# Tag Info

## Hot answers tagged potential

69

Here is a circuit representing the system. $R_{wire}$ is the resistance of the section of wire between the bird's legs. $R_{bird}$ is the resistance of the bird (which you can measure by sticking the two probes of the multimeter to the bird's two feet - if the cable is insulated, you will have to add the resistance of the insulation as well). When the bird ...

30

Perhaps I can clarify what I'm trying to get at with the famous waterwheel analogy 99 years ago, Nehemiah Hawkins published what I think is a marginally better analogy: Fig. 38. — Hydrostatic analogy of fall of potential in an electrical circuit. Explanation of above diagram In this diagram, a pump at bottom centre is pumping water from right to left. ...

28

The fact is that, in the general case $$\vec{E} = -\vec{\nabla}V - \frac{\partial\vec{A}}{\partial t};$$ (signs depend on conventions used) where $\vec{A}$ is called vector potential. You can consult for example Wikipedia. Let us consider homogeneous Maxwell equations: $$\begin{cases} \vec{\nabla}\cdot\vec{B} = 0,\\ \vec{\nabla}\times\vec{E} + \frac{\... 26 The potential difference between two points on a wire carrying a current is given by Ohm's Law, V = R\cdot I. Since wires used for long-distance power transmission have, by design, a very low resistance per unit length, and the distance between the two extremities of your hands is very small (~10cm), even for large currents the potential difference is ... 25 If you have a copy of Griffiths, he has a nice discussion of this in the delta function potential section. In summary, if the energy is less than the potential at -\infty and +\infty, then it is a bound state, and the spectrum will be discrete:$$ \Psi\left(x,t\right) = \sum_n c_n \Psi_n\left(x,t\right). $$Otherwise (if the energy is greater than the ... 25 Actually, we do! It's just that it's not the same "kind" of potential - and the reason for this is that magnetic forces work differently than electric forces. Magnetic fields, if you know, do not directly exert a force on charged particles, simply for being charged. (They would exert such on hypothetical "magnetically charged particles", but we've never ... 24 All bound states can typically be chosen to have real-valued wavefunctions. The reason for this is that their wavefunction obeys a real differential equation,$$ -\frac{\hbar^2}{2m}\nabla^2\psi(\mathbf r)+V(\mathbf r)\psi(\mathbf r)=E\psi(\mathbf r)$$and therefore for any solution you can construct a second solution by taking the complex conjugate \psi(\... 24 You are using a wrong relation. The relation is not "force equals the negative gradient of gravitational potential" but "force equals the negative gradient of gravitational potential energy":$$F=-\nabla U= -\frac{dU}{dx}$$The U here is potential energy, not potential. A potential is rather a potential energy per mass. Had you used potential energy to ... 23 The precise theorem is the following, cf. e.g. Ref. 1. Theorem 1: Given a non-positive (=attractive) potential V\leq 0 with negative spatial integral$$ v~:=~\int_{\mathbb{R}^n}\! d^n r~V({\bf r}) ~<~0 ,\tag{1} $$then there exists a bound state^1 with energy E<0 for the Hamiltonian$$ H~=~K+V, \qquad K~=~ -\frac{\hbar^2}{2m}{\bf \...

23

Great question! My answer would be that in order to get a bound state, we need to have a potential that is deeper than the kinetic energy the two particles have. We have a better chance of getting a potential of the right depth to bind two nucleons if: (1) They don't repel charge-wise. Compared to nucleon interaction the Coulomb force isn't that strong, ...

23

You are correct in your reasoning. However, that line does have a voltage equal to $0$. The trick comes in from the fact that $V=0$ doesn't mean $\vec{F}=\vec{0}$. The force depends on the change in the potential, and, as you noted, it is changing; the potential just happens to have a value of 0 along that line. Think of it like an inland hill, where the ...

21

Consider the form of the potential energy between two point charges in the case that I use a reference distance $r_0$ as the zero (written here in SI units). $$U_{r_0} = \frac{q_1 \,q_2}{4 \pi \epsilon_0} \left( \frac{1}{r} - \frac{1}{r_0} \right) \;.$$ This is quite general, but it will get to be very messy to write down and manipulate very quickly indeed. ...

20

The electrical potential has a gauge freedom because we can arbitrarily set the zero anywhere we want. This is because we can only ever measure differences in the potential and not the absolute value of the potential. Conductors that are earthed are all at the same potential (because they are earthed to the same planet) so it is usually convenient to choose ...

18

To be concrete, let us here assume that the dissipative force $${\bf F}~=~-f(v^2)~ {\bf v} \tag{1}$$ has a direction opposite of the velocity ${\bf v}=\dot{\bf r}$ of the point particle. Here $f=f(v^2)$ is a function that may depend on the speed square $v^2\equiv {\bf v}^2$. Drag is of this form (1). Linear friction/drag corresponds to a constant $f$-...

