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!

Uniformly polarized cylinder

  1. Sep 25, 2006 #1
    Hi all,

    I'd like to get your opinion on the validity of my working for the following problem. According to my lecturer the answer is incorrect, but I can't figure out where I'm going wrong.

    A cylinder of radius a is uniformly polarized in a direction perpendicular to the axis. The magnitude of the dipole moment per unit volume is P. Determine the electric field on the axis of the cylinder.

    To me this seems like a straightforward application of Gauss' law, namely

    [itex]
    \int_{\mathrm{rectangle}}\vec{E} {\cdot} d \vec{A} + \int_{\textrm{half-cylinder}}\vec{E} {\cdot} d \vec{A} = \frac{2aP\ell}{\varepsilon_0}
    [/itex]

    [itex]
    E (2a \ell) + \int_{-\pi/2}^{+\pi/2}E\cos\theta a \ell d\theta = \frac{2aP\ell}{\varepsilon_0} \Rightarrow
    [/itex]

    [itex]
    E = \frac{P}{2\varepsilon_0}
    [/itex]

    answer should by P / (4 epsilon).

    Thanks

    James
     
  2. jcsd
  3. Sep 25, 2006 #2

    Meir Achuz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Your use of Gauss is wrong.
    Dimensionally, the answer must be proportional to P/epsilonnaught,
    so any mistake will look close to right.
    You have to integrate Coulomb's law for the bound surface charge.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Uniformly polarized cylinder
  1. Polarized cylinder (Replies: 0)

  2. Polarized cylinder (Replies: 2)

Loading...