Recent content by Fernandopozasaura

Ok, so, if I have understood, the power expressed by $$\vec{S}$$ is a characteristic of the whole universe, in this case: nothing in it but the materials and the frontier and whatever source generating this fields, and the existence of this (the border) doesn't affect this quantity. And that's...

Yes, that's what I get. More precisely S_incident = S_reflected + S_transmited. Further calculation was to get S_medium1 (which is not S_incident neither S_reflected) starting from total E (Ei + Er) and total H (Hi + Hr) in that medium 1. And then S_medium2 from E and H in medium 2. And I get...

Great!, I think I needed any kind of support. What made me unsure was the fact that, somehow, in medium 1 should be more power (S) because, there, you have two powers and only one in medium 2. I think there is something counterintuitive. At least to my brain. Thanks a lot for your help.

High! I have a EM plane wave hitting normally a surface dividing universe in media 1 and 2, both without losses. So we have incident, reflected and transmitted waves. It's a simple exercise in which you are given the basic data about two media and wave incident amplitude H in medium 1. I get...

OK, I understand. As you, my first answer was what you said: free space is not lossy. But somehow I remembered have read something different, that's why I asked. I hoped someone could eliminate other possibility. Anyway, thank you very much for your help, Dale.

Well, I saw it in an exame someone showed to me. I'm sorry I don't have more details.

Recently I saw an exercise on electromagnetics (plane waves) in which it was said that a plane wave was traveling through vacuum (ε0 and μ0) BUT with losses, i.e, σ ≠0. Is that possible? I think I have read in the past that it's OK in some texbook, but now, I have searched for it in a couple of...

With such a succint question, a just correct answer is: it's not possible. Heisenberg dixit.

Good morning Tsny, I see the way. There are two ways to solve the same problem. One of them gets you to a volumetric integral that is forced in the exercise or experiment you are workin on, and the other makes you to choose a surface, at your own! and that's the good point. In this case it was...

What i think is wrong is, in part c), not making a volume integration from cylinder axis to its surface. Instead supposing all charge is placed on it and then just multiply lateral surface times charge density. There is nothing wrong with Gauss's las use. Once again, thanks Fernando

Humm, would you say the exercise is wrongly solved? I think I would. To me, most important thing is that there is no weird magic hidden that I didn't notice. Thanks again Tsny.

Thank you very much! Greg.

Thank you very much for such a quick answer Tsny, Well, first of all, let me get familiar with this editor. It's related through this expression ∫D.dS = Q (this integral is over a closed generic surface containing Q) On the other hand working out that integral you can get to Divergence of D =...

Hello, I'm happy to say hello to all Physics Formus fellows. I'm a new professor in physics (more precisely engineering in electromagnetics field) and I've just registered to solve some doubts. I hope to help others soon. Thank you for making possible this community. Fernando

Homework Statement It is known that the potencial is given as V = 80 ρ0.6 volts. Assuming free space conditions, find a) E, b) the volume charge density at ρ=0.5 m and c) the total charge lying withing the closed surface ρ=0.6, 0<z<1 Homework Equations E[/B]=-∇VThe Attempt at a Solution (this...