# Can a rotating dielectric disc be polarized?

sergiokapone
Can a rotating dielectric disc be polarized? No external electric or magnetic field .

Homework Helper
Of course it can. If it was previously polarized, the rotation will not necessarily destroy the polarization.
But I suspect this is not what you mean to ask. Do you mean if the disk can become polarized due to rotation?

sergiokapone
Do you mean if the disk can become polarized due to rotation?

Yes, I do.

Homework Helper
How do you think that polarization of a dielectric happens in presence of an electric field?

sergiokapone
I think polarisation should be appear due to electric field, which is created by
reclined free light weight charges (electrons), due to the centrifugal force. But if we have ideal dielecrtic, where we can get this FREE charges?

Or another way. Polarisation can happen due to mechanical deformation of the piezoelectric, but this is not the case.

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Homework Helper
A dielectric has no free charges. Still can be polarized in electric fields.
You don't need free charges to have polarization.

sergiokapone
Still can be polarized in electric fields.
You don't need free charges to have polarization.

Of course, but where I get this field? Assumption of the problem - is no electric field, only a rotation disk.

Homework Helper
Yeah, but do you understand that you don't need free charge?
You said that you don't understand the paper and one of the reasons is that there is not free charge in dielectric.
The paper says right in the beginning that accelerated motion of the dielectric can result in polarization. You may find it easier to think about it in terms of inertial forces, in the accelerated frame. The inertial forces will be different for the light electrons than for the heavy nuclei so it may result in a slight displacement of the centers of mass of the two types of charges. This is polarization.

sergiokapone
Now I see. Thank you. How can I calculate polarisation depend on rotational frequency?

sergiokapone
I think centrifugal force - is the some kind of electric field:
##m\omega^2 r = eE##
Then polarisation is equal to ##P = \frac{\epsilon - 1}{4\pi} \frac{m}{e}\omega^2 r##

So, I can also find charge density as ##\rho = -div \vec P##

Homework Helper
I think you should take the difference in force between electron and nucleus but in first approximation it may be OK.
The problem is that the force is not uniform and so neither is polarization.
But again, the change in r will be small at the level of few atomic distances.