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The answer to the second part of your question is yes your approach is OK. I have gone through the link you have posted and want to say that the inferences drawn from formulae written above are for non collisional case i.e. dielectric constant $\epsilon$ is real. However if $\epsilon$ is complex you can not draw the same conclusion from above formulae. ...

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Saare taate fado bhancho The dielectric will develop charges such that the net force in that direction stays the same I.e the along the positive plate the dielectric will develop some negative charges such that the net force remains same

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The dielectric constant, $\epsilon$, is also called the relative permittivity of a material. If we take the word literally, it can be seen as how much the material "permits" an electric field to come in. This basically translates to how much a material will allow itself to be polarized. Consequently, a perfect electrical conductor (PEC) would have $\epsilon ... 1 Work out how a plane wave propagates. If its intensity grows in the direction of propagation, then you need to switch the sign of the imaginary part! It depends on whether you define$e^{+i\,k\,z}$or$e^{-i\,k\,z}$as your diffraction operator in the direction of propagation. If it's$e^{+i\,k\,z}\$, then a lossy dielectric always has a positive imaginary ...

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The permittivity of the medium between two electric charges affects their electrostatic interaction because the medium can become polarized: that is, the medium has lots of neutral atoms inside it, and in each of those atoms the positive and negative charges shift to create a tiny dipole, and in the process they create thin layers of positive and negative ...

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Electrodynamics in Vacuum does not have a medium-dependence - as does gravity. If we do introduce a medium, we end up in a continuum theory of chargeable or polarisable media, which is where the permittivity comes from (for anyone not familiar with this, I would recommend reading a bit of Jackson's Electrodynamics). In gravity, there is no such thing as ...

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