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I may just need to sleep on this, but I am not able to make sense of section 4 of The Feynman Lectures Vol II 18 The Maxwell Equations. After explaining the origin and meaning of the displacement current term in the Ampere Maxwell equation

$$c^2\nabla \times \mathfrak{B}=\frac{\mathfrak{j}}{\varepsilon _o}+\frac{\partial \mathfrak{E}}{\partial t},$$

Feynman gives what he intends to be an illustrative example of an infinite charged sheet suddenly set in motion in its own plane. With another "neutralizing" plane present so there is no net charge.

The first part of the discussion makes sense. The moving plane constitutes a current, and will therefore induce a (fairly) constant magnetic field moving in a direction normal to the charged plane. He then says that the sudden initiation of current will cause a rapid change in the magnetic field, which, of course, results in an electric field, and we have light.

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But that's when things get confusing. While I agree with the basic arguments leading to his equations, I simply cannot understand why he believes the electric field will behave, more-or-less in the same way as the magnetic field.

The $E-$field does the same thing. Before $t=0$ (when we turn on the current), the field is zero everywhere. Then after the time $t$, both $E$ and $B$ are uniform out to the distance $x=vt$, and zero beyond.

That is certainly not how I interpret his applications of Maxwell's equations. The magnetic field is there because of the current. The electric field was induced by the rapidly changing magnetic field, and, etc. But for the electric field there's nothing analogous to the steady current which supports the magnetic field.

I believe the electric field will spike when the current starts, and will almost immediately return to zero after a bit of back and forth with the magnetic field. It will be more like a highly concentrated wave packet than a square wave.

enter image description here

So, do I need to sleep on this, or will it still be wrong when I wake up?

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  • $\begingroup$ Please explain why this was down-voted. If the purpose of down-voting is to communicate a quality issue, then not conveying what that issue is merely lowers the quality of the site as a whole. I have a friend who received his set of Feynman Lectures as they were handed out at Caltech. He didn't think this was a bad question. He simply told me that Feynman's development looked good to him. That's all I needed in order to resolve the confusion. $\endgroup$ Nov 5, 2018 at 15:44

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So, do I need to sleep on this, or will it still be wrong when I wake up? Apparently the answer is: within the bounds of credulity, Feynman is correct.

The confusing thing about this is that we have what amounts to a static electric field in the absence of a net source charge. I missed the significance of his emphasis on this being an "infinite" charged sheet.

If we reduce the Gaussian boundary in the $x$ direction so that it encloses virtually nothing but a section of the wave front, the $E$ field is different on either side. But between $x=0$ and $x=x_0$ there is nothing to produce a finite circulation of $E$.

I'm not sure why he presented things that way, since it is a physically impossible situation for many reasons, and there are other ways to develop the EM wave equation.

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