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-1

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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For example, consider at $t=0$ the point charge be at the origin and moving in the $z$ direction with velocity ${\bf v}$. The electric field at this moment is $${\bf E}({\bf r})=kq\frac{1-v^2/c^2}{(1-v^2 \sin^2 \theta/c^2)^{3/2}}\frac{\hat{\bf r}}{r^2}$$ Then $$\nabla \times {\bf E}=-\frac{1}{r}\frac{\partial}{\partial \theta}kq\frac{1-v^2/c^2}{(1-v^2 \sin^2 ...


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You only have the homogenious Maxwell equations there. The curl of the magnetic flux density $\vec B$ does also depend on the current density $\vec j$, see Wikipedia. The current density for your uniformly moving point charge is $$ \vec j = e \vec v \, \delta^{(3)}(\vec vt - \vec x_0) \,.$$ This therefore also creates a curl in the magnetic flux density.


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The Gauss's law in integral form is better used to find the field whenever there is a symmetry in the problem. But to make the concept clear, I'am going to use the differential form, which tells the exact meaning as the integral form. the Gauss's law in differential form reads: ∇.E=ρ/$ϵ_0$ where ρ is the volume charge density. Now observe this ...


4

Maxwell equations read $$\nabla\cdot \vec E=\rho\tag1$$ $$\nabla\times \vec E = -\frac{\partial \vec B}{\partial t}\tag2$$ $$\nabla\cdot\vec B=0\tag3$$ $$\nabla\times\vec B=\vec j+\frac{\partial\vec E}{\partial t}\tag4$$ For the sake of simplicity, I assume $\vec{j}=0$. Equations (2) and (3) form a linear first order system $$D_x {\bf X}(t,x) = \partial_t ...


2

The Maxwell's equations are the basics of EM phenomenon. Whatever be the fields you select, they shouldn't violate these fundamental 4 equations. Suppose we are provided a problem to find the electric and magnetic fields of an EM wave or a charge, or whatever be it. As you said, we have now a four component problem. But the degree of freedom is not 4 as each ...


0

If you put a positive test charge in the region of the cylinder or inside it which way would the force on the charge be? This is the direction of the electric field. If you think that it parallel to the axis can you explain why the positive test charge would move one way rather than the other? There is no difficulty with direction if you make the ...


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Lets look at the 4 equations in ED, $$\nabla\cdot \vec E=\rho\tag1$$ $$\nabla\times \vec E = -\frac{\partial \vec B}{\partial t}\tag2$$ $$\nabla\cdot\vec B=0\tag3$$ $$\nabla\times\vec B=\vec j+\frac{\partial\vec E}{\partial t}\tag4$$ which can ofcourse be written in a more compacted form, $$\partial_\mu F^{\mu\nu}=j^\nu \tag5$$ The $(2)$ and the $(3)$ ...


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Intuitively, the electric field E is regarded as a force, and the displacement D as the response of the medium. In mechanical systems, the immediate power is given by the product of the force times the velocity. In a direct analogy, the term $\mathbf E⋅(\partial\mathbf D/\partial t)$ is the immediate power density exerted by the field. And while we allow ...


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The quality of the vacuum at LHC is pretty good but particles are constantly accelerated/bent by strong electromagnetic fields along the circumference of the accelerator. Thus the situation is not the one felt by a free particle. So conceptually, in your diagram, you can simply replace the photon coming from the nucleus by a photon from the electromagnetic ...


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A conservative force is one for which the curl is zero. i.e. force F is conservative if and only if: ∇×F=0. This is true for electrostatic forces. However, if at a point in space the magnetic field is changing in time, then the electric force will, in general, be non-conservative. The changing magnetic field induces a curl in the electric field. The ...


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The causality of the response, along with the fact that it would never introduce imaginary-valued D as a response to real-valued E, stipulates that the spectra of permittivity conform to the Kramers-Kronig relations. These form a very tight relation between the real and imaginary parts of permittivity. In particular, each part is a Hilbert transform of the ...


2

I think the answer is simply: "Yes". What you should keep in mind is energy conservation: As long as there are no sources, the total energy of the electromagnetic field is conserved. But then what, in free space, would cause the initial change in the electric or magnetic field to get the oscillations going? A source, which is possibly localized ...


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Yes. Not only is it possible to find a Hamiltonian density but it is even possible to find a plain Hamiltonian. I gave the construction in this EPL ArXiv1303.6143: Euro Physics Letters, 103 (2013) 28004 This Letter is short but dense, I can only outline the method here. The reason why you can't write an hamiltonian for the electromagnetic field is that ...



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