Sunday, December 30, 2012

Electric Field Intensity due to Sheet of Charge


Field due to a uniformly charged infinite plane sheet

Let σ be the uniform surface charge density of an infinite plane sheet. We take the x-axis normal to the given plane. By symmetry, the electric field will not depend on y and z coordinates and its direction


We can take the Gaussian surface to be a rectangular parallelepiped of cross sectional area A, as shown. (A cylindrical surface will also do.) As seen from the figure, only the two faces 1 and 2 will contribute to the flux; electric field lines are parallel to the other faces and they, therefore, do not contribute to the total flux.

The unit vector normal to surface 1 is in –x direction while the unit vector normal to surface 2 is in the +x direction. Therefore, flux E.ΔS through both the surfaces are equal and add up. Therefore the net flux through the Gaussian surface is 2 EA. The charge enclosed by the closed surface is σ A. Therefore by Gauss’s law,

 where nˆ is a unit vector normal to the plane and going away from it. E is directed away from the plate if σ is positive and toward the plate if σ is negative.

Note that the above application of the Gauss’ law has brought out an additional fact: E is independent of x also

3 comments:


  1. I love reading, I love blogging, and I love comments! Thanks for sharing your thoughts and have a wonderful day!

    Zean
    www.imarksweb.org

    ReplyDelete
  2. Love it! Very interesting topics, I hope the incoming comments and suggestion are equally positive. Thank you for sharing this information that is actually helpful.


    ufgop.org
    ufgop.org

    ReplyDelete
  3. Had a great time reading your article, please read my article also imarksweb.net
    imarksweb.net . GBY :*

    ReplyDelete