1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper


    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
    [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


    User Avatar
    Homework Helper

    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

    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


    User Avatar
    Homework Helper

    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
    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Gaussian pillbox
  1. Gaussian surface (Replies: 4)

  2. Gaussian spheres (Replies: 1)

  3. Gaussian Surfaces (Replies: 16)

  4. Gaussian cylinder (Replies: 1)