Rising fluid between conducting cylinders

AI Thread Summary
The discussion revolves around calculating the height of fluid that rises between two coaxial conducting cylinders when a voltage is applied. Participants emphasize understanding the physics behind the phenomenon, particularly the behavior of capacitors in series and parallel configurations. The dielectric constant's role in affecting the electric field and charge distribution is highlighted, with concerns about the assumption of even charge distribution in the presence of a dielectric. A connection is made to the problem of calculating the force on a dielectric slab in a capacitor, which aids in understanding the fluid's behavior. The conversation underscores the importance of visualizing the setup and applying relevant equations to derive the solution for the height of the fluid.
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Homework Statement



Two long, hollow, and coaxial conducting cylinders, with radii a and b>a, are lowered into a tub of fluid with dielectric constant \kappa. A voltage V is applied between the two cylinders. The fluid is observed to rise up some height h into the volume between the cylinders. Calculate h.

Homework Equations



The dielectric constant is \kappa = 1 + \chi_e = \epsilon/\epsilon_0 in linear media.

The Attempt at a Solution



I am not sure I understand the physics behind the phenomena. This is a PhD quals question that requires only undergraduate E&M knowledge, so it shouldn't be too farfetched, but I do not know where to start.
 
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Hi,

First thing you should do is make a sketch of what is going on. Then, write down the relevant equations.

If you just had the two cylinders in air, with voltage V between them, you would have a capacitor of length L, where L is the total length of the cylinders. Now, part of the cylinder (length h) is in the fluid, and length L-h is in air. These two capacitors are in series with each other, both at voltage V. By picturing it that way, and putting your variables into the appropriate equations, you should be able to solve for h in terms of the other variables.

Warren
 
I'm still not sure why the fluid will rise once you apply the external voltage.

For capacitors in series, \frac{1}{C_{eq}} = \frac{V_1+V_2}{Q} = \frac{1}{C1} + \frac{1}{C_2}.

On one hand, assuming V(b) = 0 and V(a) = V, I get that V(r) = V \frac{r-b}{a-b} solves the Laplace equation, at least in the case where there's no dielectric. On the other hand, we know that the dielectric constant relates D and E through D = \epsilon E = \epsilon_0 \kappa E.

I also assumed that Q would distribute itself evenly and give an electric field E = \frac{\lambda}{2 \pi \epsilon r} and therefore V(a)-V(b) = V = \frac{\lambda}{2 \pi \epsilon} \log(b/a). However, that assumption of evenly distributed charges in the presence of the dielectric bugs me.

That's as far as I have gone. I know I'm missing something, i.e. what's the conserved quantity that I can use to relate the situation before and after the fluid rises. I can only think that the charges on the conductors will be conserved, but I don't know how to apply that idea.
 
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The two capacitors are actually in parallel, not series.

You might be able to relate this problem to the fairly well-known problem of calculating the force on a dielectric slab that is partially inserted between the plates of a parallel plate capacitor. See for example
http://www.pas.rochester.edu/~dmw/phy217/Lectures/Lect_25b.pdf
 
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That's exactly it - thank you!
 
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