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Dielectric between cylinders

  1. Nov 5, 2013 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    A long cable consists of two coaxial conducting cylindrical shells of radius b and 3b. The region with radius ##\delta## between b and 2b is filled with a material of relative permittivity ##\epsilon_r \neq 1## and relative permeability ##\mu_r = 1##; the remaining space between the cylinders is empty.
    (a) Suppose the cable carries charge per unit length ##\pm \gamma## on the inner and outer cylinders. Find ##\underline{D}## and ##\underline{E}## everywhere within it, where ##\underline{D}## is the electric displacement vector.
    (ii) Hence calculate the potential difference between the inner and outer shell, and obtain an expression for the capacitance per unit length of the cable.

    2. Relevant equations
    Gauss' Law, ##\underline{D} = \epsilon_o \epsilon_r \underline{E}##, potential on surface of inner cylinder is ##-\int_{\infty}^{b} \underline{E} \cdot ##d##\underline{r}##. Similarly for other cylinder.

    3. The attempt at a solution
    (a)Because of the dielectric in the space ##b < r < 2b##, the E field in the material will go down by a factor of ##\epsilon_r##. In the space ##2b < r < 3b##, the E field is unchanged. So the E field for the region [b,2b] is $$\underline{E_2} = \frac{1}{\epsilon_r} \frac{\gamma}{2 \pi \epsilon_o r_1} \underline{e}_r $$ and that in [2b,3b] is $$\underline{E_1} = \frac{\gamma}{2 \pi \epsilon_o r_2}\underline{e}_r,$$where r1 between b and 2b and r2 between 2b and 3b. ##\underline{D}## is then these expressions multiplied by ##\epsilon_r##.
    (b) Potential diff = potential at inner cylinder - potential at outer cylinder:
    Consider potential at inner cylinder first: $$V_{inner} = -\int_{\infty}^{b} = -\int_{3b}^{\delta}\underline{E} \cdot d\underline{r} - \int_{\delta}^{b}\underline{E} \cdot d\underline{r}$$ where E1 and E2 are the electric fields in the non-dielectric and dielectric areas respectively. Subbing in, I get $$V_{inner} = \frac{\gamma}{2\pi \epsilon_0} \left(\ln\left(\delta^{1/\epsilon_r -1} \right) + \ln(3) \right)$$ The potential of the outer cylinder is $$V_{outer} = -\int_{\infty}^{3b} \underline{E} \cdot d\underline{r} = 0$$ since E is zero outside the cylinder. Hence ##\Delta V = V_{inner}##

    The capacitance/length = γ/ΔV. Is it okay? Many thanks.
     
  2. jcsd
  3. Nov 5, 2013 #2

    rude man

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    Look again at your equation for D. Start with Gauss' theorem.

    Your answer for D (or εE) is dimensionally incorrect, so you know it can't be right.
     
  4. Nov 5, 2013 #3

    CAF123

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    Yes, I meant to say D is E multiplied by ##\epsilon_r \epsilon_0##. In the region [b,2b], I obtain ##\underline{D} = \frac{\gamma}{2 \pi r_1}\underline{e}_r## for b<r1<2b. I am not really sure how to obtain the D vector in the region [2b,3b] using Gauss (i.e how to define the Gaussian surface appropriately). Using D=εrεoE, I get that D = εrγ/2πr2.
     
  5. Nov 5, 2013 #4

    rude man

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    Sorry, I thought gamma was a surface charge. Your expressions for E are all correct.

    Oops, your D is not.

    The nice thing about using D is that D is continuous across the vacuum-dielectric layer. So I suggest you always use D, then convert to E at the last minute, with E = D/ε.
     
    Last edited: Nov 5, 2013
  6. Nov 5, 2013 #5

    CAF123

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    In media, it states $$\oint_S \underline{D} \cdot d\underline{S} = (Q_f)_{enc}$$ with ##Q_f## the free charge (in this case ##\gamma l##, l the length of the cylinder.)
     
  7. Nov 5, 2013 #6

    CAF123

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    Should that not be E = D/εoεr (**)? In the region [b,2b], I get D = γ/2πr1 and in [2b,3b] I get D = εrγ/2πr2. This is just by applying the formula (**). I managed to get the eqn for D in the region [b,2b] via Gauss thm as well, but not the D vector in [2b,3b]. I am not sure how to define my Gauss surface in this case.
     
  8. Nov 5, 2013 #7

    rude man

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    Right. Sorry again.
     
