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!

How would one approach such problem?

  1. Sep 8, 2013 #1
    1. The problem statement, all variables and given/known data

    Uniform plane of charge. Charge is distributed uniformly over a large square plane of side ##\ell##, as shown in the figure from the uploaded file (see file). The charge per unit area (C/m[itex]^2[/itex]) is [itex]\sigma[/itex]. Determine the electric field at a point ##P## a distance ##z## above the center of the plane, in the limit ##\ell \rightarrow \infty##.

    The hint is as followed:

    Divide the plane into long narrow strips of width ##dy## and determine the electric field due to each of the strips. Then, sum up the fields due to each strip to get the total field at ##P##.

    2. Relevant equations

    ##dE = \dfrac{1}{4\pi\varepsilon_0} \dfrac{dQ}{r^2}##

    3. The attempt at a solution

    Found ##r##, which is ##\sqrt{z^2 + y^2}##.

    With trig, I found that

    ##dE_z = dE\cos(\theta)##
    ##dE_y = dE\sin(\theta)##

    By integration,

    ##\int dE_y = \int dE\sin(\theta)##
    ##E_y = 0##

    ##\int dE_z = \int dE\cos(\theta)##
    ##E = E_z = \dfrac{\lambda}{4\pi\varepsilon_0} \int \dfrac{\cos(\theta)}{z^2 + y^2} dy##

    Since ##y = z\tan(\theta)##, then ##dy = z(\sec(\theta))^2 d\theta##. Since ##\cos(\theta) = \dfrac{z}{\sqrt{z^2 + y^2}}##, then the integrand becomes ##\dfrac{\cos(\theta) d\theta}{z}##. Thus,

    ##E = \dfrac{\lambda}{4z\pi\varepsilon_0} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos(\theta) d\theta = \dfrac{\lambda}{2z\pi\varepsilon_0}##

    That is for one of the strips. I don't know how to sum up the fields.
     

    Attached Files:

  2. jcsd
  3. Sep 9, 2013 #2

    rude man

    User Avatar
    Homework Helper
    Gold Member

    If the sheet is infinite in its two dimensons then the E field is easily determined by using Gauss' theorem. You don't need to integrate.

    If the sheet dimensions are finite that's more difficult.

    The problem states that the dimensions are infinite.
     
  4. Sep 9, 2013 #3
    Since your point of interest is in the center of the plane, just use symmetry. Even for the finite case, it's not difficult. It's difficult when the point is off axis.
     
  5. Sep 9, 2013 #4
    I haven't learned Gauss' Law 100% yet. I need to use the electric field formula to work out the problem.
     
  6. Sep 9, 2013 #5

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Oh, OK. I'll keep tabs on the thread & see how it goes.
     
  7. Sep 10, 2013 #6
    Note again, this is a fairly straight forward problem; do not over complicate it (thinking too hard only makes matters worse). Use symmetry and your golden.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: How would one approach such problem?
Loading...