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Yes, we can define a magnetic scalar potential in some problems, specifically if the current density vanishes in some places. Note that the condition is not $\nabla \cdot B = 0$ since this is always true. To define the magnetic scalar potential requires that there be a quantity whose curl is zero (curls of gradients are zero), which is to say $\nabla \times ... 3 The EMF created by a changing magnetic field is not considered to arise from a potential. This can easily be seen because when there is an emf, a charge can move around in a complete circle and dissipate energy the whole way around, but a potential cannot drive a charge around in a circle, because potentials are conservative. The two pieces of the electric ... 3 There will not be any surface charge on the inside surface! The charges will all migrate to get as far as possible from each other. That puts them all on the outside surface. 2 One thing that goes wrong here is that the Hamiltonian is no longer a self-adjoint operator acting on these singular wave-functions. Let's focus on the kinetic energy portion of the Hamiltonian: $$\hat K = - \partial_r^2 - {2 \over r} \partial_r = -{1 \over r^2} \partial_r r^2 \partial_r$$ To prove self-adjointness, consider the natural inner product on ... 2 1) Why the potential at the surface? This approach is probably used because part (a) of the problem gives you an explicit expression for$V_S$and so the expression for$V(r)$is self-contained without having to consider what happens inside the shell. Basically the solution makes use of the fact that$$V_S = \int_0^{r_2}{E dr} = \int_0^{r}{E dr} + ... 2 I concur with Sam Weir, typo.$E= \frac {b}{x^2}$So$\frac {6kv/m}{(1 m)^2} =6\,\mathrm{kV}$at$1\,\mathrm{m}$and$\frac {6kv/m}{(2 m)^2} = 1.5\,\mathrm{kV}$at$2\,\mathrm{m}$I see$6\,\mathrm{kV}-1.5\,\mathrm{kV}=4.5\,\mathrm{kV}$with$x = 1$being higher potential, regardless closer to the positive charge is going to be higher potential. 2 how do you find potential in a place where we have no intuition of force and are not allow to find it. Well I think this might be your problem; I've certainly never heard it said that you are not allowed to find forces. The Euler-Lagrange equations are simply another tool to finding the dynamics, but that doesn't mean you have to start from scratch and ... 2 We don't control the allowed energies$E_i$independently of the potential: the energies must be the eigenvalues of the Hamiltonian. The "inputs" are the shape and height of the barrier between the two wells. You can kinda sorta think of the energy difference between the symmetric state (with energy$E_1$in your diagram) and the antisymmetric state (with ... 2 A double well with a high or wide barrier will have a smaller$\Delta E=E_2−E_1$than one with a low or narrow barrier. (Less coupling.) I think we can understand this intuitively as follows but first it has to be said that rob is right: the energies$E_i$are NOT inputs but the eigenvalues of the Schrödinger equation. Width, height and potential of the ... 2 Perhaps the easiest way to see that there can't be a potential difference between$A$&$B$is a symmetry argument. You're tempted to say that$A$is at a higher potential than$B$so that current will flow from$A$to$B$. But continuing along the loop, I find that current must also flow from$B$to$A$, which would lead me to conclude that$B$is at a ... 1 Here, there is a time varying magnetic field at work, and it's flux through the given loop changes. Thus there is a non-Electrostatic Field induced along the wire, proportional to the rate of change of the flux. However, it is a non-electrostatic field", for example the closed integral over a path for this field isn't zero. Also, the line integral of this ... 1 This is a very simple problem to tackle. The points A and B are connected by a conducting wire and nothing else. Therefore, there is no potential difference from A to B. The magnetic field is a red herring. We are not told otherwise, so we assume the wire is an ideal conductor. That means there is never a potential difference across it. Remember Ohm's Law ... 1 In this case, the battery is said to be "floating". Its potential with respect to earth can be suprisingly high or low. Small buildups of static electricity on the battery can easily charge it to hundreds or thousands of volts with respect to earth. The voltage difference across the battery's terminals is still$1.5\,\text{V}$, but the voltage of the ... 1 The xylem actually creates a long thin string of water from the roots to the leaves of a tree. This string remains continuous by two forces. One is the cohesion of water molecules and the other is adhesion of water with xylem walls. Now in leaves water is evaporated. That decreases the pressure of water there. Because the protoplasm of those cells on leaves ... 1 Let me arrange some information briefly. Cohesion-tension theory : phenomena that pulls water molecules at leaves producing tension + cohesion along entire stream of water molecules. Cohesion of water molecules mainly arises from high-strength hydrogen bond and the tension that presses the stream is generated from various mechanisms. On the other hand ... 1 The first sentence started with an if. When you start with an if and end with a problem a solution you should consider is that your if never happens. So if the EMF around a zero resistance loop is zero then we don't expect the total magnetic flux through to change. Is that reasonable? Yes. Since it is a zero resistance loop, it can generate any current it ... 1 In case of varying magnetic fields the electric field is considered as nonconservative. Thus, it is simply not possible to define a potential for induced electric field (See also this post with answers). Alternatively, one can think of the induced current as charges moving in the circle of radius$r$. Through every part$r \,d\phi$flows the same amount of ... 1 If I am correctly interpreting your question, you have a 1D time independent S. equation with$V$which is discontinuous in some points. You solve that equation for a fixed eigenvalue$E$separately in each continuity interval obtaining functions$\psi_E$which are$C^2$in every open interval and depends on two arbitrary constants. Finally you mach the ... 1 The problem is considering an incoming right-mover for$x<0$and asks how it scatters off a step potential into a reflected outgoing left-mover for$x<0$and a transmitted outgoing right-mover for$x>0$. The last possibility -- an incoming left-mover for$x>0\$ -- is not present in this scattering experiment. That's the answer to OP's question. ...

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Just so this doesn't slip past: Now I think if one side of the potential barrier is higher, the particle will be more likely to spend time on the left side than the right side, so the wavefunction should have higher amplitudes on the left (skewed to the left): This is incorrect. Between A and B the well is deeper, so the particle goes faster. Between B ...

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I'll somewhat expound the solution by qmechanic, and give two examples. The solution is closely related to Landau's method for period inversion, see Mechanics Sec. 12. The quantum number can be approximately computed as \begin{align} (n + \delta) \, 2 \pi \hbar &= \oint p \, dq \\ &= \oint \sqrt{2\,m\, [E - V(q)]} \, dq \\ &= 4 \int_0^{r_\max} ...

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