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1

I'll briefly address several of your questions but do consider breaking up these questions into separate posts if you want greater detail. The very basic question that I have is whether positive ions flow in a battery? Inside a battery (cell) is an electrolyte that reacts with the electrodes, removing electrons from one and adding electrons to the ...


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1) Different batteries may operate differently. But yes, it may be much different from a wire. Wires are solid materials. In a solid, the atoms (and therefore the positive nuclei) don't move. So the only charge carrier in the wire are the negatively charged electrons. But a liquid electrolyte can have positive ions inside, and those ions can move in ...


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Regarding questions one and two, I recommend that you search up how a battery works. But in essence, ions flow in the battery, whilst electrons are routed through the wire to do useful work. This is possible through the chemical energy contained in the battery. Instead of uttering "potential difference" as one phrase, I find it helpful to emphasize ...


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OK, I will try to say something a bit more useful than my other response. I am still on the skeptical side of getting a closed-form, simple solution to this problem, specially using separation of variables. I think that the problem is that the potential outside of a square with that border conditions cannot be attacked using separation of variables. My ...


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In the region between two spherical plates, we have, $$\begin{array}{l} \nabla \times E = 0\,\,\, (1)\\ \nabla .\left( {\varepsilon \left( \theta \right)E} \right) = 0\,\,\, (2) \end{array}$$ The first equation leads to the definition of scalar potential $\Phi$, i.e. $E = - \nabla \Phi $. Therefore, from Eq. (2) we have, $$\nabla .\left( {\varepsilon ...


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If we ignore the inner section, we have a box with 3 sides held at V = 0 and the top edge at V = V1. I'm pretty sure this is easily solvable by separation of variables using an oscillatory solution in x with a decaying solution along y. Using superposition we can then treat the inner box as a separate problem of similar geometry/boundary conditions. The ...


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You have quite a discontinuity in the potential in two of the corners of each of the squares. I am not speaking only of the shape, but the value of the potential is discontinuous. We usually assume that the fields can have discontinuities, but the potential is always continuous. I suspect there is no solution.


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I think you are confused between potential difference and energy. The energy used up can vary, if the current flowing through the resistor varies. First of all, a battery is 'not' a constant energy source. It's a constant potential source. Secondly, if more current flows, more energy is dissipated, even though the per capita energy is constant. Now, to ...


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Happy birthday. Once upon a time, there was a king and he had a sister who also liked coffee... Or let me omit this portion of the answer. A battery is motivating the electrons to buy a round trip ticket around the circuit by the voltage $V$. The voltage is nothing else than the energy $E$ per unit charge $Q$, i.e. $V=E/Q$; the unit 1 volt is nothing else ...


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Consider for a moment, a cell that is not connected to a circuit, i.e., there is no path for current external to the cell. The chemical reactions inside the cell remove electrons from the cathode and add electrons to the anode. Thus, as the chemical reactions proceed, an electric field builds between the anode and cathode due to the differing charge ...


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You can think about "solving X equations in Y unknowns". When $X>Y$, you generally expect infinite solutions, when $X=Y$ you generally expect a unique solution, when $X<Y$ you generally expect no solutions. This kind of statement is not always mathematically rigorous but you can usually argue it rigorously in specific circumstances. Pick an energy $W$ ...


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For 1D potentials, the sequence of bound state energy eigenvalues $E_n$ cannot grow faster than what happens in the case of an infinite well, i.e. $E_n$ cannot grow faster than $n^2$.


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Since Hamiltonian is a linear operator, one could always choose real eigenfunctions. This is because if f is a complex eigenfunction of any linear operator O with eigenvalue e, then its both real and imaginary parts are also O's eigenfunction with the same eigenvalue e. It does not require the operator to be Hermitian. Moreover, this is true for all ...



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