, 2012 #11

    WannabeNewton

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    Thank you for clearing that up for me!
     
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