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!

Polarized cylinder (again)

  1. May 11, 2007 #1
    1. The problem statement, all variables and given/known data
    Suppose we have a solid cylinder of radius a and length L and carries a uniform polarization paralell to its axis. Find the electric field everywhere.

    Suppose we have a cylinderical cavity inside a large dielectric material which has electric displacement D0 and electric field E0 and polarziation P such taht
    [tex] \vec{D} = \epsilon_{0} \vec{E_{0}} + \vec{P} [/tex]. Then find the electric field and electric displacement within the cyldrical cavity
    2. Relevant equations
    Bound surface charge
    [tex] \sigma_{b} = \vec{P}\cdot \hat{n} [/tex]

    Bound volume charge
    [tex] \rho_{b} = - \vec{\nabla} \cdot \vec{P} [/tex]

    3. The attempt at a solution
    uniform polairztaion so there is no bound volume charge
    Since the polarization is parallel to the axis there is no surface charge on teh curved part of the cylinder. However there is a bound charge on the caps on the cylinder.

    If the cylinder is aligned parallel to the Z axis and the polaraation pointsi nteh Z axis then the lower part of the top cap is [itex] -P [/itex]

    so now we have two circular caps which are parallel with opposite charge densities. SO the electric field is just like that of a parallel plates each wit ha cahrge density of [tex] P [/tex]

    sp the electric field inside if [tex] E = \frac{P}{\epsilon_{0}} (-\hat{z}) [/tex]

    The electric field outside... Well there is no free charge so the electric displacement is zero
    [tex] D = \epsilon_{0} E + P [/tex]
    [tex] 0 = \epsilon_{0} E + P [/tex]
    [tex] E= \frac{P}{\epsilon_{0}} (\hat{-z}) [/tex]

    is this correct??

    Now for the part where this cylinder is the cavity. We assume that the polarization of the cylindrical cavity is opposite to that of the dioelectric
    THe electric field inside the cylinder is thus [tex] E = \frac{P}{\epsilon_{0}} \hat{z} [/tex]
    the average electric field inside the cylinder is thus

    [tex] E = (E_{0} + \frac{P}{\epsilon_{0}}) \hat{z} [/tex]

    The electric displacement is [tex] D = \epsilon_{0}E + P = \epsilon_{0}(E_{0} + \frac{P}{\epsilon_{0}}) \hat{z} -P = \epsilon_{0} E_{0} [/tex]

    is that correct??
    thanks for the help
    Last edited: May 11, 2007
  2. jcsd
  3. May 12, 2007 #2


    User Avatar
    Homework Helper
    Gold Member

    If the cavity is very small, you could treat is a parallel plate. Else, you could treat is as two discs.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Polarized cylinder (again)
  1. Polarized cylinder (Replies: 0)

  2. Polarized cylinder (Replies: 2)

  3. Cylinder with piston (Replies: 16)