Paralle plate capacitor, LIH dielectric, fringing field

AI Thread Summary
When a linear, isotropic, and homogeneous dielectric slab is partially inserted into a charged parallel plate capacitor, it experiences a force drawing it into the dielectric. This force is derived from energy considerations, which assume a uniform electric field perpendicular to the dielectric. However, the fringing electric field outside the capacitor, which is not included in the initial energy calculation, plays a crucial role in generating this force. The energy calculation focuses on the uniform field within the plates, allowing the force to be calculated without accounting for the fringing field's influence on the edge of the dielectric. Thus, while the fringing field affects total energy, it does not alter the energy change due to the infinitesimal displacement of the dielectric slab.
Metaleer
Messages
123
Reaction score
0
Hey, all.

If we partially introduce a linear, isotropic and homogeneous dielectric slab in a charged, isolated, parallel plate capacitor, we know that it experiences a forces pulling it into the dielectric, and we can obtain the expression of this force using energy considerations. However, when the energy calculation is done, we assume a uniform E field in the capacitor, always perpendicular to the dielectric, so in theory it looks like the E field can't pull on any charge that's in the dielectric to pull it in, yet the energy calculation still reveals a force. Books then say it's the fringing E field outside the capacitor that's pushing the dielectric in, but this fringing field wasn't taken into account when getting the energy!

What's going on? How does this method self correct itself?

Thanks in advance. :biggrin:
 
Physics news on Phys.org
That is a very good question. In fact, this was carefully addressed in an AJP paper many years ago:

"Force on a Dielectric Slab Inserted into a Parallel-Plate Capacitor", S. Margulies (Am. J. Phys. v.52, p.515 (1984)).

In it, he wrote this:

For example, how can the force act to pull the slab into the volume between the plates when the electric field there is perpendicular to this direction? If this is explained - the force is, of course, due to the fringe field - an apparent paradox arises: How can the virtual-work calculation yield an answer when it is explicitly based on the assumption of a uniform electric field existing only in the region between the plates, and so does not include the fringe field at all?

Sounds familiar? :)

Zz.
 
ZapperZ said:
That is a very good question. In fact, this was carefully addressed in an AJP paper many years ago:

"Force on a Dielectric Slab Inserted into a Parallel-Plate Capacitor", S. Margulies (Am. J. Phys. v.52, p.515 (1984)).

In it, he wrote this:



Sounds familiar? :)

Zz.

Woa, that is exactly what I needed! It's a shame I can't access that article, though. :cry:

But it's kind of incredible that what I asked turned out to be something that only a research article could answer. :eek:

Thanks for the help, ZapperZ. :biggrin:
 
For some reason, my download from AJP is not working, but there is a simple answer to your question.
In the energy calculation, the end of the dielectric slab moves an infinitesimal distance, but is well within the plates where the field is uniform. The calculated force then does not depend on what happens at the edge of the plates. The energy is given by an integral of E.D over all space, so the fringing field would affect the total energy, but it does not affect the change in energy caused by the infinitesimal displacement
If the energy calculation were attempted for the edge of the dielectric in the fringing field, the calculation would be more difficult, and the force would be different.
 
Thread 'Motional EMF in Faraday disc, co-rotating magnet axial mean flux'
So here is the motional EMF formula. Now I understand the standard Faraday paradox that an axis symmetric field source (like a speaker motor ring magnet) has a magnetic field that is frame invariant under rotation around axis of symmetry. The field is static whether you rotate the magnet or not. So far so good. What puzzles me is this , there is a term average magnetic flux or "azimuthal mean" , this term describes the average magnetic field through the area swept by the rotating Faraday...
Back
Top