Energy transfer with current in a dielectric

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SUMMARY

The discussion focuses on energy transfer in dielectrics compared to conductors, emphasizing that energy transfer occurs through fields rather than electron movement. When a dielectric is subjected to a current below its breakdown threshold, it behaves like an insulator, allowing energy transfer via the electric field. Upon exceeding this threshold, the dielectric acts like a conductor, leading to energy propagation as electromagnetic (EM) waves. The presence of a dielectric alters electric field distribution and impedance, affecting energy propagation characteristics.

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  • Understanding of dielectric materials and their properties
  • Familiarity with electromagnetic wave propagation
  • Knowledge of displacement current versus conduction current
  • Basic principles of capacitors and electrical breakdown
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mcdonneldouglas
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TL;DR
Will the situation with energy transfer at current flow in a dielectric be similar to the case with a conductor?
It is known that when an electric current flows through a conductor, energy transfer occurs not by the movement of electrons, but by means of a field near the conductor.

In this case, will the situation with energy transfer when current flows in a dielectric be similar to the case with a conductor?
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Maybe I am not fully understanding your question, but I'll try to answer it.
As far as I understand it depends on the charge/current you put on the dielectric. If you put a current on the capacity that is below the break through threshold, then the molecules will only slightly adjust themselves to your current and act like an insulator, so energy transfer should happen via a field in your picture, not charges.

Once you cross the threshold, the dielectric will behave like a conductor. So yes, in both cases you get energy transfer via fields.
 
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Welcome to PF.

Energy propagates as EM waves in the space between ideal conductors. If that space contains an ideal dielectric, the energy will propagate through that dielectric, and will remain outside the conductors.

The presence of a dielectric will change the electric field distribution, so the impedance, and the velocity factor, of energy propagation will change slightly, when compared to free space.

Any electric or magnetic field that enters an imperfect conductor will be lost as heat in the conductor. The electric field that propagates through an imperfect dielectric, may heat the dielectric, and so also become an energy loss.
 
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mcdonneldouglas said:
In this case, will the situation with energy transfer when current flows in a dielectric be similar to the case with a conductor?
The electric charge within a dielectric, cannot flow continuously, but instead the charge is displaced elastically, in proportion to the electric field. That makes a capacitor. The virtual current in a dielectric is called a displacement current, to distinguish it from a conduction current, where a continuous flow of charge carriers is possible.

Energy can be guided and delivered to a port without involving conductors. An EM wave can be refracted or guided within a dielectric by local changes in the dielectric constant. Optic fibres are an example. Without controlled geometric changes in the dielectric constant, the energy would not be guided, but would be radiated in all directions.
 
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Baluncore said:
The electric charge within a dielectric, cannot flow continuously, but instead the charge is displaced elastically, in proportion to the electric field. That makes a capacitor. The virtual current in a dielectric is called a displacement current, to distinguish it from a conduction current, where a continuous flow of charge carriers is possible.

Energy can be guided and delivered to a port without involving conductors. An EM wave can be refracted or guided within a dielectric by local changes in the dielectric constant. Optic fibres are an example. Without controlled geometric changes in the dielectric constant, the energy would not be guided, but would be radiated in all directions.
Thank you for your answer!
So, can I ask you, which way(s) the energy flows during the dielectric's electrical breakdown when there actually is a continuous flow of charge?
 
mcdonneldouglas said:
So, can I ask you, which way(s) the energy flows during the dielectric's electrical breakdown when there actually is a continuous flow of charge?
If you want to study the energy propagation when the dielectric fails, you will need to specify a circuit schematic, and show the capacitor that will break down.

When a dielectric fails, it becomes an ionised path, a resistor. The energy then stops propagating along the transmission line, and is either reflected back to the source, or heats the dielectric in the region of breakdown.
 

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