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Conductor sphere floating on a dielectric fluid

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1. Homework Statement

A conductor sphere of radius R without charge is floating half-submerged in a liquid with dielectric constant ##\epsilon_{liquid}=\epsilon## and density ##\rho_l##. The upper air can be considered to have a dielectric constant ##\epsilon_{air}=1##. Now an infinitesimal charge ##\delta Q## is added to the sphere. Find how much it will submerge or raise to the lowest order in ##\delta Q##.

2. Homework Equations

Poisson equation

$$\nabla^2 \Phi=\rho / \epsilon_0$$

Boundary conditions

$$\vec{n}\cdot (\vec{D_1}-\vec{D_2})=\rho_s$$

$$\vec{n}\times (\vec{E_1}-\vec{E_2})=0$$

Energy of electromagnetic fields

$$U=\frac{1}{2} \int \vec{D} \cdot \vec{E}$$

3. The Attempt at a Solution

Using Archimedes principle we can easily find the density of the sphere ##\rho_s## before adding the charge:

$$\rho_s (\frac{4}{3}\pi R^3)-\alpha_{before}(\rho_l (\frac{4}{3}\pi R^3))=0$$

Where ##\alpha_{before}=1/2## is the percentage sumerged before adding the charge. Thus:

$$\rho_s=\rho_l$$

After adding ##\delta Q##, it induces a field on the sphere. On Jackson's, there's an exercise which suggests that the electric field in the half-sumerged sphere is radial, hence we can posit:

$$E=E(r)$$

Since the charge is small, we can neglect it and use Laplace's equation for the potential:

$$\nabla^2 \Phi=0$$

And apply the boundary conditions assuming the constitutive equation for an homogeneous, linear, and isotropic liquid:

$$\vec{D}=\epsilon \vec{E}$$

From there I could find the electric and displacement fields ##E## and ##D## in the sphere to find the energy:

$$U(\delta Q)=\frac{1}{2} \int \vec{D} \cdot \vec{E}$$

Now since the energy is related to the force as,

$$\delta F=-\frac{\partial U (\delta Q)}{\partial dz}$$

I could find the force, reintroduce it in the equation of mechanical equilibrium as:

$$\rho_s (\frac{4}{3}\pi R^3)-\alpha_{after}(\rho_l (\frac{4}{3}\pi R^3))+\delta F=0$$

Where now ##\alpha_{after}## is the new percentage that the sphere is sumerged after adding ##\delta Q##. However, I'm stuck on whether this is a good idea or I'm just putting equations that aren't related. Moreover, if I apply the boundary conditions, how do I deal with the part where the sphere is half-sumerged in the liquid? Do I apply a boundary condition on the liquid surface too?

I actually found a similar problem here but it seems it was never properly solved: https://www.physicsforums.com/threads/conducting-sphere-floating-in-dielectric.303125/
 

Answers and Replies

56
5
Well I checked similar procedure and I managed to advance the following:

While I don't know if it's really useful, if we apply mechanical equilibrium before adding the charge, it's straightforward to find that ##\rho_s=\rho_l##, where ##\rho_s## and ##\rho_l## are the volumetric densities of the sphere and the liquid, respectively.

After adding the charge, we can suppose that the sphere moves upwards (the direction doesn't matter, since at the end the sign will determine the correct one). Assuming the charge distributes uniformily over the surface, we can apply Gauss Law to find that:

$$\int_{air} \vec{D}\cdot d\vec{A}+\int_{liquid}\vec{D}\cdot d\vec{A}=\delta Q$$

Letting ##\delta y## be the vertical distance that the sphere moves up (the area is a small ring of perimeter ##2\pi R## and height ##\delta y##), and since the electric field (and thus the displacement) of a submerged sphere is radial:

$$D_{air}(2\pi R^2+2\pi \delta y)+D_{liquid}(2\pi R^2-2\pi \delta y)=\delta Q$$

9QNdR.png


Assuming an isotropic, homogenous, linear dielectric, we can apply the constitutive relationship ##\vec{D}=\epsilon \vec{E}## to find that:

$$\vec{E}=\frac{1}{2\pi a}\frac{\delta Q}{(a+\delta y)+\epsilon(a-\delta y)}$$

And thus:

$$\vec{D_{air}}=\frac{1}{2\pi a}\frac{\delta Q}{(a+\delta y)+\epsilon(a-\delta y)}$$

$$\vec{D_{liquid}}=\frac{\epsilon}{2\pi a}\frac{\delta Q}{(a+\delta y)+\epsilon(a-\delta y)}$$

Now, for these kinds of materials, the energy of the fields is:

$$U_E=\int d^3 r \vec{E}\cdot\vec{D}$$

Substituting:

$$U_E=\frac{(\delta Q)^2}{4\pi^2 a^2}\left ( \int_{air} \frac{d^3r}{((a+\delta y)+\epsilon(a-\delta y))^2} + \epsilon\int_{liquid} \frac{d^3r}{((a+\delta y)+\epsilon(a-\delta y))^2} \right )$$

To first order, we can simplify the denominator since ##\delta y## is small:

$$((a+\delta y)+\epsilon(a-\delta y))^2=a^2(1+3\epsilon)+2a\delta y(1-\epsilon^2)$$

Therefore:

$$U_E=\frac{(\delta Q)^2}{4\pi^2 a^2}\left ( \int_{air} \frac{d^3r}{a^2(1+3\epsilon)+2a\delta y(1-\epsilon^2)} + \epsilon\int_{liquid} \frac{d^3r}{a^2(1+3\epsilon)+2a\delta y(1-\epsilon^2)} \right )$$

Now, if we could just express ##\delta y## as a function of spherical coordinates we could perform the integral, so I got stuck here.

However, assuming we can do the above integral, we can easily find the force as:

$$F_E=-\left ( \frac{\partial U_E}{\partial R} \right )_{\delta Q}$$

And perform a new mechanical equilibrium equation:

$$\rho_l V_{submerged} g - \rho_s V_{total} g +F_E=0$$

And theoretically find the submerged volume to find the percentage that the sphere moved after adding the charge (don't forget we calculated ##\rho_s=\rho_l/2##).

However, I'm not sure if my procedure is correct, or how to perform the integral. Do you have any suggestions?
 

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