1. Not finding help here? Sign up for a free 30min 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!

Gauss's Law problem

  1. Feb 22, 2008 #1
    [SOLVED] Gauss's Law problem

    1. The problem statement, all variables and given/known data
    An infinite slab of charge parallel to the yz-plane has density p for -b < x < b, 0 otherwise. Find the electric field at all points.

    3. The attempt at a solution
    I am able to do the electric field outside the slab. But I am off by a factor of 1/2 for the electric field inside. I made a Gaussian cylinder that starts at the axis to x. I'll call the area of the circular faces [itex]A = \pi r^2[/itex]. Since there are two of these faces, net flux is [itex]2EA[/itex]. The charge enclosed in the surface is [itex]\rho A x[/itex].

    [tex]2 E A = \frac{\rho A x}{\epsilon_0} => E = \frac{\rho x}{2 \epsilon_0}[/tex]

    My book says the solution is just [itex]\rho x / \epsilon_0[/itex]. So, I know the error probably came from saying the flux was 2EA. But, for the electric field outside the slab, I used 2EA for the flux, and since the charge enclosed in a cylinder running from -x to x, for |x| > b, was 2b, the factors of 2's cancel out. Here they don't. I can't figure out what I'm doing wrong.

    I suppose I could just say the flux was one EA because the field at x = 0 is 0. But, for the next problem, I have to find the electric field for a slab whose density is [itex]\rho(x) = \rho_0 e^{-|x/b|}[/itex], and I still have that extra 1/2 factor in there, without the field at x = 0 being 0.
     
    Last edited: Feb 22, 2008
  2. jcsd
  3. Feb 22, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    By having your Gaussian surface extend from the origin to a distance x from the origin you lose the symmetry needed to apply Gauss's law. You must have your Gaussian volume centered on the origin--that way you know that the electric field is the same at each end. Have it go from -x to +x.
     
  4. Feb 22, 2008 #3
    Ah, that makes sense. Thanks.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Gauss's Law problem
  1. Gauss's Law Problem (Replies: 2)

  2. Gauss' Law Problem (Replies: 1)

  3. Gauss's Law Problem (Replies: 8)

  4. Gauss's law problem (Replies: 1)

Loading...