5

Your reasoning is correct. If you store energy in a system, mass-energy equivalence provides the conversion for calculating the matter equivalent which, if you had a weighing device sensitive enough, you could (in principle) measure. For things like charged capacitors and stretched springs and pressurized gas tanks and cans filled with gasoline, the mass ...


4

Quoting Griffiths' Introduction to Electrodynamics: Where is the energy stored? Equations 2.43 and 2.45 offer two different ways of calculating the same thing. The first is an integral over the charge distribution; the second is an integral over the field. These can involve completely different regions. For instance, in the case of the spherical shell (Ex. ...


3

No. The definition of the dimension of a physical space is, informally speaking, the degrees of freedom a particle placed within has to move around. Introducing an electric field does not add more degrees of freedom; it simply associates a vector with each point of space. Assigning vectors on $\mathbb{R}^2$, for example, does not allow a particle to move ...


3

The cross term represents interference. It is the term that makes it so that the energy density is reduced when the two fields are in opposite directions and so that the energy density is increased when the two fields are in the same direction.


2

When computing the force on particle 1, you only take into account the effect of particle 2. This is true for any number of charges: if you had several charges, the force on particle 1 would be the vector sum of the individual forces caused by particle 2, particle 3, particle 4, etc. In essence, particle 1 does not exert a force on itself due to its own ...


2

Dipoles can be understood as a vector quantity located in the middle point between the charges, therefore only useful when you place that vector there and forget about the charges, but that is only possible and useful if you are far away from that point. That is you set a $\vec{p}$ and observe from $\vec{r}$ with $\vec{p} = q \vec{d}$ with $|\vec{r}| \gg |\...


2

There are two forms of energy in this problem: the kinetic energy of the slab, and the electrical (potential) energy of the capacitor. You can view the latter as being stored in the electric field. I'm not sure what you mean by "the potential of the slab". The E-field through the slab contributes to the electrical energy, but it is still part of ...


2

Discussing microscopic properties of wires within a context of circuit theory is not a good idea: the circuit theory is based on the lumped elements description, reducing the circuit to point-like resistances, inductances, and capacitors connected by ideal wires. The wires are there only to show the connections. Microscopically there are many levels on which ...


2

First of all the potential due to proton at some distance $R$ is given by $\cfrac{Kq}{R}$ not $-\cfrac{Kq}{R}$ as you have mentioned. Now since the proton would attract the electron towards itself, you could see $R$ would decrease it means, the electron moves from low to high potential


2

TLDR: the brief answer to your title question is: No, the Poynting vector represents the flow of energy in all cases including a DC circuit. There is nothing magical about DC that changes the meaning of the Poynting vector. Neither field is delivering energy outside the wire. The Poynting vector represents potential energy outside the circuit Unfortunately,...


2

The reason why we do not take the sin component and integrate is because there is a direction factor too that the integral does not take care of. It just adds up all the magnitudes, however it does not take into account that due to opposite directions they cancel out. But for the horizontal component, all the directions are the same. So you can proceed ...


2

There is no net E field inside the conductor. If there were charges would move until it was gone. The potential inside the conductor is constant, but not zero unless you choose the conductor as your reference point for zero potential.


2

It's certainly not the case that nobody has ever checked them. The issue is that by manipulating the Maxwell equations, one arrives at Poynting's theorem: $$\mathbf E \cdot \mathbf J = -\frac{\partial u}{\partial t} - \nabla \cdot \mathbf S$$ where $$u = \frac{\mathbf E \cdot \mathbf D + \mathbf B \cdot \mathbf H}{8\pi}\qquad \mathbf S = \frac{1}{4\pi} \...


2

The only way to verify the EM energy formula would be to measure its gravitational field. Feynman states this in Part II, 27.4, 2nd paragraph. Such a measurement seems very much impossible even today. Feynman has confidence in the formula because it is consistent with the Maxwell equations and the Lorentz force, and describes a conserved quantity. Feynman ...


2

In classical electrodynamics there is no way to verify experimentally that the vacuum energy density is $$\varepsilon =\varepsilon _e +\varepsilon _m=\frac{1}{2}\mathbf{E}\cdot\mathbf{D}+\frac{1}{2}\mathbf{H}\cdot\mathbf{B} \label{1}\tag{1}$$, or more generally $$\delta \varepsilon = \mathbf{E}\cdot \textrm{d}\mathbf{D}+ \mathbf{H}\cdot \textrm{d}\mathbf{B} ...