18

Since the energy spectrum does not depend on the absolute position $\vec{r}=\vec{a}$ of the delta potential, we may assume that $\vec{a}=\vec{0}$. Therefore, in its current formulation (v1), OP is effectively saying that The attractive 1D delta potential $V(x) = -A\delta(x)$, $A>0$, has exactly one bound state. The same is true for the 3D delta ...

18

Loosely speaking, the gradient of a scalar field (such as the electrostatic potential) points in the direction of that field's greatest change. Since no change occurs in the field when you go along the surface, the gradient shouldn't have a component in that direction. Here is another intuitive explanation: Imagine for a moment that the electric field was ...

17

In the language of vector calculus: The word potential is generally used to denote a function which, when differentiated in a special way, gives you a vector field. These vector fields that arise from potentials are called conservative. Given a vector field $\vec F$, the following conditions are equivalent: $\nabla \times \vec F=0$ $\vec F= -\nabla \phi$ $\... 16 The$\bf E$and$\bf B$fields viewed as independent quantum oscillators contain too many DOFs, if that's what you mean. But I'm getting ahead of myself. Here is one line of reasoning: It is reasonable to expect that for a consistent quantum theory of E&M, it should have a classical limit$\hbar\to 0$where the classical electromagnetic fields are ... 15 The "simplest" classical explanation I know is the van der Waals interaction described by Keesom between two permanent dipoles. Let us consider two permanent dipoles$\vec{p}_1$(located at$O_1$) and$\vec{p}_2$located at$O_2$. Their potential energy of interaction is: U(\vec{p}_1,\vec{p}_2,\vec{O_1 O_2}) = -\vec{p}_1\cdot \vec{E}_2 = -\... 15 Consider the following cylindrical container with liquid to be rotating at uniform$\omega$: Consider an infinitesimal liquid element$\mathrm{d}m$at height$h$above the minimum of the parabola. The forces acting on it are: 1) gravity: $$g\mathrm{d}m$$ 2) the centripetal force: $$\mathrm{d}F_c=\omega^2r\mathrm{d}m$$ Consider the angle$\alpha$: $$\tan\... 15 If the charge comes from a point of V_3=0 and gains kinetic energy, then it is because the value of potential energy at the new point is lower. Then potential energy can be converted into kinetic energy. Don't worry about the actual potential energy values - only the differences in value matter. Think of having a box on the floor. You might say that there ... 14 1) OP wrote (v1): [...] and thus this leads me to believe that {\bf A} should be somehow connected to momentum, [...]. Yes, in fact the magnetic vector potential {\bf A} (times the electric charge) is the difference between the canonical and the kinetic momentum, cf. e.g. this Phys.SE answer. 2) Another argument is that the scalar electric potential ... 14 I usually find it easier to use model multipoles that are surface charges on a sphere, rather than point charges on some polyhedron's vertices. These charge densities are given in general by$$\sigma_{lm}(\theta,\phi)=N\cos(m\phi)P_l^m(\cos(\theta)),$$where$P_l^m$is an associated Legendre function (so$\sigma_{lm}$is just a real-valued spherical ... 14 In normal usage, a gauge is a particular choice, or specification, of vector and scalar potentials$\mathbf A$and$\phi$which will generate a given set of physical force fields$\mathbf E$and$\mathbf B$. More specifically, a physical situation is specified by the electric and magnetic fields,$\mathbf E$and$\mathbf B$. A set of potentials$\mathbf A$... 14 This is a standard "issue" when one solves differential equation with a separation of variable. To be rigorous, the case$\psi=0$should be handled separately. However, you can check that$\psi''=0$when$\psi=0$, and that$\psi''/\psi$is well behaved there, so you are in fact allowed to divide by$\psi$on the whole interval. But don't say that to a ... 13 Have you seen the helicopter crews that work on overhead lines approach a live wire. They extend a conducting pole and equalize their voltage with the line. After than it is "safe" to work on the line. As long as there is a conductor connecting them to the wire it is ok. Of course the initial equalization process would kill you if the arc went through your ... 13 Hanging from a power line you should be as safe as a bird. The voltage difference is between the lines (e.g. in a 3-phase system) and between the line and ground. This voltage difference exists across the insulators and pole, as well as through the air to ground. These voltage differences are obviously small enough to avoid striking an arc, hence no current ... 13 There are various ways to decide which of the assumptions are primary and which of them are their consequences but$E=VQ$may be most naturally interpreted as the definition of the potential. The potential energy is a form of energy and the potential (and therefore voltage, when differences are taken) is defined as the potential energy (or potential energy ... 13 Earth's zero potential is just an arbitrary point similar to (0,0) of co-ordinate system. It has been chosen for Engineering practices because it has very very low theoretical potential (in light with charge at Infinity) and it's easily accessible to everyone and adding charge to it doesn't change it's theoretical potential. With reference to this arbitrary ... 13 Let me try this more clearly than the other answers, which aren't wrong. You ask: So, can someone please elaborate what this EM field is with respect to$\vec E$and$\vec B$in the context of Helmholtz decomposition? There is no "EM field in the context of Helmholtz decomposition". Helmholtz just says that every vector field$\vec V\$ is decomposable as ...

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