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How does dielectric heating really work?

Physics Asked by Experience111 on September 2, 2021

I’m currently working on a research project about dielectric heating of some materials and I’m trying to understand the physics of this phenomenon before going any further. I’m especially trying to understand the concept of complex dielectric constant and loss factor. I’m completely lost because I keep coming across two contradictory explanations:

  1. The dielectric heating is due to the movement of rotating dipoles (aligning with the alternating electric field) which creates heat

  2. "As the frequency increases further a point is reached where the reorientation polarization fails to follow the applied field and contributes less to the total polarization. The fall of the effective polarization manifests itself as a fall in the [real part of the] dielectric constant and a rise of the loss factor. Energy is now drawn from the system and dissipated as heat into the material". This is from "Industrial Microwave Heating" by Metaxas and Meredith.

So on one hand I’m told that it is the movement of the dipoles that causes heat, and on the other hand I’m told that it is the failure of the dipoles to follow the electric field that causes heat which doesn’t make much sense to me.

Could somebody be kind enough to help me make sens of it all?

2 Answers

The losses are being created because of the friction between rotating molecules. But if the field is changing slower than the molecules can rotate, then it means that their polarization is on average lagging behind the polarization of the field less, than in the case of where the changing magnetic field is so fast that they can not follow it. So if the magnetic field is faster than they can follow it, the force on the particles is greater because of bigger difference in orientations of molecule dipole moments and magnetic field, which creates more heat.

Answered by MaDrung on September 2, 2021

They are not really contradictory. Imagine the applied electric field changes so slowly that the dipoles in the dielectric are able to follow the field oscillation in real time. This is the so called adiabatic limit. Here at any given moment in time, the dipole points in the direction parallel to the applied field and is at the ground state. No energy is dissipated. After one cycle, the energy in the system is exactly the same as one cycle ago.

Next, imagine instead the frequency of the applied field is much faster than the dielectric dipole is able to keep up. This practically means if the dipole starts pointing down when the field points down, half a cycle later, the field is now pointing up but the dipole hasn't followed. It's still pointing down. This anti-alignment now means the dipole is in an excited state, not its ground state. Excited states tend to relax, so once in a while, the dipole does scatter with the environment and flips back to its ground state, pointing parallel to the field again. This allows it to temporarily "keep up" with the external field, but half a cycle later it's out of sync again and needs to give up the energy to the environment again in order to keep up. This excited state relaxation is what's causing the heating up.

Alternatively, we can represent both the dipole and the electric field as complex numbers. If the dielectric constant is real, the dipole is in phase with the electric field. If it falls out of phase because the field is rotating too fast to keep up, its direction begins to "lag behind" the field direction and this manifests as a complex phase in the ratio between the two. This is why a complex dielectric constant implies dissipation.

Answered by Gaberber on September 2, 2021

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