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Polarized cylinder

  1. May 10, 2007 #1
    1. The problem statement, all variables and given/known data
    Griffiths Problem 4.13
    A very long cylinder, of radius a, carries a uniform polarization P perpendicular to its axis. Find the electric field inside the cylinder. Show that the field outside the cylinder can be expressed in the form
    [tex] \vec{E}(\vec{r}) = \frac{a^2}{2\epsilon_{0}s^2} \left[2(\vec{P}\cdot\vec{\hat{s}})\hat{s}-\vec{P}\right][/tex]


    2. The attempt at a solution

    Now i was wondering if i could use Laplace's equation here... since there is not volume charge.

    If i were to use this equation to solve for the potnetial

    [tex]V(\vec{r}) = \frac{1}{4 \pi\epsilon_{0}} \oint_{S} \frac{\sigma_{b}}{r} da' + \int_{V} \frac{\rho_{b}}{r} d\tau'[/tex]

    Suppose we were solving for points inside then bound surface charge is zero and the volume charge cannot be zero

    SInce the polarization is perpendicular to the axis, then the polarization is +P for theta from 0 to pi
    but for theta from pi to 2pi then the polarization is -P correct??

    But how can the bound volume charge be non zero if the polarization is constant??
     
  2. jcsd
  3. May 10, 2007 #2
    well i tried to solve it using Laplace's equation

    the solution in cylindrical coords is

    [tex] V(s,\phi) = \sum_{k=0}^{\infty} (A_{k}s^k+B_{k}s^{-k}) (C_{k}\cos k\phi + D_{k} \sin k\phi) [/tex]

    Since [itex] \sigma_{b} = P\cos\phi [/itex]

    Obviosuly Dk is zero

    For the inside we need the potential to be finite so
    [tex] V_{in} (s,\phi) = \sum_{k=0)}^{\infty} A_{k} s^k \cos k\phi [/tex]

    [tex] V_{out} (s,\phi} = \sum_{k=0}^{\infty} B_{k} s^{-k} \cos k\phi [/tex]

    We need [tex] V_{in} (a,\phi) = V_{out} (a,\phi) [/tex]
    Using the above conidtion we get [tex] A_{k} a^{2k} = B_{k} [/tex]

    we also need the potnetial to be smooth across the boundary so
    [tex] \frac{d V_{out}}{ds}\right|_{s=a} - \frac{d V_{in}}{ds}\right|_{s=a} = \frac{-\sigma}{\epsilon_{0}} [/tex]

    [tex] \sum_{k=0}^{\infty} \left(-kB_{k} a^{-k-1} - kA_{k}s^{k-1}\right) \cos k\phi = \frac{\sigma}{\epsilon_{0}} = \frac{P\cos \phi}{\epsilon_{0}} [/tex]

    obviously k =1

    and we get [tex] A = \frac{P}{2\epsilon_{0}} [/tex]

    so i get
    [tex] V_{in} (s,\phi) = \frac{P}{2\epsilon_{0}} s \cos \phi [/tex]

    [tex] E = - \nabla V = -\frac{P}{2\epsilon_{0}} (\cos \phi \hat{s}-\sin\phi \hat{\phi} [/tex]


    the answer is supposed to be uniform though.. not have a phi dependance...

    where di i go wrong?

    Is there a way to do this using Gauss Law because after all the bound charge can be treated like a real charge right??
     
    Last edited: May 10, 2007
  4. May 11, 2007 #3

    Meir Achuz

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    Gold Member

    Because you are in 2D, div E=0 inside the cylinder, so the only bound charge is a surface charge sigma~P.s. Use your first equation to apply the BCs that
    V is continuous and Delta E~sigma.
     
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