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Capacitor, Dielectrics and height movement

  1. Oct 21, 2008 #1

    TFM

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    1. The problem statement, all variables and given/known data

    Two long coaxial cylindrical metal tubes (inner radius a, outer radius b) stand vertically in a tank of (dielectric) oil (suceptibility X-e, mass density rho). the inner one is maintained at potential V, the outer one is grounded. to what height (h) does the oil rise between the tubes.

    2. Relevant equations

    F = (1/2) V squared dc/dx

    x = (v squared C)/(2F)

    C = epsilon-r epsilon-0 A/d

    Q = CV

    3. The attempt at a solution

    I'm not quite sure where to start - I've put a few equations that I believe may be relevant.

    TFM
     
  2. jcsd
  3. Oct 22, 2008 #2

    alphysicist

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    Hi TFM,

    I'm not sure if this equation would be correct here. Where did this come from?

    This would not be true. This is the formula for the capacitance of a parallel plate capacitor; you need the capacitance formula for a cylindrical capacitor.

    The oil will rise until it is in equlibrium. By setting the sum of the forces equal to zero, I think you should be able to get the height.
     
  4. Oct 22, 2008 #3

    TFM

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    I rearranged

    F = (1/2) V squared dc/dx

    to give

    Fdx = (1/2) V squared dc

    which I presumed would give

    F*x = (1/2) V squared C

    and then rearranged?

    TFM
     
  5. Oct 22, 2008 #4

    TFM

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    Is the formula for a cylindrical capacitor:

    C/L = (2*pi*k*epsilon-0)/(ln(b/a))

    [tex] \frac{C}{L} = \frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})} [/tex]

    where L is the lenght of the capacitor?

    TFM
     
  6. Oct 22, 2008 #5

    alphysicist

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    I don't believe this is valid. Intead, I think you need to leave it in the original form, find an expression for the capacitance, and once you take the deriviative you will have the force.

    The bottom equation looks fine to me:

    [tex] \frac{C}{L} = \frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})} [/tex]

    for calculating the capacitance where there is no oil.

    The top one:

    C/L = (2*pi*k*epsilon-0)/(ln(b/a))

    is right and you'll need it to calculate the capacitance in the area where there is oil. However, you have k (the dielectric constant) in the equation, but in the problem statement they give the susceptibility [itex]\chi_e[/itex]. How are these related? You probably want to rewrite this with the susceptibility instead of the dielectric constant?
     
  7. Oct 22, 2008 #6

    TFM

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    So I need to use:

    [tex] \frac{C}{L} = \frac{2*\pi*k*\epsilon_0}{ln\frac{b}{a}} [/tex]

    and

    [tex]X_e = \epsilon_r - 1[/tex]

    I'm assuming k = [tex] \epsilon_r [/tex]

    Where
     
  8. Oct 22, 2008 #7

    alphysicist

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    Yes, that's true in this case, and so the capacitance is:

    [tex]
    \frac{C}{L} = (1+\chi_e) \frac{2*\pi*\epsilon_0}{ln\frac{b}{a}} [/tex]
     
  9. Oct 22, 2008 #8

    TFM

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    So the Capacitance is:

    [tex] C= L(1+\chi_e) \frac{2*\pi*\epsilon_0}{ln\frac{b}{a}} [/tex]


    So do I now somehow insert it into:

    [tex] F = \frac{1}{2}V^2 \frac{dc}{dx} [/tex]

    ???

    TFM
     
  10. Oct 22, 2008 #9

    alphysicist

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    You need to find the formula for the capacitance of this capacitor (for example, say it has length L, and the oil has risen to a height x), and then the derivative dC/dx is what goes in the force formula.

    To find the capacitance, remember that oil is only in part of the capacitor, so you'll need to treat them separately and then combine them.

    Once you find this force, this will be the electric force pulling upwards. What does this have to equal, for the oil to be at an equilibrium?
     
  11. Oct 22, 2008 #10

    TFM

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    So the capacitance for the part above the oil is

    [tex]
    C = l_1\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})}
    [/tex]

    and the part for the capacitor in the oil is:


    [tex]
    C= L_2(1+\chi_e) \frac{2*\pi*\epsilon_0}{ln\frac{b}{a}}
    [/tex]

    TFM
     
  12. Oct 22, 2008 #11

    TFM

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    So if you combine the two, do you get:

    [tex] C = (l_1\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})}) + (L_2(1+\chi_e) \frac{2*\pi*\epsilon_0}{ln\frac{b}{a}}) [/tex]

    ??

    TFM
     
  13. Oct 22, 2008 #12

    alphysicist

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    That's the right idea; but remember that you will have to take the derivative of this with respect to the height (h in the problem statement, x in your formula), so you need to write L1 and L2 in terms of the height. (Hint: One of them is just equal to the height.)
     
  14. Oct 23, 2008 #13

    TFM

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    So it should be:

    [tex]
    C = (hx\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})}) + (h(1+\chi_e) \frac{2*\pi*\epsilon_0}{ln\frac{b}{a}})
    [/tex]

    Since I assume that the part in oil is the height h, and the part out of oil is some unknwon part of h, which I have called x?

    ??

    TFM
     
  15. Oct 23, 2008 #14

    alphysicist

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    Instead of using hx as the part out of oil, call then entire length of the capacitor something like L. Then if the length of the capacitor is L, and the length in oil is h, what is the length out of the oil?
     
  16. Oct 23, 2008 #15

    TFM

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    So say the capcitor is length L,

    Capcitor in Oil is h

    Part of Capacitor out of water is L-h

    so:

    [tex] C = ((L-h)\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})}) + (h(1+\chi_e) \frac{2*\pi*\epsilon_0}{ln\frac{b}{a}}) [/tex]

    TFM
     
  17. Oct 23, 2008 #16

    alphysicist

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    That looks right to me. The derivative of that with respect to h is what appears in the force formula, so you can find the force.

    Then this force has to be cancelled by what other force at equilibrium? Find out what the value of this other force is for the oil at height h, and set the forces equal to each other, and you can find h.
     
  18. Oct 23, 2008 #17

    TFM

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    Right then, so we have the formula:

    [tex] F = \frac{1}{2} V squared dc/dx [/tex]

    and we have

    [tex] C = ((L-h)\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})}) + (h(1+\chi_e) \frac{2*\pi*\epsilon_0}{ln\frac{b}{a}}) [/tex]

    I assume we put them together like so,

    [tex] F = \frac{1}{2} V squared \frac{d\left(((L-h)\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})}) + (h(1+\chi_e) \frac{2*\pi*\epsilon_0}{ln\frac{b}{a}})\right)}{dh} [/tex]

    and I need to find another force which will equal this?

    TFM
     
  19. Oct 23, 2008 #18

    TFM

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    Also, would the other force be mg?

    TFM
     
  20. Oct 23, 2008 #19

    alphysicist

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    That's right (of course when you do the derivative it will be a lot cleaner).

    Yes, but they don't give the mass, they give the density. So you'll need to rewrite m in terms of the density.
     
  21. Oct 23, 2008 #20

    TFM

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    I am assuing that I can break up the derivative:

    [tex] F = \frac{1}{2} V^2 (\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})}(\frac{d(l-h)}{dh}))+(\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})})(\frac{d(h(1+\chi_e))}{dh})) [/tex]

    Does this look right?

    Also, the mass of the oil will be:

    [tex] \rho * \left(\pi*b^2*h - \pi*a^2*h\right) [/tex]

    TFM
     
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