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Quite a lot of questions, but I think they can all be answered by understanding the main argument: 'electric field lines can not pass through a conductor'. I will illustrate with the spherical shell-question. The statement that electric field lines cannot pass through a conductor, is simply wrong. The reason that your flashlight works, is that the battery ...


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Electric field is zero in that point because the sum of electric field vectors have same intensity and direction, but are opposite. That point is halfway between two like charges.


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Is it simply that the repulsion of the equidistant point charges creates a dead zone? It has nothing to do with repulsion or attraction. To find the electric field at some location due to a set of point charges, you have to add the electric field contribution due to each of the point charges. You also have to remember that the electric field is a ...


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Are you speaking specifically about currents in wires? If you look at (the simplest version of) Ohm's law you will see that $\mathbf{J} = \sigma \mathbf{E}$, where $\sigma$ is the conductivity (technically it's a tensor, but we'll assume a constant scalar for now). In this case, $\eta$ = $\sigma^{-1}$, which is the resistivity. Thus, we can show that ...


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Yes it is right to say that electric field by a finite plate is not constant and zero at infinity. But in case of capacitors,the separation between the plates is so small as compared to dimensions of plate that with respect to the separation between the plates the plate itself can be considered as infinite. It is just a relative assumption to simplify ...


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Well the derivation of the Snell's law in optics can be done by using Fermat's principle of least time, have a look at Feynman's lecture series volume 1. You can derive an analogous equation for charged particles using the principle of least time by using the potentials in two consecutive areas ( which take the place of the refractive index) this is will ...



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