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Lorentz Transformations Suppose we call the lab frame the K-frame and a frame moving at velocity, $\mathbf{v}$, relative to the K-frame called the K'-frame. Then we can express the electromagnetic fields in the K'-frame in terms of the K-frame fields as: $$ \begin{align} \mathbf{E}' & = \gamma \left( \mathbf{E} + \boldsymbol{\beta} \times \mathbf{B} ...


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I think the answer is clearer if you consider the equipotential as shown in the diagram below. Forgive their straight line nature as they were easier to draw that way. Given that $\vec E$ must be perpendicular to an equipotential surface then in your computation of potential difference $\displaystyle V_{AB}=-\int_B^A\vec{E\,}\cdot\mathrm d\vec{r\,}$ the ...


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You are confused due to the vectorial form of the equation, so you should write it by components (I will use cartesian coordinates), and I will use $\partial_x = \dfrac{d}{dx}$ for comfort, that being said, your equation can be written as: $$ F=-\int{(dr)}{(\vec{\nabla} \cdot \vec{P}) \vec{E} }=-\int{(dr)}{(\partial_x P_x +\partial_y P_y + \partial_z P_z ) ...


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Higher energy gamma and longer wavelength radio? Keep in mind that the different 'kinds' are merely human labeling conventions for a spectrum that is continuous in the mathematical sense. There is no feature of "radio" that distinguishes it objectively from microwaves. We just pick a boundary on the basis of some technological limitations that apply when ...


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But I don't understand the mechanism of the force creation But the concept of electric charge and electric field is, by definition, the mechanism of the force creation - that humans have invented to model that which has been observed. Never forget that the observed is the metaphysically given. It is up to us, as beings possessing a rational faculty, ...


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This sort of calculation, especially when the speed of the electrons from an observer's point of view is close to $c$, has to be done using special relativity, in that sense that the transformation between reference frames is determined by Lorentz rather than Galileo transformations. As you mentioned, you can put your reference frame origin at one of the ...


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Boundary conditions of electromagnetic waves are well looked into because of the nature between polarization and reflection. However in this particular case it is rather simple, it is the evaluation of a contour integral over the boundary. This is done by summing the length elements of the pill-box shape that is typically used in this analysis in the correct ...


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The observer moving with the CM will measure that the force of repulsion the electrons is given by $F=\frac{e^2}{4 \pi \epsilon_0 d^2}$ ($d$ is their separation), he can only make measurements in his reference frame (that is moving with speed $v$), and will not be able to be determine this speed.


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On a local level, when you measure the electric field in matter (conductor or otherwise) there is a rapid variation both in time and space. Semiclassically, the electric field get's very large as you approach the nucleus, then drops off, the electrons are moving around the nucleus at very large speeds. So, yes, you are correct when you say there will be ...


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The principle of relativity says that there is no experiment that can determine absolute motion. So all observers, regardless of relative motion, need to agree on the outcome of any experiment. Because to the relativity of observers' measuring devices, they may not numerically agree on the measurements. By applying the laws of relativity they will be able ...


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A few comments before doing the calculation: In the CM frame, there is only an attractive force, while in the given frame, there is both an attractive and a repulsive force. This is no more mysterious than the fact that a vertical object in my frame can look tilted to somebody with rotated axes. Going from the CM frame to your frame mixes the electric ...


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The electric field between the conductors is due to both sets of charges. however when finding a value for the electric field using Gauss's law it is only the charges inside the surface which are of interest and it is easier to choose the charge on the centre conductor and the red Gaussian surface $S_+$ which would be a cylider. You could find the ...


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Potential refers to a particular point - or set of points which are "equipotential". So you can talk about the potential of one of the capacitor plates (because each is an equipotential surface) but not the potential of the capacitor (because when charged the $2$ plates are at different potentials). When talking about a capacitor, potential usually means ...


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A very obvious answer is just to heat the air enough (flame). You can also accelerate electrons with electric field and inject them into air (plasma needles and a lot of commercial and lab plasma setups). Microwaves can be used to create and sustain ionization (think of microwave plasma experiments). The same goes for other wavelengths that resonate with ...


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Too long for a comment: Consider the Maxwell equations: $$\nabla \cdot {\bf E}=\rho/\epsilon_0 \qquad \nabla \cdot {\bf B}=0$$ $$\nabla\times {\bf E}=-\dfrac{\partial {\bf B}}{\partial t} \qquad\nabla\times {\bf B}=\mu_0 {\bf J}+\dfrac{1}{c^2}\dfrac{\partial {\bf E}}{\partial t}$$ According to Heras (when commenting on a paper by Griffiths and Heald) the ...


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The RHS side of Gauss's Law, that is the charge enclosed should remain the same is indeed true. The apparent confusion if any, should be in the LHS of the equation, the integral of the 'dot product' of field and area vectors. Consider the diagram, Now, when we take the dot product of the field vector with the area vector in the initial case, the field and ...


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The answer is that it depends on the field on the outside (boundary conditions) and the dielectric constant of the insulator. For a imaginary insulating sphere of vacuum in a vacuum, it should be obvious that the sphere does not affect the electric field at all. Inside a dielectric, the field will be weaker than on the outside. For a dielectric sphere in a ...


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I have found your question and the diagram a little difficult to interpret. I have redrawn you diagram to show a charge of $+Q$ on the outer shell and a charge of $-Q$ at the centre together with two conducting shells shaded grey. What else the electric field inside the conductors is zero. If there was an electric field then the mobile charge carrier ...


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There is the concept of the self energy of the electron that can be described by its electric field. The model, which goes back to classical physics, can be described as a spherically distributed charges (that equal the charge of the electron) at infinity and the work required to bring the charges to the radius of the electron. The calculation would show the ...


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Equipotential lines are like contour lines on a map which connect land which is at the same height above sea level. Finding the electric field is like finding the slope on the map. The slope (the steepest line of ascent or descent) is always perpendicular to the contour lines, and it is steepest where the contour lines are closest together (provided that ...


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Kirchoff's laws are the electric equivalents of more fundamental laws. Kirchoff's Law of Current: Is a formulation of the Charge Conservation Law, in a node there is no net charge, so from the continuity equation the net current going through the node is 0. Kirchoff's Law of voltage: Corresponds to the energy conservation, altough you need current's law to ...


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This is a homework-like question, so I will not provide a full answer. Here are a couple of good things to think about on your way to the answer: What is special about the velocities (or momenta or kinetic energies) of the particles at the instant of minimum separation? What quantities are conserved throughout the interaction?


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I think your confusion is coming from the fact that you are actually using theta for two different things here. Let's use phi for the angle between the velocity of the rod and the magnetic field, and use theta as it is depicted in the diagram. Then your expression should be written as $$qvBsin\phi$$ This comes from the fact that $qvBsin\phi$ is derived from ...


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In addition to the answer by dmckee and to answer the question how high in energy you could get a photon it might be worth thinking about 'Gamma Ray Astronomy' where the highest energy photons are detected. The record highest photon energy observed is apparently currently 80 TeV, which corresponsds to a wavelength of $1.5 \times10^{-20}m$ wavelength (if I ...


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The electric field due to the outer cylinder has no contribution inside. One way to view it using Gauss's law, the other way is that if you took a slice from that cylinder, and considered a point inside it other than the center, you'll find a point producing electric field in the near side of the point (small charge, small distance) and a corresponding arc ...


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Is there an electric field around the poles of the battery before the circuit is attached, and is there still, after the circuit is connected? Yes. However adding the connecting wires is likely to change the distribution of the field. And why is the field equally strong everywhere in the wire, no matter the shape? Usually we design our circuits ...



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