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This is a bit of an old question but I stumbled upon it and I believe it a great learning experience and quite a beautiful question. What you got right (most of it) I'll begin by pointing out that you are entirely right! your configuration creates a perfectly uniform field in the ball! (I'll use ball to refer to the volume, while sphere would be used for the ...

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If this is a linear, isotropic medium, then $\vec{P} = \epsilon_0 \chi_e \vec{E}$. So if the polarisation field is solely in the x-direction, then $E_z=0$.

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The following analogy may help intuition, although that's very subjective. Remember that a discontinuity in the electric field at a surface is a sign of the presence of a surface charge density, which is given by $$\sigma=\varepsilon_0[{\bf E}_{\rm above}-{\bf E}_{\rm below}]\cdot\hat{\bf n},$$ where $\hat{\bf n}$ is the normal vector to the surface (...

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It is very simple to understand For rainbow formation we need white light , water droplet, and air Here LED/LCD screen are designed by polariser which act like water droplet and helmet lid is polymer which is another medium where light can enter. So here we got our light--> TV's screen(droplet) -->helmet another refrective medium and here we got ...

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The TV display is an lcd display, its light is polarized. The helmet's visor is birefringent because of strains in the polymer material. That will rotate the plane of polarization, different for different wavelengths. The reflection is different for different polarizations, it acts as the analyzer here. The colors are not really the colors of a rainbow, but ...

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As the third charge is moved closer, if the orientation of the bonded charges can change it would do so in a way that the like charge will tend to move farther, and the unlike charge will tend to move closer, to the third charge. Now as the third charge is close to the pair, the difference in the distances becomes significant and the repulsion of the like ...

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In a simplified picture, which often tends to give the wrong idea but may serve to illustrate some aspects, every photon spends some time as a combination of a virtual electron plus its antiparticle, the virtual positron, which rapidly annihilates each other shortly thereafter. The combination of the energy variation needed to create these particles, and the ...

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Firstly, $\epsilon_0$ is an artifact of using SI system of units. In many other system of units $\epsilon_0=\mu_0=1$ and both permittivities are dimensionless, whereas permittivities of other materials are measured in respect to that of vacuum. Electric permittivity, along with fields $\mathbf{D}$ and $\mathbf{H}$, is an object that belongs to the ...

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The medium has a relative permittivity >1 and can be treated as a dielectric. So yes to your first question. The vacuum can be polarised, the polarisability depends on the applied field which itself is a function of distance. Therefore, the field you might measure in order to determine the charge of an electron will vary with distance due to the different ...

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If we work in the circular basis for light: $|+\rangle$ and $|-\rangle$, then linear polarization is a linear combination: $$|\leftrightarrow\rangle = \frac {-i} {\sqrt 2}[|+\rangle - |-\rangle]$$ $$|\updownarrow\rangle = \frac 1 {\sqrt 2}[|+\rangle + |-\rangle]$$ I used double arrows to indicate that it is a tensor polarization. The orthogonal $\pm 45^{\... 2 It depends on the medium. For a medium with scalar refractive index, the situation is as described by Roger Wood. Pure s-/p-polarization will keep their polarization. This is in some sense by construction, since they are chosen as the "eigenpolarizations" of the reflection response. Since s-/p-polarization typically have different reflection ... 0 The measurement of changes in polarization upon reflection is called ellipsometry. Waves polarized exactly in p (parallel to the surface) or s (perpendiular to the surface) will be unchanged in polarization though they may reflect (and refract) at different magnitudes. Light polarized at some angle in between p and s will generally have both magnitude and ... 1 Given that I solved this problem about a week ago in this other answer, I still have it more or less fresh in my mind, so I'll try to clarify some misconceptions. The dipole moment$\mathbf p$does not only make sense for the case of two charges$+q$and$-q$a distance$\mathbf r$apart. The dipole moment is a more general quantity defined as$\$\mathbf p=\...

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Another answer is Miles Padgett, Graeme Whyte, John Girkin, Amanda Wright, Les Allen, Patrik Öhberg, and Stephen M. Barnett, "Polarization and image rotation induced by a rotating dielectric rod: an optical angular momentum interpretation," Opt. Lett. 31, 2205-2207 (2006)

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The key here is that this expression for the electric field is valid only inside the sphere. The field on the exterior of the sphere depends on the radius of the sphere. So based on the exterior fields, the spheres could be distinguished. As for a sphere of uniformly polarized material with a hole punched in the center: this is no longer a sphere of ...

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