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Electric Field of a Long Charged Ribbon

  1. Oct 1, 2008 #1
    1. The problem statement, all variables and given/known data
    A long uniformly charged ribbon is located in the x-z plane, parallel to the z axis, occupying the region -[tex]\infty[/tex] [tex]\leq[/tex] x [tex]\leq[/tex] [tex]\infty[/tex] and -a/2 [tex]\leq[/tex] x [tex]\leq[/tex] a/2. The charge per unit area on the ribbon is [tex]\sigma[/tex]. a) Determine [tex]\vec{E}[/tex] at (x,0,0) where x > a/2. b) Determine [tex]\vec{E}[/tex] at (0,y,0) where y > 0.


    2. Relevant equations
    Gauss' Law


    3. The attempt at a solution
    I'm split on part (a). My intuition tells me that symmetry should make the electric field point only in the [tex]\underline{+}[/tex]y direction, with no component in the [tex]\underline{+}[/tex]x direction. But is this wrong? Should I make a Gaussian surface (say, a box or a cylinder pointing in the x direction) that straddles the edge of the ribbon to test for flux in the x direction?
     
    Last edited: Oct 1, 2008
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  3. Oct 1, 2008 #2

    Redbelly98

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    There will be an E-field in the x-direction for part (a).

    Have you looked at the field of a uniformly charged line or wire? You could use that. You'll also need to do an integral.

    p.s. welcome to PF!
     
  4. Oct 1, 2008 #3

    gabbagabbahey

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    Gauss's law is only useful for planar geometries when the plane either extends to infinity, or you are only looking for the field far from the edges of the plane. Does the ribbon qualify?
     
  5. Oct 1, 2008 #4
    Ok, so I guess I can treat the ribbon like a line of charge when I'm looking at it from a point on the x-axis... And is it safe to assume I should integrate across the width of the ribbon as well, because of the contributions of all the charge "behind" the edge I'm looking at? I would think I should do the same for part (b), i.e. treat the ribbon as a collection of infinitely long wires and integrate across the width?

    Also, good point about Gauss' law. Thanks for the help guys.
     
    Last edited: Oct 1, 2008
  6. Oct 1, 2008 #5

    gabbagabbahey

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    Well it's actually a surface charge instead of a line charge; so yes you will have to integrate over both x and y. Remember, for a surface charge [itex]\sigma(\vec{r'})[/itex] you have:

    [tex]\vec{E}(\vec{r})=\frac{1}{4 \pi \epsilon_0} \int_{\mathcal{S}} \frac{\sigma(\vec{r'})da'}{|\vec{r}-\vec{r'}|^2} \widehat{|\vec{r}-\vec{r'}|}[/tex]
     
  7. Oct 2, 2008 #6

    Redbelly98

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    Yes, exactly. For both (a) and (b).

    edit:
    This method assumes you know E for a charged wire ...
     
  8. Oct 2, 2008 #7
    Easy enough, thanks for the help!

    Edit: Haha not so easy I guess... I set up the following integral (for part (a)):

    [tex]\vec{E} = \frac{2\sigma}{4\pi\epsilon_{o}a}\int^{-a/2}_{a/2}\frac{dx}{x}\hat{x}[/tex]

    When I integrate the 1/x from a/2 to -a/2, I of course get zero. Is my integrand incorrect? Note that the integrand I picked is the electric field of an infinite wire (with sigma/a as the charge density).
     
    Last edited: Oct 2, 2008
  9. Oct 2, 2008 #8
    Scrap that last post, I treated this problem with sort of a surface integral and got an answer (an admittedly strange-looking one):

    First I used the equation:

    [tex]\vec{E} = \frac{1}{4\pi\epsilon_{o}} \int_{S}\frac{\sigma(\vec{r'})}{r^{2}}\hat{r}da'[/tex]

    Part (a): [tex]\vec{E} = \frac{1}{4\pi\epsilon_{o}}\int^{a/2}_{-a/2}\int^{\infty}_{-\infty}\frac{\sigma dz'dx'}{z'^{2}+(x-x')^{2}}cos(\theta)\hat{x}[/tex]

    where [tex]z'^{2}+(x-x')^{2}[/tex] is r2, and

    [tex]cos(\theta) = \frac{x-x'}{\sqrt{z'^{2}+(x-x')^{2}}}[/tex]

    In this case, x' is the distance from the center of the ribbon (x=0) to my area element, and x-x' is the distance from the area element to the x-coordinate of the point in question, while z' is the z distance from the origin to the area element.

    I took a similar approach to part (b):

    [tex]\vec{E} = \frac{1}{4\pi\epsilon_{o}}\int^{a/2}_{-a/2}\int^{\infty}_{-\infty}\frac{\sigma dz'dx'}{x'^{2}+y'^{2}+z'^{2}}cos(\theta)\hat{y}[/tex]

    and

    [tex]cos(\theta) = \frac{y'}{\sqrt{x'^{2}+y'^{2}+z'^{2}}}[/tex]

    where y' is just the y-distance from the ribbon.

    Anyway you can imagine what the integration looks like... and in my final answers I get some logs, which I thought was strange. Do you think these approaches are sensible?
     
    Last edited: Oct 2, 2008
  10. Oct 2, 2008 #9

    Redbelly98

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    Looks reasonable to me.

    The line charge idea will work too, but your integrand should be:

    dx' / (x-x')

    And there is no "a" term in the denominator out front (linear charge density is σdx)
     
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