1. Limited time only! Sign up for a free 30min personal 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!

Capacitance and electric fields

  1. May 6, 2008 #1
    [SOLVED] Capacitance and electric fields

    1. The problem statement, all variables and given/known data
    I want to find the capacitance of a parallel-plate capacitor consisting of two metal surfaces og area A held a distance d apart.

    First I find the electric field, which I know is [tex]\frac{\sigma }{{2\varepsilon }}[/tex]. Then I use that the potential is the line integral from minus-charge plate to positive-charge plate of electric field, so

    [tex]V = \frac{\sigma }{{2\varepsilon }}d[/tex]

    But in my book they do not divide by 2. This is because they have used superposition, which is OK - I agree with that.

    Next, we consider the following:


    Here they have used the electric field for |one| cylinder, they have not used superposition. This I do not agree with - we are still between the cylinders, so why do we not multiply by 2 here?
    Last edited: May 6, 2008
  2. jcsd
  3. May 6, 2008 #2
    Oh, right.. because the charge inside our gaussian surface is the charge coming from the "inside-cylinder"...

    I get it now. Thanks
  4. May 6, 2008 #3
    Actually the electric field is not between two cylinders, but it is in between a cylinder and a pipe, so it is not the same case as two parallel capacitors.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Capacitance and electric fields
  1. Electric fields (Replies: 1)

  2. Electric field (Replies: 8)

  3. Electric field (Replies: 1)