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The Electrostatic Potential Energy we talk about is an energy stored in Electrostatic Field. Field is a reality and it has momentum, energy etc. stored in it. How to imagine it ? Consider that some space has a field in it. Then that space has stored energy in it in the form of field present, such that if you change the configuration of the charges creating ...

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The heat of a solid is related to the energy contained in the solid. Hotter solids have more energy in them. Where is the energy in the solid? It's in the atoms and charge carriers moving/bouncing around. If something is hotter, the atoms and charge carriers are moving/bouncing around more and faster. Say you have a bar, and one end is hotter than the ...

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Well, it is kind of complicated, and not high school material actually :) But, electrons/holes do transfer heat. They move around in a solid, bounces into stuff, and from that give of heat - if we have to keep it really simple. Mostly though, the basic heat contribution (At least in your scenario with the Peltier effect an such) comes from lattice ...

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The short answer is that you are basically correct; you just need to be more careful with your notation and your minus signs. Here's the long answer. By the definition of a conductor, the sphere is at some constant potential. Additionally, the potential of the system goes to zero as $r\rightarrow \infty$. There are a lot of different ways to think about ...

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Axial charge'' refers to the (isovector) axial coupling constant $g_A$ of the nucleon $$\langle p|A_\mu^a|p\rangle = g_A \bar{u}(p)\gamma_\mu\gamma_5\tau^a u(p)$$ where $A_\mu^a=\bar{\psi}\gamma_\mu\gamma_5\tau^a\psi$ is the QCD axial current, $|p\rangle$ is a nucleon state with momentum $p$, $u(p)$ is a free nucleon spinor, and $\tau^a$ is an isospin ...

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If we know the position of the charged particle at any time, that is, if we know the value of the function $\mathbf r(t)$ for all $t$, one way to find the EM fields is based on knowledge of the fields at one time instant $t_0$. Let the field be such that $$\mathbf E(\mathbf x, t_0) = \mathbf C_E(\mathbf x),~~\mathbf B(\mathbf x, t_0)= \mathbf C_B(\mathbf ... 2 Maxwell's equations are linear, which means that if you have a (particular) solution, you can add any solution of the homogeneous Maxwell equations (i.e. electromagnetism in vacuum, without source, aka light) to it and get another solution. Furthermore, any solution is obtained that way. The way to pick one solution over another is by choice of initial ... 3 Your teacher is correct that the mass of an object if it is moving with very high energies appears to increase according to the formula , it is called the "relativistic mass" . Where E is the energy of the particle and c the velocity of light. But each elementary particle ( these are concepts that apply to elementary particles to start with) is ... 1 Well, the particles won't always follow circular paths (for instance, the particles in this video). But, if you apply a constant magnetic field across the chamber, charged particles moving in the field will be deflected according to the Lorentz Force Law. The centripetal acceleration for a particle moving in a circle is a=\frac{v^2}{r}, where v is the ... 0 It boils down to balancing the centripetal force,$$\vec{F}=\frac{mv^2}{r}\hat{r}$$with the magnetic force$$\vec{F}=q\vec{v}\times\vec{B}$$Equating these and considering the perpendicular velocity, we get$$\frac{mv_\perp^2}{r}=qv_\perp B$$Which can easily be solved for q/m:$$\frac{q}{m}=\frac{v_\perp}{rB}Thus, if you know the strength of the ... 0 A dielectric is not a conductor, thus there are no electrons that are able to flow through it. However atoms or molecules within may be able to be polarised making an electric dipole, which can align to enhance or anti-align to reduce the applied field. This is bound charge. In a metal or in free space the electrons flow and are, in a sense, free. They are ... 2 This is an experimentalist's answer and yes, accelerated charged particles either in stable circular orbits or in linear acceleration do radiate. Classically, any charged particle which moves in a curved path or is accelerated in a straight-line path will emit electromagnetic radiation. Various names are given to this radiation in different contexts. For ... 1 Specific charge is indeed the ratio of charge and mass, but since an atom is made up of neutrals and charged particles, you need to account for them. Thus, you'd use \eta=\frac{q\left(n_p-n_e\right)}{n_pm_p + n_nm_n + n_em_e}  where $\eta$ is the specific charge (my own variable, don't believe it's standard), $m_i$ is the mass of $i$ (neutrons, ...

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In the old days, most plasmas were created with discharges in gases or such. This can stay a plasma for a long time (at least seconds). If it were not quasi-neutral, the accumulated charge would strongly pull ions away from electrons and make the plasmas explode. That's why books make this claim, and it is still true in many cases. Nowadays, plasmas can be ...

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The simplest answer is that the horizon doesn't hold the charged particles, but rather, the information about the charges of particles that have fallen into the horizon lives on the horizon itself. In simpler terms, the electric monopole moment of all of the matter inside of the horizon can be inferred from the geometry and field in a neighborhood of the ...

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I believe that the "roughly" term is applied because of the associated experimental error when measuring its charge. The same cannot be said to the electron because "we" decided to make the electron the reference charge. So, the reference charge is definitely -1. However the muon charge must be measured. According to this paper, Muon Mass and Charge ...

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Assume the contrary,suppose a point exists such that the local charge density is positive,say point A. Now from Gauss' law the total charge on the inner surface is negative.So there must exist a point B at which the local charge density is negative(otherwise the net charge will be positive). Now consider the field line from point A.It will originate from ...

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The charge accelerated by the Earth gravity does not emit any radiation,follows from transforming to a frame of reference in which the charge is stationary and applying relativistic requirement that the behavior of the charge including whether or not it radiates,cannot depend on the frame of reference from which it is viewed.

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In fact, an electric charge at rest on the Earth's surface is accelerated and this actually poses a challenge to the idea that uniformly accelerated charge radiates. I believe this is still an open question. For example: One of the most familiar propositions of elementary classical electrodynamics is that "an accelerating charge radiates". In fact, ...

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