mutual capacitance?

(snip, I wrote)

Yes. The one I was actually trying for, Energy=Q*Q/(2*C) is still right. Somehow I got that one wrong.

-- glen

The reason it can't exist is because of the way a transformer changes the ratios of currents and voltages. This is why a mutual capacitance is really only a "special case" and not a true dual of a transformer. Somewhere "lost in my house" I know I've got a network theory book that describes the dual of a transformer. I believe the name they used for it was "capacitive transformer". Anyway, that's what it is. One can derive the equations by taking the dual of a transformer.

As a general passive real device the capacitive transformer doesn't exist as far as we know, but as I noted the mutual capacitance case is a special case, and as I understand it certain piezo devices (usually RF or IF bandpass filters used to replace the high frequency (magnetic) transformers in radios) can be modeled pretty well with a capacitive transformer. But mostly it's just a theoretical calculation aid sort of like Thevenin's or Norton's equivalents rather than a real circuit element.

Lessee, since Schrodinger's wave functions are "probability waves" that means you end up knowing nothing for sure! You did notice that QM is the "science of ignorance", right? We don't know what the actual forces are that flip a coin, so we calculate some average probabilities so we can say at least we know SOMETHING about the operation! Of course Physicists take that one step further and proclaim that anything beyond what THEY know is "unknowable". Right. Sure.

Actually, we can prove that a long straight wire has no inductance!

1. The B field along the axis of a straight wire due to a current IN that same wire is zero by the Biot-Savart law (Sinx =3D 0)

1. Since curl E =3D -dB/dt of Maxwell says there can be no E field within this straight wire opposing any rising currents etc.

2. Hence, as we all know, the self-inductance of a long straight wire does not exist.

Right? :-)

I am not surprised that you do not understand the situation. If you have a charge q on an isolated sphere, you can find its potential operationally. You take a small test charge and bring it in from infinity to the charge. That work for a unit charge is the charge's potential, V = q/r in esu. The capacitance C = q/V will the radius of a spherical charge of that radius and have the unit dimension of cm. You do not, and indeed cannot cannot connect voltmeter leads voltmeter leads to the charge and to infinity. That is why I said all the capacitances we ordinarily work with are mutual capacitances.

If you rake a network of resistors and self and mutual inductances, you will end up with a network of resistors and capacitors. It is not clear to me that the capacitors that will arise from the duals of self and mutual inductance will have the same meaning of self and mutual capacitance as described in Smythe and other places.

The only way I would know would be to write down the equations for the inductor-resistor network to find out. I have not done that yet and do not know if I ever will. The point is, just because names are similar, does not mean the underlying items similar.

I did the almost unthinkable thing. I looked at the entry for capacitance in Wikipedia. It had a useful section. It also mentioned that almost all capacitance people deal with are mutual capacitances. It also gave the van de Graaf generator as an example of an application for self capacitance.

While it was true for the original van de Graaf devices, it is not true for the modern versions. These are usually in pressurized envelopes.

On Jun 19, 1:54=A0am, Salmon Egg wrote::

I'd guess pretty much not true for the originals either which were all inside buildings with probable metal framework. And even for a van de Graff outside, it still sits on a ground plane which is the "other plate" of the capacitor and not a "free space" lone sphere.

In fact a "free space lone sphere" capacitor is not such a simple thing to get! I'd say probably the best example might be the earth itself. Way back when, Tesla talked a lot about the charge of the earth which has mostly been pooh-poohed. It would be an interesting calculation to compute the capacitance of the planets. Should be easy to do. The next question would be the amount of charge collected on the earth from the sun. From that one could compute the voltage of the earth and other planets and hence the electrostatic forces acting there. Interesting.

(snip on van de Graaf and self capacitance)

I suppose, but it seems closer than many other cases.

As I understand it, the earth is charged by lightning, likely enough that you couldn't detect solar sources.

-- glen

The next spheres change the self capacitance of the first.

