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1

Thanks laying out your work so neatly in the question. I think the solution is the following $$\Delta KE= \int_{r_a}^{r_b}{ KQq \over r^2} dr$$ where $r_a$ is the initial position and $r_b$ is the final position (and I have added $q$ as the charge of the point charge). so, for example, if the point charge goes from $r$ to $2r$ we have two positive ...


-1

Even in Classical Mechanics the vector potential is defined in the whole space, i.e., it is a field, because it should be known wherever the charge in question goes. The charge may go everywhere we direct it with our initial conditions. In QM, where there is a wave function equation for a wave determined everywhere in space, the vector potential should also ...


-1

I agree with both your answers, but let me add one more answer, quite direct. One of the most fundamental operators needed in the quantum mechanics is the Hamiltonian, on which we base the Schrodinger equation. It is associated to the energy of the system, kinetic + potential. For the interaction between the e.m. field and an electron (neglecting the ...


1

This seems like a reasonable homework-like question, so I'll provide a hint. Realize that, for $x>0$, both the simple and semi oscillators have the same potential $V(x)$. For the simple oscillator, draw out the first few of the energy eigenfunctions $\psi_n(x)$. Now for the semi-harmonic oscillator, think about what the boundary condition on any ...


0

I would suggest finding the electric field due to the three rods and integrating that to find the potential difference. That way you would not need a reference point, since all you are finding is a potential difference. Keep in mind that the wires have a diameter $a$, so that when integrating the electric field, you should approach each wire no closer than ...


0

What you describe is usually called regularization, as distinct from renormalization, although the terms are related. It could help to cite the paper that you are reading, but in any case, it often happens that long wavelength physics do not depend exactly on the details of short distance physics. For example, if you are scattering a particle off of a ...


0

1) If the n and p doped regions are externally connected using a perfectly conducting wire, why will not any current flow? In thermal equilibrium no current can flow if one connects the two sides of the junction using a perfectly conducting wire. The built-in potential existing at the junction will remain the same, drift and diffusion currents will ...


2

In the simplest sense of it, the Free Energy is the heat of the system minus the compulsory heat loss due to entropy. So, in short, it is the amount of "energy" left over in the system, after we consider losses due to entropy. So basically some amount of heat is wasted, and the remaining amount is useful. And this remaining amount is the Gibbs Free Energy. ...


0

What's your problem? How do you pass from a certain vector |Q> expressed according to some base of vectors {|$B_i$>} to the same vector |Q> expressed according to another base of vectors {|$C_j$>}? Let's see it together. Take for example (1) |Q> = $∑_i$ $a_i$ |$B_i$>, where $a_i$ are the amplitudes. For passing to the new base you project |Q> on each ...


5

Sanaris's answer is a great, succinct list of what each term in the free energy expression stands for: I'm going to concentrate on the $T\,S$ term (which you likely find the most mysterious) and hopefully give a little more physical intuition. Let's also think of a chemical or other reaction, so that we can concretely talk about a system changing and thus ...


2

There are two forces inluencing the spontaneity of a reaction: (1)The tendency of a system to attain a state of minimum energy and maximum orderedness, or stability. (2)The tendency of a system to attain a state of maximum energy and minimum orderedness , or entropy. If a system attains maximum stability, it attains mininum entropy; and if it attains ...


9

Short answer: Gibbs free energy $G = U + PV - TS$ combines internal energy $U$, pressure $P$, volume $V$, temperature $T$, and entropy $S$ into a single quantity that measures spontaneity. With that, I mean that processes that lower the Gibbs free energy of your system will spontaneously occur, and equilibrium is reached when the Gibbs free energy reaches ...


12

First, you have system with some energy, named $U$ by physicists. You think you have all the information you need to characterize the system but then some guy comes near and says: "Whoa, that's bad, the volume of your system can change." You say: "No problem, we just add here $pV$. Our new energy is $H=U+pV$." "But hey," they say, "your temperature can ...


1

is a thermodynamic potential that measures the "usefulness" or process-initiating work obtainable from a thermodynamic system at a constant temperature and pressure (isothermal, isobaric). Just as in mechanics, where potential energy is defined as capacity to do work, similarly different potentials have different meanings. The Gibbs free energy is the ...


1

Your boundaries are at $r=a$ and $r=b$. Notice that the potentials at these two surfaces are independent of $\theta$ (they are spherically symmetric). Look at a list of the first few Legendre Polynomials $P_{l}(\cos{\theta})$. For what value of $l$ does $P_{l}(\cos{\theta})$ not depend on $\theta$? Further, notice that $V(a) = V(r=a,\theta) = V_{0}$, and ...


2

Consider the following proof by contradiction: Suppose you find a state with $E < -V_0$, then its kinetic energy becomes negative at every point $x$ (classically the velocity is imaginary at every point), which means the whole wavefunction (at all $x$) is an evanescent (exponentially decaying) wave, but such a solution is not a physically stable solution ...



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