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08:04
@JohnRennie: Hi sir. Good morning :-)
@GuruVishnu hi :-)
I have credited your statement:
> Any time you want to understand what happens with a zero resistance you need to start with a small but non-zero resistance and see what happens in the limit of $R\to0$
in my answer. Even thought it was for a different question, it had a good fit in the answer :-)
08:53
@JohnRennie: Hi. Are you free now, sir?
@GuruVishnu hi, yes I'm free
Ok sir. If possible, could you answer the following question? :
1
Q: Why is the direction of cathode rays independent of the position of anode?

Guru VishnuThe following statement is from the book Concepts of Physics: Cathode rays are emitted normally from the cathode surface. Their direction is independent of the position of anode. I found the explanation for the first sentence from the section perpendicular emission of the Wikipedia article ...

Give me a few moments and I'll have a look. What is the Concepts of Physics book? Who wrote it?
@JohnRennie Ok sir. It was written by H.C. Verma.
Volume 1 or 2, and which chapter/page?
09:04
Volume 2; Chapter 41; Topic 41.2 Cathode rays; Page 343
The two statements:
1. Cathode rays are emitted normally from the cathode surface
2. Their direction is independent of the position of anode
are equivalent in the sense that 2 follows from 1.
That's because if the position of the anode affected the direction of emission the electrons would not all be emitted perpendicular to the surface.
Ok sir. Now I'm thinking whether is it mandatory for the statement 1 to hold true.
I guess the question is why does the anode not affect the direction after emission and during the travel through the space inside the tube?
Yes sir.
I don't think that's true. There is an electric field inside the tube and that electric field creates a force $\mathbf F = q\mathbf E$ on the electrons.
09:13
I agree with you sir. However, the original setup has cathode rays undeviated by the presence of anode at the bottom:
I suspect the geometry of the Crookes tube is such that near the cathode the field is perpendicular to the surface of the cathode so the force acts in the same direction the electrons are travelling.
Ok sir. Then I think it's safe to put the anode out of the tube. Ok sir? :-)
As the electrons near the other electrode the force on them changes direction, but by then the electrons are moving fast enough that the force doesn't deflect them very much and they continue in approximately a straight line.
Ok sir. This reason feels good to me. I thought this was a faulty explanation as this might pull out the anode out of the tube which is not done so far (so I assume it to be not possible).
So the straight line trajectory is due to the geometry of the tube and not a fundamental property of cathode rays.
I'm not sure what you mean by pulling the anode out of the tube.
09:19
We typically use an induction coil to power the electrodes. So why not a single electrode cause dielectric breakdown inside the tube, sir? If it were a battery then the reason if obvious.
@JohnRennie Alternatively, I meant what if we have only the cathode inside and the anode outside the tube. Will we see the plasma effects?
The electrons are emitted from the cathode with a low energy so they won't do anything interesting unless they are accelerated by an electric field. I guess you could put the anode outside the tube if it still generated a strong electric field inside the tube.
However, since we use an induction coil, I think we can even place the anode at infinity, the electric field supplied by the cathode would be (I think) sufficient enough to accelerate electrons.
An induction coil isn't like a Van der Graaff Generator. It won't create a field unless there are two electrodes present. So the need the anode to be close to the cathode. If you remove it to infinity you get no field.
Ah! Ok sir. I considered it to be a big capacitor and that's where I went wrong.
Can we have cathode and anode side by side, sir? I think it's not possible.
@GuruVishnu yes, and then the electrons would move in an arc between them.
09:33
Ok sir. And we'll not observe the usual phenomenon. Am I right?
Something like that ...
Ok sir. Thank you. Then, the statement 2 is incorrect.
Yes. The electrons travel in approximately straight lines in a Crookes tube because of the geometry of the tube.
Thank you sir :-)
I'll think about this for some time.
 
2 hours later…
11:46
Why do different isotherms share the same graph at lower plate voltages in a vacuum diode valve, sir?
I had a fast read at the Wikipedia article on Vacuum tube, however, I didn't find any reason for this.
@JohnRennie: I know you're busy in a different room. After that, if possible, kindly clarify the doubt about diode valves. Thank you sir.
12:02
@GuruVishnu electrons spray out from the cathode in all directions, so if the voltage is too small not all those electrons will reach the anode.
The sloping line in your graph is the region where the diode isn't capturing all the electrons released by the cathode.
12:14
@JohnRennie Ok sir. If we have still higher temperature, can we expect a difference in current due to the increase in the number of thermions? That's why I'm confused on this fact.
I must confess it isn't obvious to me why the current in the partial capture region should be independent of temperature.
Ok sir. No problem.
Shall we discuss about energy dissipation in capacitor circuit, sir?
Wikipedia has some stuff on it here:
Space charge is a concept in which excess electric charge is treated as a continuum of charge distributed over a region of space (either a volume or an area) rather than distinct point-like charges. This model typically applies when charge carriers have been emitted from some region of a solid—the cloud of emitted carriers can form a space charge region if they are sufficiently spread out, or the charged atoms or molecules left behind in the solid can form a space charge region. Space charge only occurs in dielectric media (including vacuum) because in a conductive medium the charge tends to be...
@GuruVishnu I need to go now.
@JohnRennie Ok sir. I'll have a look at it. Thank you :-)
@JohnRennie Ok sir. Then we shall discuss it later. Good bye.

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