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Gaussian pillbox

  1. Sep 27, 2007 #1
    The question states that there is an infinite plane slab of thickness 2d that has a uniform charge density rho. Find the electric field as a fucntion of y, where y=0 at the center.

    inside plane:
    [tex] \int E da = 2EA = \frac{\rho A y}{\epsilon_o}[/tex]

    outside plane:
    [tex] \int E da = 2EA = \frac{\rho A d}{\epsilon_o}[/tex]

    The 2d thickness runs along the y axis, and the plane is infinite along the x and y axis.

    Is this the right setup?
     
    Last edited: Sep 27, 2007
  2. jcsd
  3. Sep 27, 2007 #2

    learningphysics

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    everything looks right except I think you're missing a factor of 2... if y is measured from the center... then taking a section center from the center with width 2y... you'll have

    [tex]2E_y*A = \frac{\rho A (2y)}{\epsilon_o}[/tex]

    same with outside the plane...

    [tex]2E_y*A = \frac{\rho A (2d)}{\epsilon_o}[/tex]
     
  4. Sep 27, 2007 #3
    good point!

    I dont know why the E-field would be negative then y < 0?

    should the e-field always be positive when dealing with plates? Like for an infinite plane, the E field is:

    [tex] E = \frac{\sigma}{2 \epsilon _o} [/tex]

    the E-field cant be negative (unless the charge on the plane was negative). So why does my expression give a negative E when i go to negative y?
     
  5. Sep 27, 2007 #4

    learningphysics

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    The field at [tex]E_{-y} = -E_{y}[/tex] (where y>0). We took this into account when we did gauss law... that's why we got 2*Ey*A.... It was Ey*A + (-Ey)*(-A)... (this is just integral of E.dA)

    We were actually taking y to be positive and using the fact that [tex]E_{-y} = -E_y[/tex].

    This makes sense by symmetry... if the field at y is upward... then the field at -y should be downward...
     
  6. Sep 27, 2007 #5

    learningphysics

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    No.

    The field can be negative for that case... one side the field is upward... the other side the field is downward. You have to choose a particular direciton as positive... the other side is negative.
     
  7. Sep 27, 2007 #6
    So in my case:

    [tex] E = \frac{\rho d }{\epsilon_o} y[/tex] when y>d
    and
    [tex] E = \frac{-\rho d }{\epsilon_o} y[/tex] when y<d

    where y is the unit vector in the y direction
     
  8. Sep 27, 2007 #7

    learningphysics

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    yeah. just to be sure though, you didn't use y as a unit vector here right: ?

    [tex]2E_y*A = \frac{\rho A (2y)}{\epsilon_o}[/tex]

    here y is the distance from the center...
     
  9. Sep 27, 2007 #8
    yes.

    it seems like when I do the same thing for the opposite side, i get the same equation. is it the negative y unit vector that makes the E field negative?
     
  10. Sep 28, 2007 #9

    learningphysics

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    yes... By symmetry the field is in the opposite direction. Hence the vector needs to be negative...
     
  11. Oct 22, 2008 #10
    I have a similar exercise and I'm finding some difficulty in understanding it.
    So from Gauss' Law I have
    [tex]\epsilon_{0}\oint EdA=q_{enc}[/tex]
    I know
    [tex]q_{enc}=\rho\,V=\rho\,y[/tex]
    But I don't get where do those "2" come from as in 2EA and 2y.
    I also fail to understand the difference between the inside and outside equations. The y gets replaced by d but the only connection I know of is that when y=2d it's the end of the plate
     
    Last edited: Oct 22, 2008
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