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Bound Charges

  1. Oct 29, 2009 #1
    1. The problem statement, all variables and given/known data
    A conducting wire carrying a charge [tex]\lambda[/tex] per unit length is embedded along the axis of the cylinder of Class-A dielectric. The radius of the wire is a; the radius of the cylinder is b.

    Show that the bound charge on the outer surface of the dielectric is equal to the bound charge on the inner surface, except for sign.


    2. Relevant equations

    [tex]\int \vec{E}\cdot \vec{da}=\frac{Q_{enc}}{\epsilon_0}[/tex]

    [tex]\vec{P}=\epsilon_0 X_{e}\cdot \vec{E}[/tex]

    [tex]\sigma_{b}=\vec{P}\cdot \hat{n}[/tex]

    3. The attempt at a solution

    Using Gauss' Law, we get that [tex]\vec{E}=\frac{\lambda}{2\pi r} \cdot \hat{r}[/tex]

    Therefore, [tex]\vec{P}=\frac{\epsilon_0 X_{e}\lambda}{2\pi r} \cdot \hat{r}[/tex]

    Therefore, [tex]\sigma_{b}=\frac{\epsilon_0 X_{e}\lambda}{2\pi b}[/tex] on the outer surface and [tex]\sigma_{b}=-\frac{\epsilon_0 X_{e}\lambda}{2\pi a}[/tex] on the inner surface....

    These are not equal in magnitude!!! Can someone explain where I went wrong?
     
    Last edited: Oct 29, 2009
  2. jcsd
  3. Oct 29, 2009 #2

    ehild

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    [tex]
    \sigma_{b}=\vec{P}\cdot \hat{n}
    [/tex]

    is the surface charge density. Multiply it with the inner and outer surface to get the bound charge.

    ehild
     
  4. Oct 29, 2009 #3
    Ahhh!! Thank you!!! :)
     
  5. Oct 29, 2009 #4

    gabbagabbahey

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    No, Gauss' Law for [itex]\textbf{E}[/itex] involves the total (free and bound) charge enclosed by the Gaussian surface. Since you don't know what the bound charges are, you'll want to use Gauss' law for [itex]\textbf{D}[/itex] instead.
     
  6. Oct 29, 2009 #5
    Another valid point

    [tex]\vec{D}=\frac{\lambda}{2\pi r} \cdot \hat{r}[/tex]

    and therefore, [tex]\vec{E}=\frac{\lambda}{2\pi \epsilon\cdot r} \cdot \hat{r}[/tex]
     
  7. Oct 29, 2009 #6

    gabbagabbahey

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    That's better.:approve:

    P.S. To write [itex]\chi[/itex] in [itex]\LaTeX[/itex], use \chi
     
  8. Oct 29, 2009 #7
    Cool and thanks for your help
     
  9. Oct 29, 2009 #8
    Quick final question:

    For the bound charge (on both the inner and outer surface) final answer I get [tex]Q_{bound}=\frac{\epsilon_0\cdot \chi_{e}}{\epsilon}\cdot \lambda L[/tex]

    but doesn't [tex]\lambda L=Q_{free}[/tex] , where [tex]Q_{free}[/tex] is the free charge of the configuration?
     
  10. Oct 29, 2009 #9

    gabbagabbahey

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    Isn't the length of the cylinder and wire infinite?
     
  11. Oct 29, 2009 #10
    Doesn't say that it is so I just assumed it was length L
     
  12. Oct 29, 2009 #11

    gabbagabbahey

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    But if the length is finite, the field isn't cylindrically symmetric and you can't use Gauss' Law...
     
  13. Oct 29, 2009 #12
    urgh... so should I assume that its finite or assume that it is infinite? lol... is there some way of calculating the inner and outer surface bound charge density so that it doesn't matter?
     
  14. Oct 29, 2009 #13

    gabbagabbahey

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    I'd assume it's infinite, and calculate the bound charge per unit length...
     
  15. Oct 29, 2009 #14
    sounds good to me, I'm actually asked later to calculate the net charge per unit length so I'll need that.

    And since we are on the topic... :) to get the net charge per unit length at r=a, [tex]\frac{Q_{net}}{L}=\frac{Q_{free}}{L}+\frac{Q_{bound}}{L}[/tex]? I'm told that the answer is [tex]\frac{Q_{net}}{L}=\frac{\lambda}{\epsilon_r}[/tex].


    However...

    With [tex]\frac{Q_{free}}{L}=\lambda[/tex]

    and...

    [tex]\frac{Q_{bound}}{L}=\frac{\epsilon_0\chi_{e}}{\epsilon}\cdot \lambda=\left(1-\frac{1}{\epsilon_{r}}\right)\cdot \lambda[/tex]

    the only way I can get the given answer is if I subtract the net bound charge per unit length from the net free charge per unit length... what am I missing?
     
  16. Oct 29, 2009 #15

    gabbagabbahey

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    I thought the bound charge at [itex]r=a[/itex] was negative...:wink:
     
  17. Oct 29, 2009 #16
    That's it!! I always miss the small things, thanks
     
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