I am currently studying Maxwell equations and I learned that copper wires are essentially just wave guides for EM waves. Why do we not use an insulator to guide the wave and transport the energy which is in the fields? Because a conductor is lossy wouldnt it be better to use a dielectric as wire?


3 Answers 3


Dielectrics are lossy too. And wires are good enough already.

Indeed, it is possible to transport energy via dielectric, e.g. using light in optical fibers. But there is still some energy loss due to absorption there.

It would be hard to transport lots (megawatts) of energy efficiently this way. Electric energy produced in power plant would have to be converted to light energy and back with high efficiency(very hard to do and expensive; lasers or diodes are not very effective at this), and a thick enough dielectric fiber would have to be created and installed (expensive). All while the electric energy losses in long distance transmission are already solved by using very high voltage.

Transport via dielectrics would make sense if for some reason metals can't be used. There is a related practical example in transmission of audio information (not transmission of energy): it is sometimes transported using optical fibers and optical fiber connectors (see TOSLINK) to prevent pollution of a high-quality audio signal with unwanted EM noise.

Why do we use wires/conductors to transport energy?

Both conductors and non-conductors can guide EM energy where we want it to go. At least one of these is needed; without a material guide, we can't transport EM energy except in a straight line via radiation (and that we can do only with substantial energy losses due to conversion at both ends).

  • $\begingroup$ To complement this answer, in order to make dielectric waveguides for short wavelengths would require dielectric slots of several hundreds of meters in order to fit the wave inside it. While it is true that metals are lossy media, it is more related to thermal losses due to the energy transport (which would also be a problem if the energy transported across large distances were in the range of power plant generation). $\endgroup$
    – ondas
    Apr 25, 2022 at 11:45
  • $\begingroup$ @ondas what do you mean by short wavelengths? And why are they relevant? $\endgroup$ Apr 25, 2022 at 12:38
  • $\begingroup$ Short wavelengths would be electromagnetic waves of very small frequency. On a general rule of thumb I would consider up to a few MHz, which is the range in which the skin effect starts taking higher relevance. If the generated EM fields are much larger than the physical space they occupy, the EM behavior is more suitably modeled as a current (which generates an EM field). $\endgroup$
    – ondas
    Apr 26, 2022 at 13:12
  • $\begingroup$ Oh, I think you're right, EM energy transport would not go well using wavelengths comparable or greater than cable/waveguide thickness. This is a reason to use much shorter wavelengths (milimeter waves, microwaves...). $\endgroup$ Apr 26, 2022 at 17:48
  • 2
    $\begingroup$ @ondas Short wavelengths means high frequency. $\endgroup$
    – ProfRob
    Apr 26, 2022 at 18:21

Insulators do a wonderful job at expense of significant heat gain. Since all conductors have a specific heat threshold before failing.

And we do not use wires for all energy,just electricity.... only 13-14% of all global energy consumption is electrical. The vast majority of world's energy is chemical as either heat or mechanical energy.


Too much Ve.

Why do we not use an insulator to guide the wave and transport the energy which is in the fields?

That is totally wrong. The transferred electric energy per unit of time in a wire transmission line is not the EM field in the wire but the kinetic energy of the charge carriers (i.e. free electrons) in the conductor wire.

To transfer electric energy and do work you need charge carriers. The more energy is do be transferred the more carriers you need,


where $E$ is the transferred energy, $Q$ is the total charge and $V$ the generated potential difference by the work done on the source (i.e. Electric Power Plant). As you can see from the equation, $Q$ is provided by the charges (i.e. free electrons) in wires and and $V$ by the Power plant generators. The more conductive (i.e. rich in electric charge carriers) the wires are the more electric energy can me transferred to the consumer of electric power.

A dielectric wire is very poor to non-existing, in the amount of free moving charges (i.e. free electrons) it contains and therefore not suitable for the transfer of electric power. Thus, conductor wires are the best and only choice today for this job for electric power transfer. The confusion comes because it is wrongly communed by some popular science media lately, that large electric power is synonymous with large wireless EM radiation transfer which actually is not true. Electric power transfer depends on electric fermionic charges motion and not on photonic EM radiation.

On the other hand dielectrics like air are suitable for EM radiated signal transfer where power is not so much a concern but we just need to transfer information across space at relative large distances economically.

There are currently no alternatives for demanding electric power transfer and there will be no in the future wireless alternatives (at least what electricity concerns) simply because physics don't allow it.

The best development for more effective electric power transfer would be if we could use superconductive wire technology (i.e. lossless) in the future.

  • $\begingroup$ As an example for the transfer of 1 MWh of electric energy a ~4.5A charge current would be needed at 220Vrms or ~9A at 110Vrms. Insulator dielectrics cannot generate this amount of current unless reaching their breakdown voltage which can be in the range of hundred of MVolts and which practically means the dielectric catches fire and is destroyed. $\endgroup$
    – Markoul11
    Apr 27, 2022 at 11:19

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