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## Hot answers tagged potential

15

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: $$\begin{cases} \vec{\nabla}\cdot\vec{B} = 0,\\ \vec{\nabla}\times\vec{E} + ... 6 When there is a time-varying magnetic field, the electric field is non-conservative and therefore cannot be written in the form \mathbf{E}=-\nabla V. 3 Yes, it is perfectly possible to satisfy more than one gauge condition. The easiest way to see that is to consider a static configuration of fields. In a static problem, all quantities are time independent. That means that the Coulomb gauge \nabla\cdot\vec{A}=0 and Lorenz (not "Lorentz"; it's named after a different physicist) gauge ... 2 The electric potential of a point in space is defined by the mechanical energy that it takes to get a (positive) unit charge to that point. We usually define the potential of an infinitely distant point as zero, and then the movement of our test charge is from infinity to the point for which we want to measure the potential. In case of a capacitor the ... 2 It is sometimes easier to visualise what is happening by using the idea of potential. To do this make one point in the circuit 0 V. This is a totally arbitrary choice. It is the bottom right hand corner of your circuit. Note to make the sums easier I have change the emf of the battery to 90 V so 2 A flows through the battery and 1 A through each of the ... 2 For dynamic electric and magnetic fields, there is a piece of the electric field that depends on the vector potential:$$ \vec{E} = - \vec{\nabla} V - \frac{\partial \vec{A}}{\partial t}, \qquad \vec{B} = \vec{\nabla} \times \vec{A}. $$Taking the curl of the first equation yields Faraday's Law (with the V-dependent term dropping out as you note); taking ... 2 Boundary conditions of electromagnetic waves are well looked into because of the nature between polarization and reflection. However in this particular case it is rather simple, it is the evaluation of a contour integral over the boundary. This is done by summing the length elements of the pill-box shape that is typically used in this analysis in the correct ... 1 Relation between Electric field and potential The relationship between electric field \bf E and scalar potential \varphi is given as$$\mathbf E= -\mathbf \nabla\,\varphi$$where \mathbf \nabla \equiv \textrm{gradient operator}\;. I am unable to understand from this - sign comes. It is worthy to quote from Purcell: The minus sign came in ... 1 The mass of a ball is scalar. Its potential energy is stored in its position in a gravitation field. A dipole has its potential energy in its orientation with regards to an external field. It can do work by exerting torque when orienting along the field. 1 The analogy would be voltage as the potential, so this would be analogous to height (as in the height of a rock of mass m in gravity field g). Since power (which is also energy) is Voltage x Amps, then Amperes is analogous to mg. If you want to break it down further, I guess mg = Q/sec. I'm not sure if this answers your question, but it at least gets ... 1 This time you're wrong. You are overcounting the terms when multiplying by 4. Potential energy is given for two particles at a time, so you should be multiplying by 2, which is the same as multiplying by 4 and then dividing by 2. 1 As Giorgio says, you are overcounting the terms when multiplying by 4. I would like to elaborate a little bit, just to show how this is more than just a math error. It is also a conceptual error: You mention in your reasoning "The total potential energy = the sum of the PE of each of the 4 charges." But this is false. Not only that, but there is no such ... 1 The relation is indeed correct. Let me show you why. This type of problem is solved by equating forces. Other classical mechanics problems are by equating energies or, in rare cases, momentum. The gravitational potential indeed is$$ V(r) = G \frac Mr \,. This is not directly useful for this problem. Here you rather want to look at the gravitational ...

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It's called the virial theorem: in a bound system, the average of the potential energy of a $\frac{1}{r}$-potential V is related to the average of the kinetic energy T like $\frac{\langle V \rangle}{2} = - \langle T\rangle$.

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