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Boundary Conditions - Cylinder in dielectric

  1. May 9, 2014 #1
    1. The problem statement, all variables and given/known data

    2ypkjkz.png

    Part (a): List the boundary conditions
    Part (b): Show the relation for potential is:
    Part (c): Find Potential everywhere.
    Part (d): With a surface charge, where does the Electric field disappear?

    2. Relevant equations



    3. The attempt at a solution

    Part (a)

    Boundary conditions are:
    [tex]E_{in}^{||} = E_{out}^{||}[/tex]

    [tex]D_{out}^{\perp} - D_{in}^{\perp} = \frac{\sigma}{\epsilon_0}[/tex]

    2jd3yae.png

    Part (c)

    First we find potential inside and outside.

    1. For potential inside ##V_{in}##, it is finite when r = 0, so ##b_n = 0##. When r = 0, ##V_{in} = 0##, so ##a_0 = b_0 = 0##.

    2. For potential outside ##V_{out}##, as r → ∞, ##V \propto \frac{1}{r}## (From part (b)). So only term n = 1 exists.

    3. For n=1, and ##\phi = \frac{\pi}{2}##, ##V_{in} = V_{out} = 0##, so sine term goes away: ##c_n = 0##.

    Now we have the two potentials:

    [tex]V_{in} = a_1 r cos(\phi)[/tex]
    [tex]V_{out} = \frac{d_1}{r} cos(\phi)[/tex]

    Applying first boundary condition at r = a,

    [tex]\frac{\partial V_{in}}{\partial \theta} = \frac{\partial V_{out}}{\partial \theta}[/tex]

    [tex] a_1 a^2 = d_1[/tex]

    Applying second boundary condition at r = a,

    [tex]D_{out}^{\perp} = D_{in}^{\perp} [/tex]
    [tex]\epsilon_0 \frac{\partial V_{in}}{\partial r} = \frac{\partial V_{out}}{\partial r}[/tex]
    [tex] \epsilon_0 a_1 = -\frac{\epsilon d_1}{a^2}[/tex]

    Solving them, this implies ##\epsilon_0 + \epsilon = 0##, which doesn't make sense.
     
  2. jcsd
  3. May 11, 2014 #2
    I've got a feeling that my potential inside is missing a term like the one in part (b): ##V = \frac{\lambda}{2\pi \epsilon_0}\frac{d cos \phi}{r}##

    Because it shouldn't strictly depend only on ##cos \phi##?
     
  4. May 18, 2014 #3
    I think the potential inside should be something like:

    [tex]V_{in} = Ar cos \phi + \frac{B}{r^2}cos \phi[/tex]

    That should solve it.
     
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