To calculate the self capacitance you must do the step described by Salmon: "I am not surprised that you do not understand the situation. If you have a charge q on an isolated sphere, you can find its potential operationally. You take a small test charge and bring it in from infinity to the sphere. That work for a unit charge is the charge's potential, V = q/r in esu"

But it is work "from infinity to the sphere" proportional to r. To place the charge on the sphere you must compress the charges on the sphere. This work is proportional to r squared. S*

"Salmon Egg" napisa³ w wiadomo¶ci news: snipped-for-privacy@news60.forteinc.com...

It is not enough to bring a charge from infinity to an isolated sphere. Yom must compress the charges on the sphere.

If you have two spheres (like the two planets) the self capacitance will be modified with the distance. But the calculated C is useful.

S*

"Benj" napisal w wiadomosci news: snipped-for-privacy@e35g2000yqc.googlegroups.com...

calculation to compute the capacitance of the planets. Should be easy to do. The next question would be the amount of charge collected on the earth from the sun. From that one could compute the voltage of the earth and other planets and hence the electrostatic forces acting there. Interesting.

It is done by students. The calculated voltage is 10^9V. It is the result of misunderstanding of self capacitance. In textbooks the self capacitance is proportional to r. In reality (and if properly calculateed) to r squared. S*

The Sun produces plasma. And each body in plasma is negatively charged.

In the Earth atmosphere the electrons migrate up with the water vapour and come back with lightning. S*

The only proof here is that a little knowledge can be a dangerous thing.

There was a server error so I apolgize if this is a repeat/

The only proof here is that a little knowledge can be a dangerous thing.

Conceptually, lightning does not change the charge on Earth. It merely moves the around. As clouds charge up, charge can go from one cloud to another while the net charge on Earth including its atmospher remains unchanged.

If you want to nitpick, it is possible that some charged particles can be accelerated beyond escape velocity and leave Earth.

Ah

Picture this... (something we actually have at work)..

A capacitor with one main back plate, and 2 plates on the front side. These two plates are angled ~ 45 degrees towards the back plate but don't touch each other but are rather close. Depending on the phase angle of the two independent sources at the angle plates determines the circuit's behavior.. The out of phase signals gives different effects than in phase one's

Apparently the cross over section of the two work together or against themselves..

I don't know if you would call that mutual capacitance, but it's a thought.

Jamie

(snip, I wrote)

Yes, I meant the solid earth. There is a description in the "Feynman Lectures on Physics" of the charge on the earth and the effects of lightning.

Slightly related, note that the angular momentum of the solid earth plus atmosphere is (mostly) conserved, but the length of the day depends on the rotation rate of the solid earth. There is a NASA group that measures the LOD (length of day) accurately enough to notice the difference. That is, the effect due to weather.

-- glen

Is there that the Earth has 10^9V?

If water is in air the rotation rate is slower than if water fall down. S*

And you can't bring a charge in from "infinity" (wherever that is?) except in your mind. But assuming the earth is conductive (it more or less is) and the universe is a pretty large sphere in comparison to the planet (it is), one can calculate the "self capacitance" of the earth. But determining how much charge is on the planet or finding what it's potential is is not so simple. Presumably one doesn't have to locate "infinity" but merely bring a test charge to earth from very far away and measure the work. But getting very far away from earth isn't so easy either. Didn't Velikovsky talk about this?

I'm not surprised you know nothing of network theory. But you are correct that taking the dual of a transformer does NOT in general produce the self and mutual capacitances usually found in real circuits. As I indicated capacitive transformers are models not generally found in reality. And of course you have no interest in learning anything new. It's SO uncomfortable!

(snip on measuring voltage bringing a charge from infinity.)

I am not so convinced that there isn't a dual, as a capacitive transformer, though there isn't one practical at the frequencies we usual need transformers.

Note that one can use a capacitor in place of the current-limiting inductor used for discharge lamps. That is not practical at 60Hz, but is at higher frequencies.

So, take two nearby sphere (self capacitors). Put an AC voltage one one, measure the voltage on the other. Also, measure the current on both. How does it change as a function of the sphere radii?

-- glen