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Dielectric Cylinder Polarization (Electrostatics)

  1. Apr 27, 2015 #1
    1. The problem statement, all variables and given/known data

    An infinitely long dielectric cylinder with radius R and relative permittivity ##\epsilon_r## contains a free charge density given by:

    ##\rho(s)= ks## for s<R and 0 for s>R, where k is a constant.

    Find the polarization P and any volume or surface polarization charge densities.

    2. Relevant equations

    ##D=\epsilon_0 E+P##

    ##P=\epsilon_0(\epsilon_r -1)E##

    ##\oint D.da = Q_{fenc}##

    ##D=\epsilon E##

    Gauss's law for E fields

    ##\epsilon_r = \epsilon/\epsilon_0##

    ##\rho = \rho_{bound} + \rho_{free}## with ##\rho_{bound}=- \nabla. P## and ##\nabla. D = \rho_{free}##

    3. The attempt at a solution

    Using Gauss's law we find:

    ##E_{in} = \frac{ks^2}{3 \epsilon_0} \hat{s}##
    Then we find the electric displacement D:

    ##\oint D.da = D 2 \pi s L = (ks) \pi s^2 L = Q_{fenc}##

    ##Q_{fenc} = 2\pi k l \int^s_0 s'^2 = \frac{2}{3} \pi kls^3##

    ##D= \frac{ks^2}{3} \hat{s}##
    If I use D to find E we have: ##D=\epsilon E \implies E=\frac{ks^2}{3\epsilon_0\epsilon_r}##

    Now using to find polarization I have used the E found from D instead of the E found from Gauss's law (please correct me if I am wrong):

    ##P=D-\epsilon_0 E = \frac{ks^2}{3}(1- \frac{1}{\epsilon_r})##

    Is this the correct approach?

    I believe ##\rho_{free}=0## since it is a dielectric. I'm guessing we need to have surface charges ##\sigma_{bound}=+P, \ \sigma_{bound}=-P##, but since the cylinder is infinitely long, where would be the two ends in which they are located? :confused:

    And for the bound volume charge density I tried to find the gradient:

    ##\nabla . P = \frac{1}{s} \frac{\partial}{\partial s} (\frac{ks^3}{3} (1-\frac{1}{\epsilon_r})) = ks(1-\frac{1}{\epsilon_r})##

    Is this correct? Also, shouldn't the net polarization charge per unit length be zero? How do we know (or can verify) this?

    Any help is greatly appreciated.
     
    Last edited: Apr 27, 2015
  2. jcsd
  3. Apr 27, 2015 #2
    Hello
    Well done , you find E , D and P correctly .
    As you say that we have free charge density in the dielectric it means we inject free charge and it should be non-zero value because of gif.latex?%5Cint%20pdv%3DQ_%7Bfree%7D.gif
    we have surface polarization charges on s=R location that can be obtain with gif.latex?%5Crho%20%3D%5Cvec%7BP%7D.%5Chat%7Ban%7D%3D%5Cvec%7BP%7D.%5Chat%7Bas%7D.gif and for volume charge we have gif.latex?%5Crho%20%3D-%5Cbigtriangledown%20.%5Cvec%7BP%7D.gif that you may correct it with multiply it by -1 .

    you can verify your answer with knowledge of atex?-%5Cbigtriangledown%20.%5Cvec%7BP%7D&plus;%5Cvec%7BP%7D.%5Cvec%7Ban%7D%28all-surface%29%3D0.gif
     
    Last edited by a moderator: Apr 17, 2017
  4. Apr 28, 2015 #3
    Hello!

    Thank you for the input.

    In your equation for ##\rho_{pol}##, what do "a" and "n" represent?

    So for the surface charge density we basically equate that to the magnitude of polarization vector P, and then substitute R into the equation for P?

    This is what I did:

    ##\sigma_{pol}=P=\frac{kR^2}{3} (1- \frac{1}{\epsilon_r})##​


    What unit vector do you use to indicate the component of P perpendicular to the surface? (it's not ##\hat{s}## since it's not radial)

    Now, for ##\rho_{pol}## I have added the minus sign to get: ##\rho_{pol}=- \nabla . P = -ks(1- 1/\epsilon_r)##.

    I didn't understand what your last equation was, could you please type it more clearly? What does "all" and "surface" mean? And how does that prove the net polarization charge per unit length is zero?
     
  5. Apr 29, 2015 #4
    How do I verify that the net polarization charge per unit length is zero? Do I need to show that ##Q_{tot} = (\sigma_{bound} \times Area) + (\rho_{bound} \times Volume)=0##?

    Since ##\rho_{bound} = - \nabla . P = \frac{-2ks}{3} (1-\frac{1}{\epsilon_r}) \implies \frac{-2kR}{3} (1-\frac{1}{\epsilon_r})## we have:

    ##\therefore \frac{-2kR}{3}(1-\frac{1}{\epsilon_r}) \pi s^2 h + \frac{ks^2}{3} (1-\frac{1}{\epsilon_r}) 2 \pi sh \neq 0##

    This is only equal to zero if I substitute R into all the s variables. But I thought this should only be done with ##\sigma_{bound}##. So what should I do?
     
    Last edited: Apr 29, 2015
  6. May 4, 2015 #5
    Hello
    Sorry I response late ( I had & have different exams)
    "an" is Perpendicular unit vector on your surface.as you see in this picture each surface has it's "Perpendicular unit vector" that you should calculate t%257Ban%257D%253D%255Cvec%257BP%257D.%255Chat%257Bas%257D&hash=07d4f4746d868669af314d575488f211.gif for each of them.
    7305655600_1430762731.jpg
    In simple medium(with Homogeneous permittivity) that P and E are in the same direction it is easy that you use D=eE+P equation such as in your approach.
    257BP%257D.%255Cvec%257Ban%257D%2528all-surface%2529%253D0&hash=f0e6fa13db3ecef36cd732ba7849d3c6.gif

    for each surface that you calculate t%257Ban%257D%253D%255Cvec%257BP%257D.%255Chat%257Bas%257D&hash=07d4f4746d868669af314d575488f211.gif you should define exact value of (s,phi,z) then summation of all integral of them on their surface and integral of ho%2520%253D-%255Cbigtriangledown%2520.%255Cvec%257BP%257D&hash=98ed42ec3ba7af8fb39b4cd6b809e058.gif in specific volume should be zero ( As I remember it doesn't verify the answer when the length of object is infinite
     
    Last edited by a moderator: Apr 16, 2017
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