# Tag Info

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In this case it is more accurate to say that V is proportional to Q because the potential, V, (at some distance from the charged insulator) is dependent on how much charge, Q there is on the charged insulator setting up the electric field. It is not the source (charged insulator) that acquires potential--the potential we are referring to is at a point away ...

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It is not clear to me whether the total charge is also scaled, or you want that to be conserved. So I will work under the assumption of the former case and show that the potential scales as $n^2$. The new density $\rho'$ is related to the old density by $$\rho'(\mathbf r) = \rho(\mathbf r/n).$$ Denoting by $\Omega$ the support of $\rho$, we have $$V_0 = ... 2 No, that is the simplest way to solve the problem. As mentioned in the comments, this is in the absolute scale of things a very easy problem: the spherical symmetry allows you to even have an integral to calculate, and the exponential is not only exactly integrable, but easily so. If you allow general spherically symmetric charge densities$$\rho(\mathbf ...

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Your derivation of $V$ is fine, the $E$ not so good. But for the potential you did something better. In fact, you computed that potential for any height $d$ above the rod. So compute $V$ with a $d=0$, then compute it with your actual $d$ and then use the two to get an electric potential difference. Use the charge of the small sphere and the electric ...

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In the Schrödinger equation you can introduce, in principle, whatever form of potential you like. All the question is whether it allows a physical solution. About a particle in the magnetic field you can very well use the Schrödinger equation in which you introduce the interaction term $mB$, where $m$ is the magnetic dipole of the particle and $B$ the ...

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To include magnetic field into non-relativistic Schrödinger's equation, you can consider Pauli's phenomenological equation, which is a non-relativistic approximation of Dirac equation. It includes both spin and vector potential. If you drop spin-dependent term from it, you'll get an equation for charged spinless particle in magnetic field. For a spinless ...

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As you yourself have illustrated, you gain an understanding of how the energy changes with respect to the parameter $a$. For fixed $a$ you may argue that this ground state energy is just an irrelevant constant. But if $a$ is not fixed (e.g., you cite two different values of $a$ in your example) then this formula does tell you how the energy varies with $a$. ...

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One of many good reasons to know it is that you can also use the value you get for a metal, along with the dependence on the number of the energy level (usually "n") and the fact that electrons, as fermions, "pile up" in terms of their energy states, to get an approximation of the energy distribution of electrons in metals. I would second the notion given by ...

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In the ground state, both electrons are in the state with the lowest value of "n". E.g., in the case of an infinite potential well, the lowest quantum number is n=1. In this case, both electrons have n=1, but one electron is spin up and one electron is spin down, because of exclusion. The first excited state is one in which one of the electrons has n=1 ...

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If there are no charges inside the cylinder, then the potential obeys Laplace's Equation: $$\nabla^2V = 0$$ In cylindrical coordinates, that's $$\left(\frac{\partial^2}{\partial r^2} + \frac 1 r \frac{\partial}{\partial r} + \frac{1}{r^2} \frac{\partial^2}{\partial \phi^2} + \frac{\partial^2}{\partial z^2}\right) V = 0$$ Based on your notation, it seems ...

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