# Rising fluid between conducting cylinders

1. Aug 19, 2013

### monnomestalex

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

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 $h$.

2. Relevant equations

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

3. 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.

2. Aug 19, 2013

### huelsnitz

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

3. Aug 20, 2013

### monnomestalex

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.

Last edited: Aug 20, 2013
4. Aug 20, 2013

### TSny

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

5. Aug 20, 2013

### monnomestalex

That's exactly it - thank you!