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In Griffiths' electrodynamics book in the chapter on electrodynamics he does some computations of electric fields using electrostatics methods when the charge is actually changing.

So, two examples:

  1. Two long coaxial metal cylinders (radii $a$ and $b$) are separated by material of conductivity $\sigma$. If they are maintained at a potential difference $V$, what current flows from one to the other, in a lenght $L$? The field between the cylinders is

$$\mathbf{E}=\dfrac{\lambda}{2\pi\epsilon_0 s}\hat{\mathbf{s}}.$$

  1. Imagine two concentric metal spherical shells. The inner one (radius $a$) carries a charge $Q(t)$, and the outer one (radius $b$) carries a chage $-Q(t)$. The space between them is filled with Ohmic material of conductivity $\sigma$, so a radial current flows:

$$\mathbf{J} = \sigma\mathbf{E}=\sigma\dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{r^2}\hat{\mathbf{r}}; \quad I = -\dot{Q}=\int\mathbf{J}\cdot d\mathbf{a}=\dfrac{\sigma Q}{\epsilon_0}.$$

Now in both problems we have currents. In the second we even have a time varying charge $Q(t)$. So in both problems, charges are moving around. These are not static configurations.

But it seems the fields are being found via Gauss law. I mean, the methods of electrostatics are being directly used without explanation of why they can be used.

Furthermore, there's even a proof that relies on using Laplace's equation for the potential inside a wire. But the very existance of the potential is something acquired from electrostatics, based on $\nabla \times \mathbf{E} = 0$, which we know won't hold in Electrodynamics.

Of course one possible answer could be: "it is used because it works", and I don't doubt it works, since it is being used.

But my whole point is: as the author did, he constructed the theory step by step - first electrostatics, based on Coulomb's law and superposition (the author even says that all of that is just for static configurations), then magnetostatics, based on steady currents and Biot Savart's Law.

Now when it comes to electrodynamics, I thought there would be some explanation on how the electric field is now computed with charges moving around. Why in these cases the traditional methods for electrostatics work? How can we justify it properly inside the theory instead of just saying that it works?

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  • $\begingroup$ Not clear to me what you're looking for here. The electrostatic and magnetostatic assumptions are idealizations that hold, to a good approximation, when the rates of change are small enough becoming exact in the limit as the rates of change go to zero. This is typical in physics. Are you looking for something else? $\endgroup$ – Alfred Centauri Sep 25 '16 at 21:47
  • $\begingroup$ Perhaps one example can make my point better: in the first example, we have two charged cylinders, an inner one and a outer one, the space between them filled with a material with a certain conductivity. If we forget the material between, this is a typical electrostatics problem and the field is what the author stated. On the other hand, with the conducting material in between, charges will move from one cylinder to the other. This confuses me, it is not electrostatics anymore, right? $\endgroup$ – user1620696 Sep 27 '16 at 2:06
  • $\begingroup$ It seems we should have a contribution to $\mathbf{E}$ from these moving charges, so that $\mathbf{E}$ should not, at first, be the same as in electrostatics. Still it is just the field from the traditional electrostatics configuration, it is like the current flowing, because of the electrostatic field simply doesn't contribute or affect the electric field at all. This is what I can't understand. $\endgroup$ – user1620696 Sep 27 '16 at 2:07
  • $\begingroup$ The charge distribution $\rho(\vec{x},t)$ is static. It doesn't matter that the distribution comprises many moving charges, since the overall distribution does not change. $\endgroup$ – ZachMcDargh Apr 27 '18 at 15:17
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The "dynamics" part comes in when the $d/dt$ terms start to be significant, so let's review where they come from in the theory. The differential form is easiest to consider, so we don't have to worry about $E$ and $q$ being in different places. Then

  • $d \vec{B} / dt$ and $d \vec{E} / dt$: are the fields static? That's set by whether the sources and boundary conditions are static.
  • time changes in the sources $\rho$ and $\vec{J}$, constrained by the continuity equation

The first example has none of these. It has a current, but that current (not the individual charges that make it up) is static. So it's clearly the realm of static analysis.

The second one is trickier, and it probably would have been better if there was more explanation.

For any real macroscopic setup of the second case, it seems clear that the time-varying terms are going to be small compared to everything else. To put it another way, the characteristic times put it in the "discharging capacitor" realm, where we don't really worry about electromagnetic radiation from the changing current.

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Now in both problems we have currents. In the second we even have a time varying charge $Q(t)$. So in both problems, charges are moving around. These are not static configurations.

In both cases, the objects are maintained at [something] = constant, i.e., constant potential difference in first and constant total charge in second scenario. A current flows in both scenarios, but this does not mean the current is time-varying. In fact, in both cases the current could be static. In other words, charge enters each object at the same rate it leaves each object. So if you draw your gaussian surface around each object, the net charge enclosed could remain constant in time, which means using Gauss' law would be okay.

Technically, Gauss' law does not require $\partial_{t} \mathbf{B} = 0$, as it just states that ...the total flux of $\mathbf{D}$ out through the surface is equal to the charge contained inside... [page 17 of J.D. Jackson's E&M book, 3rd Edition (i.e., blue cover one)]. So the current in the second case (as @AlfredCentauri correctly pointed out to me) can be time-varying but the use of Gauss' law is not invalidated by this condition.

Furthermore, there's even a proof that relies on using Laplace's equation for the potential inside a wire. But the very existance of the potential is something acquired from electrostatics, based on $\nabla \times \mathbf{E} = 0$, which we know won't hold in Electrodynamics.

First, a constant current will result in a time-independent magnetic field that would satisfy $\partial_{t} \mathbf{B} = 0$, which makes the assumption $\nabla \times \mathbf{E} = 0$ okay. Technically, one can argue that $\nabla \times \mathbf{E} = 0$ if $\mathbf{E} = -\nabla \phi$ (i.e., gradient of a scalar), in which case we know from vector calculus that the curl of a gradient is zero.

Second, the existence of a magnetic field does not imply the need for the term electrodynamics, if the magnetic field is itself static.

Why in these cases the traditional methods for electrostatics work?

See my comments above...

How can we justify it properly inside the theory instead of just saying that it works?

I think the divergence theorem (which is a special case of Stokes' theorem) is a general mathematical rule that does not require anything to be static, other than the constancy of the arbitrary surface/volume over which one integrates.

As I said before, a constant current results in a static magnetic field or $\partial_{t} \mathbf{B} = 0$. In both cases, we can define $\mathbf{E} = -\nabla \phi$ because any time-varying component does not matter (i.e., $\partial_{t} \mathbf{A} = 0$) for the total electric field. We know from vector calculus that $\nabla \times \nabla \phi = 0$, where $\phi$ is any scalar quantity or function so then $\nabla \times \mathbf{E} = 0$.

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    $\begingroup$ "In fact, in both cases the current must be static to maintain those configurations." - are you sure this applies to the second case? To me, the 2nd case sounds like a charged capacitor discharging via a 'leaky' dielectric and so the current should decay to zero over time. $\endgroup$ – Alfred Centauri Sep 27 '16 at 0:00
  • $\begingroup$ @AlfredCentauri - Good catch, it appears I answered too quickly... Shoot I think I misread the second one... $\endgroup$ – honeste_vivere Sep 27 '16 at 12:09

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