crashsite said:
The link I posted earlier seems to suggest that there's some sort of (electrostatic") attraction between the plates and dielectric that must be accounted for and that takes energy. Thinking along those lines might be a step in resolving the dielectric question but, it hardly explains it.
It has been decades since I studied the physics of capacitors, so I have stayed out of this discussion. Maybe, however, I can help with the question of
how the dielectric acts.
The capacitor equation as you all know is:
C = Q/V
Q = charge
V = potential difference (volts)
C = capacitance
There are other equalities to consider as well (See: FW Sears and MW Zemansky, University Physics, 3rd edition (yes, third
), 1964, pages 594-596).
(1) V = E
l and
(2) E
l = (1/ε) (Q
l/A)
E = electric intensity or potential gradient
l = separation of the plates
A = area of plates
ε = dielectric constant
Substituting equation (2) in the capacitor equation gives:
C = ε A/
l
Thus, if one took a charged air-gap capacitor and inserted something of higher dielectric constant between the plates, the voltage would decrease as the capacitance increased. That is the same effect one would get by reducing the plate separation.
So, now for a physical model. Assume the plates of a capacitor are 1 mm apart. The charge between the plates must go from + to zero to – in that distance. What if you can decrease the effective distance over which the charge goes from + to - ? According to equation (1), you would increase the capacitance and reduce the voltage. You can do that by inserting something with a greater ability to dissipate charge, i.e., it can dissipate charge over a shorter distance. That is basically what “dielectric strength” tells you. Materials with a high dielectric strength are able to dissipate charge over a shorter distance, i.e.,
they increase the attainable potential gradient.
I find it easiest to visualize the mechanism with solutions. Polar molecules have dipole moments. The moments are not necessarily fixed charges, nor does that mean they are electrical conductors. In fact, dielectrics are non-conductors. As an example, take a simple diatomic molecule such as carbon monoxide, CO, which has a dipole moment along its axis. Oxygen is more electronegative than carbon and can be represented as pulling electrons from the carbon giving the carbon a partial positive charge and the oxygen a partial negative charge. There are lots of similar examples. Elements to the right of carbon in the periodic table are more electronegative than carbon. Thus, polychlorinated biphenyls (PCBs), polybrominated biphenyls (PBBs), acetonitrile (CH3CN), and HCN are polar with the electrons being more localized on the electronegative atoms. They have high dielectric constants.
Now, do the following thought experiment. Consider a charged plate in a liquid dielectric. At the surface of the plate, the first layer of molecules will be lined up with opposite dipoles facing the plate. The next layer will repeat the process, but show more randomization due to thermal effects and entropy, and so forth. Eventually, the molecules will be completely random, and one could say the charge has been dissipated over whatever distance it took. That distance is shorter for compounds with high dielectric constants than it is for those with lower dielectric constants. Changing to a solid phase doesn’t really change what is happening, the alignments may not change as easily, but the forces on the dipoles and the mechanism for dissipation of the energy is the same.
If you look at a list of dielectric constants, you can explain the observed order by the chemistry just described. You may ask, why is Teflon just 2.1, if fluorine is so electronegative? The perfluorination makes the molecule symmetrical. For other examples, compare Plexiglas (acrylic, polar, non-symmetrical) with a value of 3.4 with polyethylene (non-polar) with a value of 2.25. Or, compare benzene, which is very symmetrical and has a value of 2.28 with isoprene, which is much less symmetrical and has a value of 6.7.
Sorry for the length of this comment. I hope it helps.
John