Ok, got it, thanks - this solves the issue with argument (1).
I got it. Putting an extra charge in the center would repell the free electrons in the bulk outwards, such that the surface gets charged more, and the bulk becomes neutral, and ##\vec{E}=0##. This refutes the argument (3).
I don't...
Oh, so am I forgetting that only a small fraction of the free electrons (in case of a negatively charged conductor) end up on the surface, but most of them are still in the bulk of the material, but distributed, such that the net charge of the bulk is zero, and the net charge of the surface is...
I don't agree. Let's look at the derivation of the microscopic Ohm's law, e.g. here in Sec. 7.5: http://web.mit.edu/sahughes/www/8.022/lec07.pdf
E.g. the whole argument doesn't apply if the are no free charges in the bulk of the material in the first place.
Thank you for the replies
I'll think about the uniqueness theorem. I didn't think of that before.
I know. My arguments (1)-(3) all apply to the static case though.
Ok but that's again saying that since charges don't move, there must be no force, which falls into my argument (1).
That's...
They say: "But, the electric field-strength inside a conductor must be zero, since the charges are free to move through the conductor, and will, thus, continue to move until no field remains." but I think I refuted that in my points (1) and (3).
Hi, I thought that I understood why, once the free charges stop moving, ##E=0## inside a conductor, but I really don't. Can someone please help me out?
I've heard the following arguments, but I don't think I understand any of them:
I don't think ##q=0## implies ##\vec{E}=0##. I understand that...
Hi again,
Stockzahn, thanks for your replies. In my studying I got to gas mixtures (hence the delay) and while I still can't fully justify to myself the procedure shown in the book, I found this paper...
But suppose we have equal amounts ##\frac{n}{2}## of gas A and gas B separated in two equal volumes ##\frac{V}{2}##. The temperatures are the same and hence the pressures are the same. If we remove the separating wall, the gases will mix. In the final state their partial volumes will still be...
Oh I see, thanks! That might be something I have to read about.
So you're saying for either one of the gases A and B separately, one should use full pressure and partial volume. But why not partial pressure and full volume?
I agree that what you derived, follows from Eq. 2 from our discussion, and gives the right result. But I don't understand why my equation is wrong. E.g. my form of ##\Delta S## is derived here http://web.mit.edu/16.unified/www/SPRING/propulsion/notes/node39.html . It appears just above Eq. (5.3).
I'm following that, but shouldn't a tottaly equivalent approach be that nA gas molecules change there state from (nA, V0, TA) to (nA, Vf, Tf) accompanied by an entropy change ΔSA and that nB gas molecules change there state from (nB, V0, TB) to (nB, Vf, Tf) accompanied by an entropy change ΔSB...
Same as you - Eq. 2 works and Eq. 1 doesn't.
So the question is why Eq. 1 is wrong, keeping in mind that I derived it from the other equations in my original post.
I used ##R=8.31\frac{J}{mol K}##, ##c_V=\frac{5}{2}R##, ##c_p=\frac{7}{2}R## and got ##n_A=686.71## and ##n_B=264.34##. This gives their answer, when plugged into the last equation.
They use ##R'=\frac{R}{\mu}##, where ##\mu## is the molar mass, and then ##m=\mu n## so it's all consistent.
And...
Do you know then where my mistake is?
I'm using the mole number, but when I plug in values to the last equation in my post, I get the answer from the book, so I don't think this is where the problem is.
Hi again Physics Forums! Last time I was here, I was an undergrad student. Now I almost finished a PhD in quantum information and became a high school teacher.
I've never properly learned thermodynamics. I'm now trying to get it on the same level, that I understand the other topics in classical...