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$$\Delta E=\Delta P+\Delta K$$ Where the change is from the past/present to the theoretical future. The only assumption is that "Least Action" is taken and nothing blacks the Kinetic/Potential Energy from changing. In other words, the change is over all interactions. That's why potential energy can be set at an arbitrary reference. We are looking for the ...

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Complex analysis is very useful in potential theory, the study of harmonic functions, which (by definition) satisfy Laplace's equation. One way to see this connection is to note that any harmonic function of two variables can be taken to be the real part of a complex analytic function, to which a conjugate harmonic function representing the imaginary part of ...

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I think what Griffiths meant was that "Since Laplace's equation requires, on the contrary, that the sec­ond derivative is zero, it seems reasonable that solutions should exhibit no extrema." But it does not necessarily mean that a function cannot have a null second derivative in an extreme, since the $x^4$ function has a minimal value at the same point ...

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( I turned my comment into an answer to signal that this is question is answered) if in doubt (and unexperienced) work using Cartesian coordinates. That being said: You simply miscalculate the gradient. $F_c$ is not directed in the direction $r$ but $r−x$. If you want to compute the Gradient in spherical coordinates you have to expand the absolute value in ...

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The atom has some charge distribution $\rho(r)$. We don't don't know what form the function $\rho(r)$ has, but we do know it depends only on $r$ because an atom is spherically symmetric. When you have a spherical charge distribution the potential at a distance $r$ is simply due to the total charge inside the distance $r$: $$V(r) = ... 0 You are mistaken about your fundamentals. Your initial assumption was that the pd across the resistors was nE not the whole circuit. You implicitly assumed the difference across the whole circui was 0 when you connected the components in a closed circuit. One of Kirchoffs law says the sum of the voltages in a loop in 0. 2 A particle subjected to only a conservative force field, without other restrictions, will move in a direction which reduces the potential energy of the system. The particle itself does not have potential energy; the particle-field system has potential energy (some may say the field-field system, but let's not nit-pick that yet). So what's important is not ... 0 Yes the diagram in the book is not to scale. It should look like this: Each transition increases by 1V. The outer edge would have a potential of 1V, the transition between green and yellow would have a potential of 2V. The final transition has a potential of 6V. 0 Should it? Lets say the 6V mark is at distance at 1, then 4V is at 1.5 and 2V is at 3 meaning that 4V should be in in the middle between the origin and the 2V mark, which could work as far as i can tell on my screen 1 However it did pass within Δx of the electron. The Δx is not the difference in space with the electron, as the electron is bound to a nucleus with a potential simulated by "an infinite potential well" . The Δx is related to the whole system, from the center of its mass as a possible location to start with. So the problem is : "photon + atom" as a ... 1 Both the zeroes are "real" to answer the question. There is absolutely no problem with multiple points in a space having the same electric potential. Note that we have ASSUMED the potential to be zero at infinity and based upon that assumption we have found out the potential to be again zero at the equidistant point between two opposite charges. So there ... 2 You are imagining the particle in the well as a classical system i.e. a point particle moving to and fro in the well. However this is not a good description of the system. A quantum particle does not have a position. By this I mean that it is meaningless to ask what the position of the particle is because position, in the sense we normally use the term, is ... 0$$\int_{-\frac{d}{2}}^{\frac{d}{2}}\psi^*(x)\psi(x)dx$$is the probability of finding particle between -\frac{d}{2} and \frac{d}{2}. Expectation value is :$$\langle x\rangle=\langle \psi|\hat{x}|\psi\rangle$$|\psi\rangle is the summation of probability amplitude times given basis kets$$|\psi\rangle = \sum_i c_i |{e_i}\rangle$$... 4 The definition of the expectation value of an operator A is $$\langle A\rangle=\int{\psi^* (x) A(x) \psi (x) dx}$$ (because it represents "the value of the variable" A(x) times "the probability of being in that configuration" P(x)=\psi^* (x) \psi (x)) and for the particular case of the expectation value of the position ... 2 So the principle that systems find themselves at an energy minimum needs to be heavily qualified: it is a consequence of ever-present friction/drag forces having a sign opposite to your velocity, so that the power wasted due to friction is terminally negative. The principle therefore states, "when a system only has two forces, one due to drag and the other ... 0 Use the definition that, if a potential exists, then it is given by \phi(x,\,y) = -\int_{(0,\,0)}^{(x,\,y)} \vec{F}\cdot\mathrm{d}\vec{x} and the result must be independent of the path between the endpoints in the path integral. So, let's integrate along the straight line \vec{x} = t (x\,\hat{i} + y\,\hat{j}), then, with \vec{F}=\hat{i}+2\hat{j}, we ... 1 You have to integrate (i.e. find the primitive) in dy as well, and then compare derivatives of both equations:$$ f(x,y) = \int 1 d x = x + g(y)  f(x,y) = \int 2 d y = 2 y + h(x) $$now compare derivative terms$$ \frac{\partial f}{\partial x} = h'(x) = 1 $$and$$ \frac{\partial g}{\partial y} = g'(y) = 2 $$therefore$$h(x) = x + c_x  ...

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The amount of work done by unit charge between any two nodes of current carrying circuit is called the potential difference between those nodes. The amount of work done against the electric field by displacing (without acceleration) a unit test charge from one terminal to other terminal in an open circuit is called the electromotive force. Obviously when ...

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