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Permittivity $\varepsilon$ is what characterizes the amount of polarization $\mathbf{P}$ which occurs when an external electric field $\mathbf{E}$ is applied to a certain dielectric medium. The relation of the three quantities is given by $$\mathbf{P}=\varepsilon\mathbf{E},$$ where permittivity can also be a (rank-two) tensor: this is the case in an ...


19

Just because the material doesn't conduct currents on a macroscopic scale, does not mean it doesn't contain any movable charges at all. In fact, as the very name “dielectric” suggests, such a material contains charges which can be to some degree separated – electrons move a bit to one side or the other, never actually bidding their parent atom farewell but ...


18

If $\varepsilon$ or $\mu$ are tensors (read, matrices), then so is $c_m$: $$ \overbrace{\varepsilon}^\mathrm{matrix} \underbrace{\mu}_\mathrm{matrix}=c_m^{-2}\ \leftarrow\ \text{matrix as well} $$ In other words, if the permeability and/or permittivity are matrices, then the speed of light is a matrix as well. In this case, the $\_^{\color{red}{-1}}$ is ...


18

You cannot totally avoid quantum mechanics, but it may suffice to say that reflections by free electrons are not the only way to prevent transmission. Any situation where light can promote an electron from a low energy state to a higher one will cause absorption, regardless of DC conductivity. Or even with little absorption, massive scattering of light by ...


16

What you are describing is anomalous dispersion. This happens when a material becomes strongly absorbing, typically near an absorption line, and the refractive index becomes complex. In these circumstances the phase and group velocities are different and indeed the group velocity can be greater than $c$. This isn't a problem since the group velocity is no ...


14

The explanation of the force as due to oscillating dipoles is only an approximate description and should be regard only as a guide for students. The actual explanation is that the wavefunctions of the two atoms overlap and the system now needs to be described by a larger wavefunction that includes the electrons in both atoms. In principle this new ...


11

A dielectric effectively behaves as if it was thicker than it is. If the dielectric constant is $K$ and the thickness of the dielectric is $t$, then for calculating the force it behaves as if the thickness was $t\sqrt{K}$. To see this let's take the example we know about where the dielectric fills the space between the charges: In (a) the thickness of the ...


9

At sufficiently high voltages almost everything conducts due in part to quantum tunneling of electrons. An insulator has a breakdown voltage which is the field strength required before it will start conducting. Related to the breakdown voltage is the dielectric strength which is the minimum voltage over distance ($\mathrm{V}/\mathrm{m}$) before a material ...


9

The permittivity of a conductor is infinite. Let the value of an external electric field in free space (relative permittivity = 1) be $E$. If this is applied to a material of relative permittivity $\epsilon_r$ then the electric field in the material is $\dfrac {E}{\epsilon_r}$ Inside a conductor the electric field is zero hence its relative ...


8

The Kramers-Kronig relations are the expression, in the Fourier frequency domain, of the fact that the linear susceptibility $\chi(\tau)$ is a causal function, i.e. that the dielectric response of the signal $f$ to a forcing $F$ has the form $$ f(t) = \int_0^\infty \chi(\tau) F(t-\tau) \mathrm d\tau = \int_{-\infty}^\infty \theta(\tau)\chi(\tau) F(t-\tau) \...


7

$D$ is the electric displacement field or commonly the flux density and $E$ is the field intensity. There is a fundamental difference between them which will be understood to certain extent as you go through the following answer. Consider a point charge of $Q$ coulombs. This means that the number of flux lines emitted by the charge is $Q$ coulombs. . Let ...


7

First make a parallel plate capacitor with plates of area A and spacing d. Fill the space between the plates with the dielectric whose complex permittivity $\epsilon(\omega)$ you wish to measure. The formula for this capacitance is a complex function of frequency because the permittivity is a complex function of frequency. $$ C(\omega)={{\epsilon(\omega)A}\...


6

Imagine a blob of liquid water. Each molecule is polar because the electrons are closer to the oxygen than the hydrogens. Without a large external electric field, the water is moving around bumping this way and that way with basically random orientations. Now while in orbit make a very large parallel plate capacitor,charge it up and put your blob of water ...


6

There are two contributions to the electric field in a dielectric: The field generated by the 'free' charges, i.e the ones on the capacitor plates. Call it $E_0$ $E_0$ polarizes the dielectric, which in turn adds to the total electric field. Call that polarization $P$. The total electric field is $$E=E_0-\epsilon_0^{-1}P$$ (The factor of $\epsilon_0^{-1}$ ...


