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Gauss's Law: a nonconductive wall

  1. Feb 15, 2009 #1
    1. The problem statement, all variables and given/known data

    A nonconducting wall carries charge with a uniform denisty of 8.g µC/cm2. What is the electric field 7.00cm in front of the wall? Explain whether your result changes as the distance from the wall is varied.

    2. Relevant equations

    I'm still really new, I just registered, and trying to find the integral sign and stuff is proving to be pretty difficult

    E=k[tex]\frac{q}{r2}r[/tex]unit vector

    So Gauss's Law and the equation for the electric field, I think, are relevant. The relationship [tex]\sigma[/tex]=Q/A I think is important too. Let's throw in Coulomb's Law for good measure.

    F=kqq0/(r2)r unit vector

    3. The attempt at a solution

    Okay, so from my understanding of Gouss's Law, you can use it to calculate the electric field around a closed symmetrical surface that surrounds the thing you want to find the electric field of. Rather, Gauss's Law is a way of counting up the electric field lines. I think I understand how to use Gauss's Law to find the electric field at a point around a sphere of charge; you can just make a Gaussian Sphere around it.

    Anyway, I guess my major source of confusion is that you can't really encapsulate a wall, like if it was a rod, or if we could interpret it as a long line of charge, then we could put a cylinder around it. I think I gathered from another post that

    I think that this is the key, but I don't know how this came about. How can this be derived from Gauss's Law or the equation for electric field?

    I'm generally pretty confused, and I think I'm going to try to firm up my understanding of applying Gauss's law while I hope that someone replies.
  2. jcsd
  3. Feb 15, 2009 #2


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    Welcome to PF,

    Perhaps this will help?

    I think the key is to what distance your Gaussian cylindrical surface can be taken and how close to en edge you are at such that the walls of the cylinder can be taken not to have any contribution to flux through the end caps.

    For instance at 100 m from the wall? How big is the wall?
  4. Feb 16, 2009 #3
    That link was super helpful! I now understand why the flux = E2A, but am still a little fuzzy how that equals [tex]\sigma[/tex]A/[tex]\epsilon[/tex]0 I was thinking because flux=Qenclosed/[tex]\epsilon[/tex]0, and maybe you could say that the charge density is equal to the charge enclosed? Am I getting the ideas mixed up? Because if Qenclosed = [tex]\sigma[/tex], then you could say

    and since the integral of dA=1, it would be then simplified to


    Okay... yay! I'm pretty happy now. Sweet!! Excellent!!! Am I right? Is my thought process right? Is it okay to assume that when they are talking about charge density they are talking about Qenclosed?

    Oh, and they didn't give me any dimensions of the wall, I think the wall is highly similar to the flat sheet of charge depicted in the picture in the link. From this calculation, I think, the distance from the wall doesn't matter.

    Bah, that seems weird to me. I mean, what about the inverse square relationship? Like, do things really give off fields that are not dependent upon distance? Like field lines, shouldn't field lines be tapered or something? Maybe my understanding of field lines is off.

    Thank you so much for the help with the problem!!!! Thanksthanksthanks :)
    Last edited: Feb 16, 2009
  5. Feb 16, 2009 #4


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    Close but not quite. [itex] \sigma [/itex] is a surface charge density so it must be multiplied by an area to give a charge. So Q enclosed = [itex] \sigma A [/itex] where A is the area of the material that is enclosed by your gaussian surface

    No, the integral of dA is just A, the area enclosed.

    Now, the A will cancel out and you will recover the result you had before. It was correct but the two steps above were incorrect (basically, you made two small mistakes that canceled one another by pure luck)

    To adress your other question: for an infinite plane surface (uniformly charged), the E field is indeed constant, it does not depend on the distance. But of course, this is an unphysical situation. In real life, planes are always finite in which case the answer you get is only approximative and is a good approximation only for points at distances much less than the dimensions of the plane (points where the plane effectively looks almost infinite)
  6. Feb 16, 2009 #5
    crap... the integral of dA is A. bahhh well...I think I was on the right track. woahhh I think I was thinking about that integral way incorrectly. It's the integral of E2AdA right, so that gives E2A2/2 so EA2. bahhh I'm back at square one. It's alright I think I've got some sort of handle on something or other :-/
  7. Feb 16, 2009 #6
    Bah! Willickers I forgot that [tex]\sigma[/tex]=Q/A ! Okay okay hmm okay. Sorry about my last response, I was working on it and wasn't paying attention when you replied. Thanks for catching my mistakes!!

    Woah! Okay it works now!! disregard that crazy stuff with the integral bit, I don't know what that was.
  8. Feb 16, 2009 #7


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    No, it is the integral

    [tex] \int \vec{E} \cdot \vec{dA} = \int E \cos \theta dA [/tex]

    In your case, theta is 0 so you get ismply [itex] \int E dA = E \int dA [/itex]

    You were completely correct except that you did not have the factors of A where they should have been. But they cancel out.
  9. Feb 16, 2009 #8
    That makes complete sense. I should've just looked and relied on the definitions instead of extrapolating weird junk, yeah that makes total sense now. Thanks so much!
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