# Tag Info

2

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).

2

$\newcommand{\ket}[1]{\lvert #1 \rangle}$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, $$\ket{n_e} = ... 2 For the unboundedness see ACuriousMind's answer. About the associated projections, for unbounded operators there is the notion of affiliation. An unbounded, closed and densely defined operator A is affiliated with a von Neumann algebra M if all its spectral projections are in M, that is$$A = \int\lambda\text dE(\lambda), with the spectral measure ...

1

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 ...

1

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 ...

1

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 ...

1

There is no need for taking the mod of the charge. The voltage has the same sign as the charge. So if you start out with +20 µC on one capacitor (they give the + sign for a reason - so that's the side where we will put the positive charge) and +60 µC on the other (from $Q=CV$) then it follows that the total of redistributed charge is 80 µC as you correctly ...

Only top voted, non community-wiki answers of a minimum length are eligible