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I don't think this is the case since grounding only defines the potential of the plate to be 0. There's more. An ideal ground can supply / absorb an arbitrary amount of charge while remaining at zero potential. "A grounded conductor is a special type of equipotential: infinite amounts of electrical charge (± Q) can flow from / to ground to / from ...


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I would say- in ideal static case - conductor has no charge gradient inside (and total charge is 0, as you say). The electric field of both systems, grounded and not-grounded, will be the same, assuming that the plate is in both cases on zero potential (Q=0).


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It depends where and how you ground your conducting plate. If it is grounded at infinity, where the potential of your charge is zero, it should not change anything over an already (net) uncharged plate. However, if you tie any bit of the plate that would otherwise be at a nonzero potential to zero, you will cause the plate to become charged as a consequence, ...


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A single charged object is sufficient to produce an electric field. Following Coulomb's law: $\textbf{E} = {Q \over 4\pi \epsilon_0\textbf{r}^2}\hat{\textbf{r}}$ where $\textbf{E}$ is the vector electric field, Q the charge of the object in question, $\epsilon_0$ the permittivity of vacuum or the electric constant, $\textbf{r}$ the vector position of the ...


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A point charge is given at a single, specific location in space. Think of it as an idealized source of charge. At the point where r = (0, 0, 0), electric field is zero and V(r) = $\infty$. As the value of r approaches 0 from any direction, V increases towards infinity but it is never discontinuous. For any other surface, the concept is well-explained by ...


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Consider a point charge $q$ at $(0,0,0)$, the potential at $\bf{r}$ is given by $V(\bf{r})$ $= \frac{q}{4\pi\epsilon_0r}$. If you consider a path through $(0,0,0)$, you encounter a discontinuity in electric field and the direction of field changes. Is the potential for the field of point charge discontinuous at the location of the point charge? ...


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The capacitor plates are made of some kind of conductor having non-zero thickness. The charges on the plate will reside on the surface of the conductor that faces the opposite plate. We say that these are surface charges. These are the only free charges in the system. We model them as a "sheet of charge". So the same description actually applies to both ...


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Think of charges as a bunch of $+1$s (protons) and $-1$s (electrons). It doesn't matter how many $0$s (uncharged particles) we have because $0 + 0 + 0 + 0 +... +0 = 0$. Now in most cases it's pretty difficult to change the number of $+1$s a body has, so we can see if it has more $-1$s than $+1$s to know if it's negatively charged or not. Let's call the ...


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If you connect two lossless capacitors in parallel, charge will flow back and forth. The loop formed by them will store energy in the form of magnetic energy (due to the current flowing) and you will end up with a resonant circuit. The frequency of resonance will be determined by the series capacitance and the inductance of the loop. So there will not be a ...


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A book on semiconductor devices in general, like this one (e.g. section 2.7), may help. Also, just look at the diagram at the right of this Wikipedia page. In general, imagine that we're looking at a graph of different energies (usually shown on the $y$ axis) in this semiconductor. The big thing that you'll notice in a semiconductor is that the density of ...


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The answer lies in a thermodynamic argument. The diffusion is a spontaneous process that occurs when particles with random motions are not uniformly distributed. The electrons in the conduction band in the n region are much more numerous that in the p junction. They will naturally tend to balance the concentration from the n junction to the p junction since ...


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For the unboundedness see ACuriousMind's answer. About the associated projections, for unbounded operators there is the notion of affiliation. An unbounded, closed and densely defined operator $A$ is affiliated with a von Neumann algebra $M$ if all its spectral projections are in $M$, that is $$A = \int\lambda\text dE(\lambda),$$ with the spectral measure ...


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$\newcommand{\ket}[1]{\lvert #1 \rangle}$The charge and baryon number operators are not bounded because you can create states of arbitrary charge and baryon number: Let $a^\dagger$ be any creation operator that creates a bosonic charged particle state (let's say with unit charge), and let $\ket{\Omega}$ be the vacuum. Then, $$\ket{n_e} = ...


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charge is an intrinsic property of any particle. we in principle cannot change the intrinsic property of any particle. photons are the carriers of electromagnetic interaction(action at a distance).


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If you knew the voltage at c and at d, then you would know the voltage everywhere, and then find the energy in each capacitor, and so you could add it up to get all the energy. I'd also try a simple circuit with two capacitors and recall a technique that works there and see if it simplifies your problem at all. Then try three. If you can't find an ...


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The point is that a Dirac field isn't really a solution for an electronand positron but instead an electron and the conjugate of the positron. This means that the Dirac electron field does indeed obey, \begin{equation} e ^{ i \alpha (x) Q } \psi = e ^{ - i \alpha (x) } \psi \end{equation} Such misconceptions are dissolved if one works in the Weyl ...



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