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Calculating the height of dielectric oil between two coaxial cylindrical metal tubes

  1. Oct 30, 2012 #1
    1. The problem statement, all variables and given/known data

    Fromm Griffiths
    Two long coaxial cylindrical metal tubes (inner radius a, outer radius b) stand vertically in a tank of dielectric oil (susceptibility [itex]\chi_{epsilon}[/itex], mass density [itex]\rho[/itex]. The inner one is maintained at potential V, and the outer one is grounded. To what height (h) does the oil rise in the space between the tubes?

    So E=0 when r<a. And the potential from a to b is -V, right?

    2. Relevant equations

    [itex]F=-\frac{dW}{dx}[/itex]
    [itex]-\frac{dW}{dx}=\frac{1}{2}V^{2}\frac{dC}{dx}[/itex]
    [itex]\frac{dC}{dx}=-\frac{\epsilon_{0} \chi_{e} \omega}{d}[/itex]
    [itex]F_{gravity}=F_{electric}[/itex] on the dielectric oil
    and [itex]F_{electric}=F_{air}[/itex] where the latter is the electric force in the air above the oil.

    3. The attempt at a solution

    There are two separate sections between the tubes, one air filled and the other oil filled.

    C in the air section is [itex]C= \frac{\epsilon_{0} A}{d} = \frac{2 \pi (b-a) (l-h)}{b-a} = 2 \pi (l-h) [/itex]

    In the air, [itex]W=\frac{1}{2} CV^{2}[/itex].
    [itex]F=-\frac{dW}{dx}=\frac{1}{2}V^{2} -\frac{\epsilon_{0} \chi_{e} \omega}{d} [/itex], where [itex] \omega= 2 \pi (b-a), F=\frac{1}{2}V^{2} -\epsilon_{0} \chi_{e} 2 \pi [/itex]

    I think I'm going wrong somewhere, but ideally with the capacitace and voltage you can calculate forces and then use mass density to find the height at which the forces are equivalent.
     
  2. jcsd
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