Conducting Sphere Floating in Dielectric

[a_h]

Hello fellow physicists!

This is a problem out of my graduate E&M course, and I was hoping some of y'all might have some ideas on it.

We have an uncharged conducting sphere. It is floating half-submerged in a bath of oil (with dielectric constant $$\epsilon$$). Now we put a charge Q on the sphere. The question is, does this cause the sphere to go up or down? How far does it go up or down?

Earlier in the course, we did the problem (out of Jackson, I believe) of two concentric charged shells, with only half of the region between filled with dielectric. We found that the electric field inside the region is radial and independent of $$\theta$$ and $$\phi$$. If we let the outer shell go to infinity, this becomes our problem, and so we can conclude that in our problem the field is radial. This part I am sure about (we talked about it some in class).

From here, I am of the impression that we should use an energy approach; that is, find the energy for a virtual displacement, and determine which sign (up or down) that virtual displacement should be to minimize the energy. I am having problems incorporating this displacement into the energy, though. The energy of the fields I am fine with, using

$$\int_{\mbox{all space}}\!\vec{E}\cdot\vec{D}\,dV\mbox{,}$$

but how does this displacement come into play?

Sorry for the long post, but I hope some of you have some tips.

Thanks,
Austin

 

[JoeZ99]

Austin
I happen to have a slightly different version of your problem. I´m sure the solution would be the same.

Did you, by any chance, find out the solution?? If so, I would really appreciate if you could tell me. I´ve digged the physics forum and found nothing about int.

txs.

 

[kanato]

I think the problem with incorporating the displacement is that you've started with the assumption that you have two concentric shells, but if the floating shell was displaced then they would no longer be concentric. I think you either need to start from scratch and allow for a displacement so the two spheres are not concentric, or you should use a force approach and calculate the force on the sphere to see which direction it will move.

I think there is probably a clever image charge solution to this problem. One charge of magnitude Q inside the sphere with a displacement in the z direction, and one charge of some other magnitude Q' directly below the sphere in the medium. Meet the boundary condition of constant potential on the sphere and you should be able to calculate the force easily.

Edit: oh I forgot; when using image charges I think you will have to use a different image charge distribution for the part of the sphere inside the dielectric than outside. It's been a long time since I've done this.

 

[Born2bwire]

Take the derivative of the energy with respect to the direction of force in terms of the spatial coordinates representing the position of the object. Basically I think it's related as
$$\mathbf{F} = -\nabla E$$
We do this with the Casimir force calculation. We calculate the energy of the system with respect to our desired configuration and take the derivative with respect to the properties describing a shift in the position of the desired object. An alternative though would be to use the Maxwell Stress Tensor. I think Jackson talks about that too but the problem is that I think you need to integrate the tensor over the surface of the object. It's another way I have seen used for Casimir force but I have never used it myself, I always took the derivative of the energy.

 

[Bob S]

Look at pgs 59-60 in Smythe "Static and Dynamic Electricity", 3rd edition. This is conceptually similar to two conducting plates of area A, with an air-gap separation d, and with charge +/-Q on each plate.
Bob S

 

[JoeZ99]

I'm finally going for the stress-tensor approach comments, amendments welcome.

The expression form the stress-tensor. given a l.h.i media with permitivity $$\epsilon$$ is:

$$T_{ij} = \frac{1}{2}\epsilon \left( E_i E_j - E^2 \delta_{ij} \right)$$

and it's related to the force density by:

$$\vec f = \vec \nabla \tilde T$$
$$\int{\vec f dV} = \oint {\tilde T \vec n dS}$$

So, in the sphere we look for the z component of the force:
$$F_z = \int_{S_I} {T_{3i}n_i dS} + \int_{S_{II}}{T_{3i}n_i dS}$$
$$F_z = \int_{S_I}{\frac{\epsilon_0}{2}(E_3 E_n - E^2 n_3)} + \int_{S_II}{\frac{\epsilon}{2}(E_3 E_n - E^2 n_3)}$$

knowing that
$$E_n = E_r cos\theta$$
and , since $$cos \theta < 0$$ if $$\theta < \pi / 2$$ , the region II part of the force substracts from the region I part.

Then we just need to force an equilibrium between electrostatic force, gravitational and the "archimedes law" force (sorry, I don't know the english term for that).

again, comments and amendments welcome / expected.

 

[Magnesium]

Ignore this post.

 

[Magnesium]

Austin
I happen to have a slightly different version of your problem. I´m sure the solution would be the same.

Did you, by any chance, find out the solution?? If so, I would really appreciate if you could tell me. I´ve digged the physics forum and found nothing about int.

txs.

(I'm the same Austin...just changed my user name.)

Heh...this thread sure got busy all of a sudden! :tongue2:

I never thought about the image solution...that might very well be the easiest. But here's what I did:

From that Jackson problem I mentioned (two concentric spheres, half filled with dielectric), we can find that the electric field is radial, and equal to:

$$E=\frac{Q}{2\pi \epsilon_0 (1+\epsilon_r)r^2}$$

Also, if we divide up the surface of the sphere into little dS's, we have, at each surface element:

$$d\vec F = \frac{\sigma^2}{2\epsilon_0} d\vec S$$

Finally, we use:

$$\sigma_{free}=D_n=\left\{ \begin{array}{lr} \epsilon_0 E & : \mbox{Portion out of Oil}\\ \epsilon_0 \epsilon_r E & : \mbox{Portion submerged} \end{array} \right.$$

Then we can integrate $$d\vec F \cdot \hat z$$ over the surface (if z points normal to the oil surface). You'll get something proportional to

$$1-\left(\frac{\Delta x}{a}\right)^2$$.

Equating that to the buoyancy force (from Archimedes' Principle) gives you a cubic equation, that you can numerically solve.

If anyone has any other questions, I'll be happy to provide more details.