2

You're right! While Gauss's law always holds true, it is precisely for this reason that it isn't always useful to calculate the Electric Field. It is only useful when the problem possesses some symmetries that greatly simplify it. In general, before you start using Gauss's Law you must first enumerate the symmetries of the problem. The trick is to begin by ...


2

The frequency is, as always, the number of cycles per second of the oscillations. It is related to the spatial wavelength by $f = \frac{c}{2 \pi} |\vec{k}|$, where $c$ is the speed of propagation of free waves in the medium and $$ \vec{k} = (k_x, k_y, k_z) = \left( \frac{2 \pi}{\lambda_x}, \frac{2 \pi}{\lambda_y}, \frac{2 \pi}{\lambda_z} \right). $$ and so ...


2

Yes, the charge will move under the influence of the induced E-field, but it will not move in a circle: this would require a centripetal force, but there is none. Instead, the (positive) charge will start to move in the direction of the E-field, then spiral outward.


2

In a conductor the electric field applies a force to the conduction electrons so those electrons accelerate. The electrons then scatter off lattice vibrations (phonons) and decelerate. The current settles to an equilibrium state when the acceleration and deceleration have equal magnitudes, and when we do circuit analysis we assume that the circuit has ...


1

We know that the inductor doesn't allow the instantaneous change of current and flux through it, so the currents in L1 and L2 should remain the same immediately after the switch is opened up, but that is also not possible as the direction of current in the loop containing L1 and L2 cannot be in the same direction, so we reach to a contradiction. How do we ...


1

Think of momentum conservation. Joule effect in a resistor means infrared radiation emmited to the surroundings. That EM radiation has a momentum, proportional to its Poynting vector. For a typical cylindrical resistor, that vector comes from it in a cylindrical symmetry. But if we think only of electrons being scattering by the nucleous as a source for the ...


1

On a carefully constructed sketch in three dimensions, field lines start (or end) at charges, and the line density is proportional to the field strength. This is not true for a 2D cut through a 3D sketch. In the sketch you provide, the lines should disappear as they approach the null point.


1

The Poyting vector $\vec{S}=\vec{E}\times\vec{H}$ and Poynting's theorem require no radiation or even time-varying fields to be present: they are perfectly valid for a DC circuit. For example, integrating the Poynting vector across a closed surface $S$ surrounding a resistor will yield the power dissipated in this resistor.


1

First, here's an example of a circuit that works as you describe. We say that the switch has been open a long time which is to say that, before the switch closes at $t=0$, the circuit is in DC steady state (all voltages and currents are constant in time). Since the current through the inductor is constant, the magnetic flux threading the inductor is ...


1

The inductor cannot win! When the current changes so does the magnetic flux linked with the inductor, an emf is induced which produces a current in opposition to the changing current producing it - Faraday and Lenz. You get an endless sequence if the inductor stopped the current changing, no emf would be induced and there would be no opposition to any ...


1

Objects with mass tend to resist (due to inertia) changes in their velocity. But that doesn't mean that nothing ever experiences acceleration. It's the same story here. When we say "an inductor doesn't like changes in current" or "an inductor opposes changes in current" we don't mean that the current through the inductor can't change at ...


1

An external force doesn't need to supply energy to bring an electron closer to a proton. The electric field does the work at the expense of electrostatic potential energy. The gravity analogy is when an object is released from some height $h$ near the surface of the earth, gravity does work converting gravitational potential energy of $mgh$ to an equal ...


1

The potential you cite is defined as the work required to bring a unit of charge from infinity to within a distance,d, of a given point charge. If the distance is very small (as within an atomic nucleus), the energies involved can be quite large. The limits are still being explored.


1

Both the electric fields are equal The net electric fields inside both the dielectrics need to be same. Why? Because since electric fields are conservative, which means we can define a corresponding electric potential and thus electric potential difference. The electric potential difference between any two points, $a$ and $b$, is $$\Delta V=\int_a^b \mathbf ...


1

Since the line of charge is of infinite length, there will always be a charge on the other side that will produce an electric field whose vertical component will cancel out the vertical component of the field due to first charge. The horizontal components will add up and the direction of net electric field will be radial.


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