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!

Electric Field in a Capacitor

  1. May 11, 2008 #1
    I have a problem with electric fields in capacitors. Using Gauss' Law, we always choose the Gaussian surface that encloses the positive plate in the capacitor. When we go through field calculations, we get an equation for the electric field and we use that to find voltage, capacitance, and everything else. What I don't understand is how enclosing ONLY the positive plate takes into account the fact that the negative plate is present and that it amplifies the field between the two. In other words, how would our calculations or conclusions be any different if the negative plate were not even present?
    Thank you very much for the help
  2. jcsd
  3. May 12, 2008 #2


    User Avatar
    Homework Helper

    Hi kash25,

    I don't think that's correct. For example, for parallel plates we might use Gauss's law to find the electric field of just the postive plate (giving a result of [itex]E = \sigma/(2\epsilon_0)[/itex] everywhere), but then we would double that to find the electric field of the capacitor between the plates ([itex]E=\sigma/\epsilon_0[/itex]) and set the field to be zero outside the plates.

    (Between the plates, the fields are in the same direction and so add; outside the plates they are in opposite directions and therefore cancel; also I'm making the usual approximation that the nonuniform field at the ends of the plate can be neglected, etc.)
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Electric Field in a Capacitor