# How an electric field is absorbed by a dielectric

• Soumava
In summary: This radiation carries energy away from the dipole, resulting in a decrease of the dipole's own energy. This decrease in energy manifests as a decrease in amplitude of the dipole's oscillation, which ultimately leads to a decrease in its kinetic energy. This process continues until the dipole reaches equilibrium, where the energy it receives from the EM wave is equal to the energy it radiates away.
Soumava
New poster has been reminded to use capitalization and punctuation to make their posts more readable.
when a em waves strikes a dielectric the atoms vibrate in response to the electric field and if the frequency matches the resonant frequency the Lorentz oscillator the electric field is absorbed how a field can be absorbed we know the em field contain energy how is the em field destroyed is some kind of opposite field created what's the physical picture of fields when a em wave strike a free electron

It would be very nice to have a play-by-play description of how a classical EM wave is absorbed by atoms. Quantum-mechanically, we talk about the absorption and emission of photons, which are quanta of energy of the EM field.

The best answer that I can give without photons is that atoms are described as elastic electric dipoles, with the negative electron cloud surrounding the positive nucleus.

When the EM wave reaches the atom, the electrons get pulled by the electric field a little bit away from the nucleus inducing an electric dipole in the atom. This electric dipole would oscillate at its natural frequency, but is being driven by the frequency of the EM wave. The oscillating electric dipole radiates its own electric field (as all accelerating charges do) at the same frequency as the driving field, but slightly out of phase due to the inertia of the atoms (among other factors). When the frequency of the EM wave is equal to the natural frequency of the dipole, the dipole oscillates with maximum amplitude, and more energy from the EM wave is converted to kinetic energy of the dipole.

Depending on various properties of these atoms, this dipole radiation can be released quickly or slowly, but is released all the same, and in all directions. Because of multiple atomic energy levels (and therefore multiple natural frequencies), the emitted light need not be the same frequency as the light of the EM wave, and could even be in the infrared while the initial EM wave was visible.

This, of course is a hand-wavey description, and specific situations will vary in the details.

Soumava said:
when a em waves strikes a dielectric the atoms vibrate in response to the electric field and if the frequency matches the resonant frequency the Lorentz oscillator the electric field is absorbed how a field can be absorbed we know the em field contain energy how is the em field destroyed is some kind of opposite field created what's the physical picture of fields when a em wave strike a free electron

First of all, you need to learn how to use punctuations! Your post above does not contain even a single one, and it makes for a very confusing, difficult reading.

Secondly, in a solid, the property of a solid does NOT depend only on the atoms/molecules that make up the solid, but also the collective property of a large group of atoms/molecules attached to each other. There are not "conduction electrons" or "band gap" in individual atoms, but there are such characteristics in a solid.

One important feature relevant here is the collective vibrational mode of the solid, often called phonons. Optical phonons are "active" to certain range of frequencies of EM wave. So they can affect whether a particular frequency transmits, transmits partially, or gets completely absorbed. This behavior is exploited in many experimental methods, such as UV-VIS and Raman spectroscopy, and why we can study the phonon spectrum using EM radiation.

If the EM wave is absorbed, the energy is typically transformed into vibrational energy of the solid, i.e. heat. If the EM wave has a "high" enough energy, then other phenomena can occur that transform the EM energy into a number of other channels.

Zz.

jfizzix said:
It would be very nice to have a play-by-play description of how a classical EM wave is absorbed by atoms. Quantum-mechanically, we talk about the absorption and emission of photons, which are quanta of energy of the EM field.

The best answer that I can give without photons is that atoms are described as elastic electric dipoles, with the negative electron cloud surrounding the positive nucleus.

When the EM wave reaches the atom, the electrons get pulled by the electric field a little bit away from the nucleus inducing an electric dipole in the atom. This electric dipole would oscillate at its natural frequency, but is being driven by the frequency of the EM wave. The oscillating electric dipole radiates its own electric field (as all accelerating charges do) at the same frequency as the driving field, but slightly out of phase due to the inertia of the atoms (among other factors). When the frequency of the EM wave is equal to the natural frequency of the dipole, the dipole oscillates with maximum amplitude, and more energy from the EM wave is converted to kinetic energy of the dipole.

Depending on various properties of these atoms, this dipole radiation can be released quickly or slowly, but is released all the same, and in all directions. Because of multiple atomic energy levels (and therefore multiple natural frequencies), the emitted light need not be the same frequency as the light of the EM wave, and could even be in the infrared while the initial EM wave was visible.

This, of course is a hand-wavey description, and specific situations will vary in the details.
how energy of wave is converted into kinetic energy of the particle

Soumava said:
how energy of wave is converted into kinetic energy of the particle

I don't know if there is a good answer to this question.

It may help to imagine the process in reverse.
If you have an oscillating dipole, it will produce electromagnetic radiation because the charges are accelerating. The conservation of energy requires that the energy of the radiation released be equal to the kinetic energy lost.
A fuller description of this mechanism can be found by looking up how the Larmor formula for the power radiated by accelerating charges works.

jfizzix said:
I don't know if there is a good answer to this question.