  9. Nov 5, 2013 #8

    rude man

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    Same thing. ε = εrε0.
    You are misapplying the formula. Take out the εr.

    Again: use D and you won't get into this kind of trouble. Change to E at the end:
    E = D/ε where ε = εrε0
    and where εr = 1 in vacuo and >1 in a dielectric.

    I notice you assumed a unit length (1m) for your Gaussian cylinders. That is OK but I prefer to assume a length L, then you can do dimensional checks on all your terms.
     
  10. Nov 5, 2013 #9

    CAF123

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    I don't think I see it yet - given that I have already calculated E let's find D. ##D = εE,## where ##E = \frac{\gamma}{2 \pi \epsilon_o r_2}\,\,, 2b<r_2 < 3b##. Mulitplying this by ##\epsilon## gives ##D = \frac{\epsilon_r \gamma}{2 \pi r_2}##, no?
    EDIT: ##\epsilon_r = 1## in vacuo so remove the ##\epsilon_r##, I see your point. Thanks.

    How so? When I found the E fields via the microscopic form of Gauss, ##E(2\pi r l) = \gamma l## and the l cancelled.
     
  11. Nov 5, 2013 #10

    CAF123

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    Do you agree with the rest of the OP? In particular, my calculations of V and C?
     
  12. Nov 5, 2013 #11

    rude man

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  13. Nov 6, 2013 #12

    CAF123

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    Hi rude man,
    Did you see #10?
     
  14. Nov 6, 2013 #13

    rude man

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    So4rry, did not.

    How on Earth did you come up with the first term below:

    Subbing in, I get $$V_{inner} = \frac{\gamma}{2\pi \epsilon_0} \left(\ln\left(\delta^{1/\epsilon_r -1} \right) + \ln(3) \right)$$.

    δ doesn't even belong in that expression. You're integrating from b to 2b. The two integrations give expressions that are entirely similar, just different constants involved.

    The second term is also not right.

    Check your limits of integration. They are very simple functions of b and do not include δ. The way the problem is stated, δ is a variable radius lying in the dielectric region: b < δ < 2b.
     
  15. Nov 6, 2013 #14

    CAF123

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    Okay, that makes sense and with these corrections I obtain the potential difference between the two cylinders is $$P.D = \frac{\gamma}{2 \pi \epsilon_o} \left(\ln \left(\frac{3}{2}\right) + \frac{\ln(2)}{\epsilon_r}\right) = \frac{\gamma}{2 \pi \epsilon_o} \left( \ln \left(2^{1/\epsilon_r - 1}\right) + \ln(3)\right)$$
     
  16. Nov 6, 2013 #15

    rude man

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    Correct.
     
  17. Nov 6, 2013 #16

    CAF123

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    Is it okay to say the capacitance/length is Q/lΔV = γ/ΔV? I know that the capacitance will be increased because of the dielectric but this will not be by a factor of εr.
     
  18. Nov 6, 2013 #17

    rude man

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    Yes.
     
  19. Nov 6, 2013 #18

    CAF123

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    Many thanks, the last part asks to compute the Poynting vector giving reasons for the direction. It is clear that the Poynting vector points in the ##\underline{e}_z## direction if the E field is in ##\underline{e}_r## and B in ##\underline{e}_{\phi}##. I think this makes sense - the set up is remiscent of a cable and data flow/light/ would travel in this direction, along the wire.
    Is this the comment to be made?

    Also, I don't know whether it makes sense to define two Poynting vectors here. The E field inside and outside the dielectric is different and so I am not sure what one to use in ##\underline{S} = 1/\mu_o \underline{E} \times \underline{B}##
     
  20. Nov 6, 2013 #19

    rude man

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    CAF, put this into the right thread. I need to review the prior posts.

    rude man

    EDIT: never mind, I guess it's the right thread, but please provide the wording for this ensuing question.
    EDIT EDIT: never mind that too!
     
    Last edited: Nov 6, 2013
  21. Nov 6, 2013 #20

    rude man

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    [/quote]

    The Poynting vector always points in the direction of energy flow. So if the cable is end-excited by a time-varying voltage then the Poynting vector points (mostly) in the z direction outside the cable.

    Inside the cable conductors the situation is different. The Poynting vector actually points radially from the outside towards the axis, representing the diminishing flow of heat energy dissipated within the conductor. This is kind of an advanced concept so I would just say that the Poynting vector points basically along the z axis. If the conductivities are infinite then there is no radial internal component and the Poynting vector points exactly along the z axis.
     
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