# Tag Info

1

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 ...

0

Comments to the question (v3): I) The Gurtin-Tonti bi-local method [which OP mentions in an example; see also Section II below] of pairing opposite times $t\leftrightarrow (t_f-t_i)-t$ (hidden inside a convolution) is an artificial trick from a fundamental physics point of view, unless further justified. Why would such correlations into the past/future take ...

1

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 ...

0

Periodic potentials necessarily imply periodic fields, so this would restrict the physical situations your field would be applicable to. That said, any set of potentials will give some physically viable set of electric and magnetic fields (though the charges and currents might be difficult to arrange, and the fields may or may not have finite energy). You ...

1

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 ...

1

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 $a$. ...

0

The electrostatic potential is a potential just like any other mechanical potential. Since the force on a particle with charge $q$ is $\vec F_\text{stat} = q\vec E$, the potential $\phi'$ that gives the force as $\vec F_\text{stat} = - \vec \nabla \phi'$ is just the electrostatic potential (the "voltage") times $-q$. The electrostatic force is a perfectly ...

1

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 ...

1

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 ...

0

Let me assume for simplicity that we have some big charge(s), and we bring a positive test charge. So, there is a field between the big charge(s) and the test charge. The field encapsulates energy. (For simplicity I will also assume that the big charges are placed on massive bodies.) Now, for out test-charge there are two situations of passing from one ...

0

If you move an object, then work $W$ is done when ever you use a force $F$ to move something over a distance $d$. $$W=F \cdot d$$ The electric field that defines the potential difference is a force (per charge) that pulls (or pushes) in the test charge you put in. A force is moving the charge, so work is done. If a force does work on an object, you have ...

0

At a fixed time $t$, the wire is the line $(0,vt,z)$ where $z$ can be any value and $s$ can be your instantaneous distance to this line. Specifically, we can consider $\vec{A}(x,y,z,t)=-\frac{\mu_0I}{2\pi}\ln s\hat{z}$ where $s$ is nothing more than a shorthand for $\sqrt{x^2+(y-vt)^2}$. Thus, taking the quasistatic approximation, $$\vec{E} = ... 0 It is easier to write the charge condition as$$\frac{\partial \phi(r\rightarrow R_-,\theta)}{\partial r}-\frac{\partial \phi(r\rightarrow R_+,\theta)}{\partial r}=\sigma(\theta)$$and start by calculating the operator on the left. This is straightforward:$$ \frac{\partial \phi(r\rightarrow R_-,\theta)}{\partial r} = \frac{\partial }{\partial ...

2

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 ...

0

V is energy per coulomb, or in other units, energy per electron. Here energy is potential energy that comes from pushing electrons near other electrons against an electrostatic force. Usually V is the potential difference between two points. Potential arises from a set of charges, typically a charged object. The charges fill space with an E field. V is ...

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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 ...

-1

From http://dev.physicslab.org/Document.aspx?doctype=3&filename=Electrostatics_ContinuousChargedRod.xml, the electric field is: \begin{align}E_p&=\frac{Q}{4\pi\epsilon_0}\left(\frac{1}{b}+\frac{1}{L+b}\right)\\ &=\frac{Q}{4\pi\epsilon_0}\left(\frac{1}{0.5}+\frac{1}{2}\right)\\ &=\frac{2.5Q}{4\pi\epsilon_0}\\ \end{align} Now, for the ...

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 $$\rho(\mathbf ... 0 This comes from the idea that the action S=\int d^4x\,\mathcal{L} is a scalar. From planck's law it is easy to see that the energy E=\hbar \omega has the dimension of inverse time. From de-broglie relation you can see that momentum has the inverse dimension of space. Hence \mathcal{L} should have a dimension of energy {[\mathcal{L}]=E^{4}}. This ... 0 Everything up until you say "Physically speaking" is fine. The first equality:$$V=-\int_a^b \vec{E} \cdot d\vec{\ell},$$holds because that is the definition of the potential for electrostatics. The second equality:$$-\int_a^b \vec{E} \cdot d\vec{\ell}=\int_a^b \vec{f}_s \cdot d\vec{\ell},$$holds because (inside the battery, where we are integrating), we ... 2 It is not clear to me whether the total charge is also scaled, or you want that to be conserved. So I will work under the assumption of the former case and show that the potential scales as n^2. The new density \rho' is related to the old density by$$\rho'(\mathbf r) = \rho(\mathbf r/n).$$Denoting by \Omega the support of \rho, we have$$V_0 = ...

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