Recent content by TSny

👍 I get the same result. I found the calculation to be fairly tedious. Easy to slip up.

In your example in post #1, ##\large \frac{\partial A_y}{\partial y} \neq 0## for points along the x-axis. So, ##\nabla \cdot \mathbf{A} \neq 0## along the x-axis. You can verify this explicitly for the case where the two charges move with constant speed. Expressions for ##\mathbf{A}(x, y, z...

That's not the reason. The assumption is more general. Free charge inside the volume of a dielectric material would have to be "embedded" there by some means. So, in textbook problems, we assume no free charge inside dielectrics unless explicitly stated otherwise. Note the comment in the...

Intuitively, the reason that ##\sigma_b## is negative is because the field of the positive, external charge ##q## tends to pull negative charge in the dielectric molecules toward ##q##. So, the negative end of the polarized molecules "poke out" at the surface of the dielectric. It's really...

@kuruman has addressed your questions. But I’ll add a few additional remarks. There are two sentences in your quote. Are you asking why the force is attractive, or are you asking why there is no volume bound charge? The dielectric extends throughout the entire region below the plane ##z = 0##...

Yes. Thank you.

The flux through the spherical surface without including ##\partial_t A## is ##(q_o-I_ot)/\varepsilon_o##. The flux of ##-\partial_t A## through the cap is ##+I_o t/\varepsilon_o## , which brings the net flux to ##q_o/\varepsilon_o##. For some reason, I couldn't get Latex to work when I was...

I used Mathematica to plot the field. I definitely agree. In the case of opening the switch, you have resolved the apparent paradox by noting the charge accumulation on the wire that must be included in the charge enclosed by the Gaussian surface. In my example, it's the nonzero value of ∂tA...

This all sounds good. If the current is switched off by opening a local switch, things are quite complicated, as you have very nicely explained. However, I think we can consider a simpler scenario in which the current suddenly drops to zero instantaneously at all points along the wire. This...

Good. Yes, I think so. Your two different graphical solutions with three reflections are very insightful!

Your diagrams appear to show 3 reflections before the rays form an image at S.

How many reflections are you considering?

If the current suddenly stops, ##\partial_t \mathbf{A}## will no longer be zero everywhere. Immediately after the current stops, there will be a huge spike in ##\partial_t \mathbf{A}## in the immediate vicinity of the wire. This disturbance in ##\partial_t \mathbf{A}## will spread out rapidly...

At which point of the paper do you believe the author takes ##\delta \int L \;dt = 0## for granted? The paper starts with d'Alembert's principle in the form of equation (1) of the paper. Equation (4) is used to make a substitution in (1) to derive the "extended Hamilton's principle", equation...

You're not missing anything. When I posted #9, I was apparently missing something - my brain :oldsmile: The magnetic field in my example is static, but the electric field varies with time as given in post #5. Of course, there is no instantaneous signaling going on between the charge and the...