6

The problem here isn't a simple algebra error, but rather an issue with the physics. A medium which at rest is isotropic no longer behaves as an isotropic medium when it is moving relativistically. Instead, it behaves as a nonreciprocal bianisotropic material. In particular, the phase velocity of light at a particular frequency in a medium is no longer ...


6

Short answer: yes, the surface charges are taken into account; in fact, they're what ensures that $\vec{E} = 0$ inside the conductor. The electric field at any point in space can be viewed as the superposition of the fields from the point charge outside the sphere, and the induced surface charges: $$ \vec{E} = \vec{E}_\text{point} + \vec{E}_\text{induced} ...


6

how the direction of the force on slab in both situation differs? Recall that the energy stored in a capacitor is given by $$W = \frac{1}{2}CV^2 = \frac{1}{2}\frac{Q^2}{C}$$ where $V$ is the voltage across the capacitor and $Q$ is the magnitude of the electric charge on either plate. For the case that a constant voltage source (e.g., battery) is ...


6

I believe you are talking aboug the Goos-Hänchen shift which is described in this paper. From there, the following diagram: That link also gives a detailed mathematical description. The original paper (referenced from the above) is F. Goos and H. (Lindberg-)Hänchen, Ein neuer und fundamentaler Versuch zur Totalreflexion, Ann. Phys. 1, 333 (1947), http://...


5

You are correct: there is no free charge so $\vec{D}=0$ which means $$ \vec{E}=-\frac{1}{\epsilon_0}\vec{P}=-\frac{k}{\epsilon_0r}\hat{r} $$ But this is for $R_1\leq r\leq R_2$. Inside the shell, $r<R_1$, there are no enclosed charges, so $\vec{E}=0$ there. Outside the shell, there is also no charge. Recall that the total charge for dielectrics can be ...


5

A dielectric is not a conductor, thus there are no electrons that are able to flow through it. However atoms or molecules within may be able to be polarised making an electric dipole, which can align to enhance or anti-align to reduce the applied field. This is bound charge. In a metal or in free space the electrons flow and are, in a sense, free. They are ...


5

The only property of metals used in deriving $C=\varepsilon A/d$ is that they are perfect conductors. Ideally, all metals have this property. So even if you change the metal, it should not matter. But if you use something other than metal, then it will of course change the capacitance.


5

When you have a dividing line which is between two dielectrics parallel to the plates you have to ask yourself; is the dividing line an equipotential? That is relatively easy for diagram B as there is no change of dielectric on either side of the dividing line. In diagram A there is a change of dielectric on either side of the dividing line and so in ...


5

As you correctly observed, the electric field stays the same in the capacitor after insertion of the dielectric because the applied voltage is constant. This is accomplished by the increase in positive and negative areal charge on the plates of the capacitor which is provided by the battery. Before the insertion there is a vacuum between the plates $K=1$ and ...


5

Permittivity and permeability are not just constants, but instead are complex functions that depend on a number of other quantities, including the wavelength of the light. In fact, the refractive index $n$ is only half the story - there is also a related quantity, the extinction coefficient $k$, that describes the absorption in a medium. To put it more ...


5

. . . . why don't we just put these metal plates as close as possible without touching each other . . . . How is this going to be done? One of the functions of a solid dielectric is to keep the plates separated. Air as with other dielectric is an insulator but if the electric field is to large it "breaks down" and becomes a conductor. So for a given ...


4

Everywhere inside of the dielectric, the following (Gauss's Law inside of meadia) equation holds $$ \nabla\cdot\mathbf D = \rho_\mathrm{free}, \qquad \mathbf D = \epsilon\mathbf E $$ Inside of the dielectric, there is no free charge, so we have the equation $$ \nabla\cdot(\epsilon\mathbf E) = 0, \qquad 0<z<a $$ Now, we recall the definition of the ...


4

The energy is used to polarize the dielectric, i.e.: Moving charges inside the dielectric.


4

So, this is an old post that I came across when I had a similar question. Here's a paper where they dissolve different amounts of ions in the water and found that the ability for the microwave oven to heat the water actually reduces as more ions are introduced. So from this study, you could conclude that ion drag is not source of heating in the water.


4

There isn't really a good physical explanation - this simply arises from the conventions we choose to represent our electromagnetic fields. The electric constant $\epsilon_0$ was defined as the constant needed to make Gauss's law for electricity and Coulomb's law work for whatever units of length, charge and force you want to choose. When we add a medium, ...


4

The issue here is how much the refractive index $n$ tells you about dissipation. As you rightly said, the imaginary part of $n$, which depends on both real and imaginary parts of $\epsilon$, leads to an imaginary part in k which describes an exponentially decaying electric field. However, this doesn't necessarily correspond to dissipation (i.e. a drop in ...


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