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5

Given your question, it seems likely that your misunderstanding comes from a limited sense of vectors, fields, and partial derivatives. So there's a lot of education that we have to cover in a very short time. Multivariate functions When we transition from a function $f(x)$ to a field, which is a function of many variables $f(x, y, z)$, we suddenly have ...


1

The work-energy theorem leads us to the following result; \begin{equation} \oint \vec F\cdot d\vec s=0 \end{equation} \begin{equation} \oint \vec F\cdot d\vec s=\underbrace{\int \int }_{\text{surface}}(\nabla \times \vec F)\cdot d\vec n \end{equation} Using the rules of vector calculus there must exist some scalar function such that; \begin{equation} \vec ...


1

Let $F$ be a force field. Assuming that the force field is a conservative vector field, then it follows that the line integral of the force field is zero $$\oint_{O} F \cdot dr = 0$$ The del operator $\nabla$ is defined in 3 dimensions as $$\nabla = \langle\frac{\partial}{\partial{x}}, \frac{\partial}{\partial{y}}, \frac{\partial}{\partial{z}}\rangle$$ ...


6

On spherical coordinates, the gradient of a general function $V$ is: $$ \nabla V = \frac{\partial V}{\partial r}\mathbf e_r + \frac{1}{r}\frac{\partial V}{\partial\theta}\mathbf e_\theta + \frac{1}{r\sin\theta}\frac{\partial V}{\partial\phi}\mathbf e_\phi $$ If $V(r, \theta, \phi)$ only depends on $r$, that is $V = V(r)$, which is exactly the case of the ...


0

I think your error is in assuming that $E_{n+1} - E_{n}$ is proportional to $n$. At least, I assume you assumed it; it's the only way I can see that you could go from the statement $$ E_{n+1} - E_n \propto E_n^{-1/2} $$ to the statement $$ E_n \propto n^{-2}. $$ Really, what the first proportionality above implies is that $$ \frac{dE}{dn} \propto ...


0

The electric poetential in a place is the potential energy per unit charge that would be placed in that position. So, for a particular charge, a difference in electric potential between two places, means a difference in potential energy in those two places. Now, as everything else in this universe, a charged particle tends to minimize it's potential energy ...


0

The two spheres share the same potential if and only if they are in equilibrium, which in general, as you pointed out, does not have to always hold. In fact, if you take any two spheres and connect them with a wire, they will not share the same potential. They eventually will when all the charge carriers have flown from one to another according to the ...


1

Think of a gas. We will ignore gravitational potential, and we'll consider the situation to be at steady state. Compressing the gas takes energy, so we conclude that the entire container of the gas will be at a constant level of compression - in this case, constant density as well. You may intuitively understand that this is true no matter how oblong the ...


1

The E field inside the conducting wire is 0, so what is it really doing? The potential difference between two points is related to the electric field along the path between them: $$V_{ba}=\int_a^b \vec{E}\cdot{}\vec{dl}$$ So the fact that the high-conductivity material forces a (near-)zero electric field is exactly why the two ends of the conductor ...


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See. If there is some potential difference between the 2 ends of a wire the electrons will flow in a direction where there is less potential difference. Hence the potential difference will be uniform throughout the conductor.


2

I don't want to be too much precise, but the Schrödinger equation ($i \dot{\psi}= H\psi$ to avoid confusion) has at most an unique solution under very general assumptions on the Hamiltonian operator $H$, even if you see it as a liner equation in the more general setting of Banach spaces. In particular, it is not a priori necessary that $H$ is self-adjoint ...


0

EDIT: Based on the comments below I agree that my original answer is incorrect, I wrote it too quickly. I'll leave the original post beyond the horizontal line. I'll make some corrected comments, but maybe it's better if someone else writes an answer. Corrected answer: The basic idea of the method of images is to find a solution to the original BVP by ...


0

A potential is essentially an integral of work to carry a particle (magnetic or electric one) from an infinite distance up to the point you need to know the potential value. If the force making the work is not a Dirac's Delta, then the integral (i.e. the potential) must be a continuous function.


0

This equation is supposed to be used when the boundary integral goes to $0$ at infinity. See Helmholtz decomposition. In your case you are implicitely considering that your $\vec{B}$ field goes faster to $0$ than $\frac{1}{r}$, but you get a divergence, thus basically showing that this is not the case. In conclusion, you cannot use this formula.


0

Generally speaking, bound states in a system give rise to resonant behaviour. The number of states defines the number of different resonances that can be observed. For example, consider the absorption of photons by an electron trapped in a quantum well: If the well only contains one bound state then it could absorb any photon with any energy greater than ...


0

Gigi Butbala's answer in terms of sines and cosines is perfectly correct. But just to prove that it can be done, here's how you'd proceed from your two equations in terms of $A$ and $B$: The first equation shows that you must have $$ B = - e^{-ikL} A $$ and plugging this in to the second equation yields $$ A e^{ikL/2} - A e^{-3 ikL/2} = 0 \quad ...


1

You should look at an electric circuit a like a river or a water slide attraction in an amusement park (see my little artist impression below). The resistors are the steep parts: that's where the potential energy is lost. The wires are the horizontal parts, so there no potential energy is lost. But as the water is already moving, it doesn't stop moving in ...


1

There is a small misprint in the third line (the 3rd term). If you try to differentiate the second part, you will get exactly part of the force: $\frac{d}{dt}\frac{\partial}{\partial \dot{r}}\left(\frac{1}{r}+\frac{\dot{r}^2}{c^2 r}\right)=\frac{d}{dt}\left(\frac{2\dot{r}^2}{c^2 r}\right)=\frac{2 \ddot{r}}{c^2 r}-\frac{2 \dot{r}^2}{c^2r^2}$ And also your ...


0

Technically you are correct and there is a voltage drop with distance but since the voltage drop is current * resistance and your resistance on a PCB track is usually measured in the milliohms it's ignorable unless your doing something really hairy with LOTS of current or ultra sensative.


0

There is always a scalar potential $V$ and a vector potential $\vec A$. The vector potential always determines the magnetic field $\vec B$ but if your magnetic fields are not changing in time then you don't need the vector potential to determine the electric field $\vec E.$ You can get $\vec E = -\vec \nabla V -\frac{\partial \vec A}{\partial t}$ and $\vec ...


0

Whenever there is a battery connected in a circuit we assume that the positive terminal of the battery is at a higher potential then the negative terminal by convention. Also by convention + charge flows from positive to negative terminal .The driving force here is the potential difference. Correct Electrons flow in the opposite direction. ...



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