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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: $$\begin{cases} \vec{\nabla}\cdot\vec{B} = 0,\\ \vec{\nabla}\times\vec{E} + ... 6 When there is a time-varying magnetic field, the electric field is non-conservative and therefore cannot be written in the form \mathbf{E}=-\nabla V. 3 A field is a field: not a force. A force describes the effect of the field on a specific "object" (see below) coupling to the field. In the current state of research, quantum fields (such as the photon field for electromagnetism, see Virtual photon description of B and E fields) are fundamental structures that cannot be decomposed or explained in simpler ... 3 As stated by Lemon, electric flux through a volume enclosed by a closed surface is zero when the volume contains no net charge. Electric flux through a closed surface \rm S is$$\Phi= \int_{\mathrm S} \,\mathbf E\cdot \mathbf n\,\mathrm d^2 \mathbf r\;.Now, according to Divergence Theorem, \begin{align}\int_{\mathrm S} \,\mathbf E\cdot \mathbf ... 2 The equation you provided is actually given byW=\frac{\epsilon_0}{2}\int E^2\,\mathrm d\tau$$which is the energy stored in an electric field. This energy is utilized by the charge to generate it's field of influence or it's electric field. It's dependent on the magnitude of charge and the distance of separation between the charge and the point of ... 2 Boundary conditions of electromagnetic waves are well looked into because of the nature between polarization and reflection. However in this particular case it is rather simple, it is the evaluation of a contour integral over the boundary. This is done by summing the length elements of the pill-box shape that is typically used in this analysis in the correct ... 2 This sort of calculation, especially when the speed of the electrons from an observer's point of view is close to c, has to be done using special relativity, in that sense that the transformation between reference frames is determined by Lorentz rather than Galileo transformations. As you mentioned, you can put your reference frame origin at one of the ... 2 Lorentz Transformations Suppose we call the lab frame the K-frame and a frame moving at velocity, \mathbf{v}, relative to the K-frame called the K'-frame. Then we can express the electromagnetic fields in the K'-frame in terms of the K-frame fields as:$$ \begin{align} \mathbf{E}' & = \gamma \left( \mathbf{E} + \boldsymbol{\beta} \times \mathbf{B} ...

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

A closed surface like a sphere encloses some volume. Anything coming out through the surface (the net outward flow which we call the flux) will be in the expense of what remains inside. If the sphere encloses some charge, then electric field diverging out from the volume containing the charge will be equal to the normal component of the electric field lines ...

1

how the direction of the force on slab in both situation differs? Recall that the energy stored in a capacitor is given by $$W = \frac{1}{2}CV^2 = \frac{1}{2}\frac{Q^2}{C}$$ where $V$ is the voltage across the capacitor and $Q$ is the magnitude of the electric charge on either plate. For the case that a constant voltage source (e.g., battery) is ...

1

I guess you mean $$V=\frac{1}{r}$$ instead of $V=1/x$ (if $\partial_y V=\partial_z V=0$ all your integrals diverge). If this is right, then $$V=\frac{1}{r}\Longleftrightarrow \rho\propto \delta(\boldsymbol r)$$ and not $\rho\propto \frac{1}{r^3}$. Anyway, your example $V=1/r$ is highly pathological: it diverges too fast at the origin (the integrals blow up ...

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

When doing electrodynamics, you don't really consider a finite distance at which field lines extend - you consider the behavior of the field as it extends to infinity. In other words, when you have a potential that's dependent upon a distance r from the origin, you take the limit as $$\lim \phi (\textbf{r})\rightarrow\infty$$ and see what happens (e.g. it ...

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The range of the Coulomb force is infinite (the force between two charges $Q_1, Q_2$ separated by a distance $r$ is given as $F = \frac{Q_1 \, Q_2}{4 \pi \epsilon_0 r^2}$), with the implication that the photon has zero (rest) mass. However if you were to suddenly create (say) a positive then the "news" about this would travel at the speed of light, so that ...

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You will get infinity because in addition to $kQ_1Q_2/d$, it also includes the self-energy of the two point charges, which is infinity.

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The electrons follow the field as far as they may and as far as there is any field. Be aware, that there is no such thing as "positive field" as you wrote. The field is a vector and has a direction. Consider first an easier setup: if the external charge would be inside the conductor, the electrons would come and surround it and cancel it completely, there ...

1

The potential energy possessed by the system of two charges is $$U=\dfrac{Q_1Q_2}{4\pi\varepsilon_0r}$$ You can rearrange the equation into$$U=\frac{1}{2}\dfrac{Q_2}{4\pi\varepsilon_0r}Q_1+\frac{1}{2}\dfrac{Q_1}{4\pi\varepsilon_0r}Q_2$$ which then becomes$$U=\frac{1}{2}Q_1V_1+\frac{1}{2}Q_2V_2$$ where $V_1$ and $V_2$ are respectively the potentials created ...

1

On a local level, when you measure the electric field in matter (conductor or otherwise) there is a rapid variation both in time and space. Semiclassically, the electric field get's very large as you approach the nucleus, then drops off, the electrons are moving around the nucleus at very large speeds. So, yes, you are correct when you say there will be ...

1

For a charge distribution $\rho({\bf r'})$, the electric field at ${\bf r}$ is $${\bf E}({\bf r})=k\iiint\frac{\rho({\bf r'})}{|{\bf r}-{\bf r'}|^2}\hat{\bf R}d^3{\bf r'}$$ where ${\bf R}={\bf r}-{\bf r'}$. I think the OP's claim is at positions where $\rho \ne 0$, the above integral is infinite because the integrand blows up at ${\bf R}={\bf 0}$. I think ...

1

Your thinking is correct. At least, if the sphere is made up of small point charges, then the field will be infinite as you approach them. There is a point you are missing when you say that this contradicts Gauss' law: Gauss' law only gives you the flux of the field. To get the field of the sphere from it in the textbook-way, you have to use symmetry. You ...

1

You just use vector addition and Newton's Second Law. For example, if you have $$\overrightarrow{F}_E=qE\hat{x},\space\space\space\overrightarrow{F}_B=qvB\hat{y}$$ then your total force $F_{tot}=F_E+F_B$ is just $$\overrightarrow{F}_{tot}=qE\hat{x}+qvB\hat{y}$$ Since $\hat{x}$ and $\hat{y}$ are totally linearly independent, these terms cannot be combined. ...

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