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1

A key to understanding this is realizing that it's not always true. In fact, at x-ray frequencies, refractive indices are typically less than 1, so that the phase velocity is faster than the vacuum speed of light. The key difference is that x-ray frequencies are well above the natural frequencies of most of the electronic excitations that are involved in the ...

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The short answer is that it doesn't matter. Whichever direction you choose to define the reflected E-field initially you will always gets its direction with respect to the incident E-field coming out the same. In the case you have in your question (where you have reversed the sign of $E_R$ and defined it to be in the -x direction and opposite to the ...

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I think that your confusion is due to a subtlety associated with the solution approach of going to a rotating frame of reference. If you're just trying to find out the polarization of a stationary rectangular, block-shaped slab of dielectric material due to a uniform electric field, then you could simply write down an electric polarization equation similar ...

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Jackson's E&M book covers this in Chapter 6, Section 8 (3rd Edition, or the blue covered book). The following is a summary of that section. Answer (not much explanation) In a lossy medium, where $\Im \left[ \varepsilon \right] \neq 0$ and/or $\Im \left[ \mu \right] \neq 0$, we can still use linear approximations but there are some modification. In ...

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$E_1 = E_2$ . since $E$ is independent of dielectric as long as potential b/w plates is constant. $$E= = -\frac{dV}{dr}$$ So, it is independent of dielectric b/w it. So, correct statement would be $$E_1d = E_2d$$ $$Ed = Ed$$

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You seems to assume both capacitors has the same plate separation $d$. So, lets assume that. Assume there is no dielectric material. Therefore, nicely $Ed = Ed$ in both capacitors. Which is nice. :). Now, I think I understand your confusion. Have an isolated capacitor with electric field inside plates of $E$. Insert dielectric $K$. Under this case, the ...

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Actually, $w_0=\frac{1}{2} \bf{D} \cdot E$ can be right only in a linear dielectric without dissipation( permittivity is a real number). In an arbitrary dielectric, $$\nabla \centerdot \mathbf{D}=\rho_0$$ $$\nabla \times \mathbf{B}=-\frac{\partial\mathbf{E}}{\partial t}$$ $$\nabla \centerdot \mathbf{B}=0$$ $$\nabla \times \mathbf{H}=\bf{j}_0 + ... -1 My guess is that the \vec{E_r} and \vec{E_i} just have a dot product equal to 0 leaving just the rest which gives you the correct answer of u=\frac{1}{2} \epsilon_r \vec{E_r} \cdot \vec{E_r}. I would say that if their dot product is zero, it is because they are perpendicular vectors only like imaginary parts are perpendicular to real parts in the ... 0 After some thought I think I have come up with a reason why the energy density is indeed:$$u=\frac{1}{2} \epsilon_r \vec E_r \cdot \vec E_r My reasoning is that the imaginary parts of $E$ and $\epsilon$ solely come about due the propagation of the wave through a medium. The energy density of the electric field, however should not depend on how it is ...

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