It may help to imagine the process in reverse.
If you have an oscillating dipole, it will produce electromagnetic radiation because the charges are accelerating. The conservation of energy requires that the energy of the radiation released be equal to the kinetic energy lost.
A fuller description of this mechanism can be found by looking up how the Larmor formula for the power radiated by accelerating charges works.
what happens to the original electromagnetic wave

Soumava, I've got to admire your fortitude - sticking to your guns and refusing to use punctuation and writing complete sentences. That'll teach us! Yesiree! Couple that with providing us clarifying information one tiny piece at a time and you can make a question that could be answered in five minutes stretch out for days.

Or, you could think about the question you are asking, write it clearly, concisely, and completely - using proper English - and get your answer quickly. Up to you.

Soumava said:
what happens to the original electromagnetic wave

This is a good question to ask, but I can only give a partial answer without talking about photons.

To get some intuition for what happens to an EM wave as it gets absorbed, you will want to look up how EM waves behave in a medium that can conduct electricity, since here, we have free charges interacting with the field.

Looking at Maxwell's equations, you can argue that for an EM wave in a vacuum, the rate of change of the electric field drives the magnetic field, while the rate of change of the magnetic field in turn drives the electric field.

In a conductor, this simple reciprocal relationship is no longer true, as the magnetic field is driven both by the rate of change of the electric field, and the electric current, which in turn is approximately proportional to the electric field (from Ohm's law).

[[]]
This is a hand-wavey way of looking at it, but...
Because the electric field is causing the charges to move, and because moving charges generate magnetic fields, and because accelerating charges generate changing magnetic fields, accelerating charges produce electric fields in the opposite direction of their acceleration, subtracting away from the original electric field, which decreases it.
[[]]

Because of the extra dependence on electric current, the EM wave equation you get by taking the curl of both sides of Faraday's law contains an extra term that depends on the rate of change of the current (i.e., the overall acceleration of these moving free charges). This acceleration term acts as a damping term in the wave equation. Indeed, when you solve Maxwell's equations for a plane-wave incident on a conductor, you see the EM wave decrease in intensity exponentially with distance through the medium.

I know this answer is very mathematical, but I hope it helps a little.

jfizzix said:
This is a good question to ask, but I can only give a partial answer without talking about photons.

To get some intuition for what happens to an EM wave as it gets absorbed, you will want to look up how EM waves behave in a medium that can conduct electricity, since here, we have free charges interacting with the field.

Looking at Maxwell's equations, you can argue that for an EM wave in a vacuum, the rate of change of the electric field drives the magnetic field, while the rate of change of the magnetic field in turn drives the electric field.

In a conductor, this simple reciprocal relationship is no longer true, as the magnetic field is driven both by the rate of change of the electric field, and the electric current, which in turn is approximately proportional to the electric field (from Ohm's law).

[[]]
This is a hand-wavey way of looking at it, but...
Because the electric field is causing the charges to move, and because moving charges generate magnetic fields, and because accelerating charges generate changing magnetic fields, accelerating charges produce electric fields in the opposite direction of their acceleration, subtracting away from the original electric field, which decreases it.
[[]]

Because of the extra dependence on electric current, the EM wave equation you get by taking the curl of both sides of Faraday's law contains an extra term that depends on the rate of change of the current (i.e., the overall acceleration of these moving free charges). This acceleration term acts as a damping term in the wave equation. Indeed, when you solve Maxwell's equations for a plane-wave incident on a conductor, you see the EM wave decrease in intensity exponentially with distance through the medium.

I know this answer is very mathematical, but I hope it helps a little.
I understand but please provide me a more layman expanation
is any kind of opposite field created as you mentioned

## 1. How does a dielectric absorb an electric field?

Dielectrics absorb an electric field by allowing the field lines to pass through their molecules. The molecules become polarized and align themselves with the electric field, resulting in the absorption of energy from the field.

## 2. What happens to the electric field when it is absorbed by a dielectric?

When an electric field is absorbed by a dielectric, the field is weakened and the overall electric potential is reduced. This is due to the energy being used to polarize the molecules of the dielectric material.

## 3. How does the dielectric constant affect the absorption of an electric field?

The dielectric constant, also known as the relative permittivity, is a measure of how easily a material can be polarized by an electric field. The higher the dielectric constant, the more easily the material can absorb an electric field.

## 4. Can a dielectric become saturated with an electric field?

Yes, a dielectric can become saturated with an electric field. This occurs when the applied electric field is strong enough to fully polarize the molecules of the dielectric material. Once this happens, the material is unable to absorb any more energy from the field.

## 5. How does the thickness of a dielectric affect its ability to absorb an electric field?

The thickness of a dielectric can affect its ability to absorb an electric field. Thicker dielectric materials have a higher capacitance and can store more energy, allowing them to absorb a stronger electric field. However, if the dielectric is too thick, it may impede the flow of the electric field and reduce its absorption.

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