The author of literature (Prog. Polym. Sci. 26 (2001) 3-65) said:' poly(tetrafluoroethylene) and some of its derivatives exhibit the lowest known dielectric constants with values of e between 1.9 and 2.1. It is well known that the incorporation of fluorinated substituents into polymers decreases the dielectric constant due to the small dipole and the low polarizability of the C-F bond as well as the increase in free volume.' Similar descriptions had been emerged in many literatures. However, the bond moment of C-F is 1.83 Debye, much higher than that of C-H (0.2). On the other hand, the dielectric constant of polyvinyl fluoride (PVF) is 8.5, much higher than that of polyethylene (2.3). So, I wonder to know, is fluorine atom increase or decrease the dielectric constant of polymer?
To start I should make it clear I am not an expert in the dielectric properties of materials.
However, the size of the dielectric constant of a material is a measure of how it responds to an electric field and its ability to store the energy of that field. Materials with high dielectric constants are able to respond to electric field by reorganising their shape, orientation, or electron density within them, for example, liquid water with a dielectric constant of about 80 can clearly respond to an applied electric field by reorienting its dipole to align with that field.
So for polyfluoroalkanes: Although each CF bond is fairly polar, I imagine the polymer is not free to reorient in response to an applied electric field (for example, to have all the fluorine atoms pointing in one direction and all the carbon or hydrogen atoms pointing in the other). In addition to this, fluorine atoms are not very polarisable so their electron density does not redistribute very much in the presence of an applied electric field.
In the example you give PVF vs PE: I imagine the key difference is the presence of a very polarisable double bond in PVF (and the large effect of the double bond overwhelms any difference between the polarisabilities of fluorine and hydrogen)
My guess is that if you have two similar materials, the one with more fluorine atoms with have the lower dielectric constant, but there are many other things that can also affect this property.
Sorry, but I cannot see any C=C (double) bond in PVF, except, maybe, at the very ends of the chains: -(CH2-CHF)n-. It is the same for all polymers made from ehtylene and substituted ethylenes.
Ah yes, my mistake, apologies!
I suppose for PVF the key difference is that the polymer chain can rotate slightly to move the CH2 groups more to to one side and the CF bonds more to the other side in the presence of an electric field (so there is probably not a very simple relationship between number of fluorine atoms and dielectric constant).
It could be a better hypothesis. By the way, I am not a specialist either. I just came across the following chapter, which should be useful:
Engineering » Electrical and Electronic Engineering » "Dielectric Material", book edited by Marius Alexandru Silaghi, ISBN 978-953-51-0764-4, Published: October 3, 2012 under CC BY 3.0 license. © The Author(s).
Chapter 1
Polymer Dielectric Materials
By Zulkifli Ahmad DOI: 10.5772/50638
A nice reference, section 2.2 does a fair job of explaining the reasons for the differences between PE, PTFE, and PVF.
- Badges
- Science topic
- Similar topics
- Engineering
- Petroleum Engineering
- Gas