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In general, $\frac{\partial L}{\partial \dot{q}}$ is the canonical (or generalized or conjugate*) momentum, and $m\dot x$, for $x$ the actual position, is kinetic momentum. Likewise, the cross product of the former with the generalized coordinate vector $q$ might be called "canonical angular momentum", and the cross product of the latter "kinetic angular ...

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The mistake you made was essentially transforming the force twice. In the frame in which the matter is stationary (I'll call this the primed frame), you correctly found: $F'=q\sigma'/2\epsilon_0$ and in the frame in which the matter is moving (unprimed) you correctly found: $F=q\sigma/2\epsilon_0 \gamma^2$ Since $\sigma'=\sigma/\gamma$ this is: ...

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The answer is already as you have given it: $m_{em}=\frac{4}{3}E_{em}/c^2$ The electromagnetic mass depends on the shape you assume for the charged object. In the case above it is assumed the object is a charged, hollow sphere. In general the electromagnetic mass for a charged object producing electric and magnetic fields $E$ and $B$ is: ...

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The radiation emitted by an accelerated charge depends on the boundary conditions on the fields at infinity. When one takes this into account properly, then accelerated observers will agree with inertial observers about the emitted radiation (after trivial transforms are applied). Any treatment which purports to show that in the accelerated observer's frame ...

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Assuming that Classical Electrodynamics (Maxwell's Equations) holds, the answer is that the inertial observer would see the radiation while the non-inertial observer would NOT. The question you are asking is basically the following paradox: https://en.wikipedia.org/wiki/Paradox_of_a_charge_in_a_gravitational_field This paradox has been analyzed and ...

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$\nabla \cdot \vec E(r) = \dfrac {1}{r^2} \dfrac {d(r^2 E)}{dr} \ne \dfrac{dE}{dr}$ in spherical coordinates.

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The differential and integral forms should in principle always lead to the same result, since they are related to each other via Gauss's theorem. (e.g. see What are the differences between the differential and integral forms of (e.g. Maxwell's) equations? ). In this case you have not applied the differential form correctly, because you have used an ...

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I'm not sure about that particular equation. There's always issues with how quantities are defined, and signs of particular values. To understand what that equation is saying exactly, you have to look at the derivation. So to solve this problem, I'm going back to some basic electromagnetic relations for a plane wave. $$E = c B$$ $$c = ... -1 There are no known transistors or oscillators yet produced that can handle 10^19 frequancies, we are still experimenting at terahertz range 10^12 or thereabouts so the simple answer is no. 0 Too long for a comment so have to provide as an answer. The relevant em equations to derive the wave equations in a conductor are: \nabla . {\bf E} = 0, \nabla . {\bf B} = 0 and \nabla \times {\bf E} = -\frac{\partial {\bf B}}{\partial t}, \nabla \times {\bf B} = \mu \epsilon \frac{\partial {\bf E}}{\partial t} + \mu {\bf J}_f. The first two equations ... 2 Well, if there were charge between the boundaries, you would be solving Poisson's equation rather than Laplace's equation. However, boundary conditions for the potential function are also crucial, because there could always be distant point charges that modify the field in the region of interest, without changing the distribution of charge on the boundary. 1 As a general note, asking notation questions without providing a reference to the original occurrence (from which we'd be able to infer the context) is an excellent recipe for an unanswerable question. In this particular case, though, it's pretty clear that it refers to the unit basis vector in the z direction,$$\hat e_z=(0,0,1).$$It arises in this ... 2 You have to think about what \vec j,\mathrm{d}\vec l and \mathrm{d}\vec s actually are: \mathrm{d}\vec l points along the flow of the current. So does \vec j. So \mathrm{d}\vec l and \vec j are parallel, and indeed$$ (\vec j\cdot\mathrm{d}\vec s) \mathrm{d}\vec l = (\mathrm{d}\vec l\cdot\mathrm{d}\vec s)\vec j holds in that case.

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In my understanding, the free charge is any charged particle that is not being restrained in the boundary, while the bound charge is in the boundary.It does not matter whether the material you currently discuss is a dielectric or a conductor.

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