mutual capacitance?

[glen herrmannsfeldt]

In sci.physics.electromag RichD <r_delaney2001@yahoo.com> wrote:

On Jun 17, "Don Kelly" <d...@shawcross.ca> wrote:

(snip)

I1 =C11(dv1/dt) +C12(dv2/dt)
I2=C21(dv1/dt) +C22(dv2/dt)
Capacitive coupling

Finally, someone gets it.

A 2-port is what you're describing, which models a
transformer (or mutual inductance, generally). Why
don't we see the capacitive form in circuit theory, or
practice?

Maybe you do in microwave electronics, though it is harder
to say. Inductance and capacitance are harder to separate.

Practical 60Hz inductors are much easier to make into
transformers than practical 60Hz capacitors.

Note that inductors are commonly used for fluorescent lamp
ballasts at 60Hz, which capacitors are usually used at 20kHz.
(The usual electronic ballast.)

It's the dual of a transformer, yes/no?
Electric flux linkage, vs. magnetic flux linkage?

-- glen

 

[Salmon Egg]

In article
<5ae7c5e6-b960-48da-96d2-0fd1511a9921@d1g2000yqm.googlegroups.com>,
Benj <bjacoby@iwaynet.net> wrote:

Lets get back to reality, shall we? The whole concept of the
"capacitance" of an isolated sphere is bogus! Please. Just HOW does
one connect their voltage source to "infinity"? Just HOW big is that
"infinite" outer sphere? Well, actually it can be ANY size so long as
it's big enough! So in reality what we are saying is that IF we have
two concentric spheres and we measure the capacitance between them, so
long as the outer sphere is much larger than the inner sphere the
capacitance of the PAIR depends ONLY on the radius of the smaller
inner sphere.

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.

--

Sam

Conservatives are against Darwinism but for natural selection.
Liberals are for Darwinism but totally against any selection.

 

[Salmon Egg]

In article <itjpl1$5ks$2@dont-email.me>,
glen herrmannsfeldt <gah@ugcs.caltech.edu> wrote:

In sci.physics.electromag Salmon Egg <SalmonEgg@sbcglobal.net> wrote:

(snip)
The mutual capacitance matrix that relates charge too potential, as I
indicated earlier, is described well in Smythe's book. It is not at all
clear to me the self and mutual capacitance described in this way is the
dual equivalent of what you get from a circuit dual of self and mutual
inductance. It may be, but I have not worked it out. I may not get to it
for a while.

I do not know of a common circuit in which self capacitance is an
important feature. The closest concept would be that of stray
capacitance. But even there, the capacitance arises mostly from
capacitance to ground rather than to infinity.

How about the van de Graaf generator? Doesn't seem so far off.

-- glen

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.

--

Sam

Conservatives are against Darwinism but for natural selection.
Liberals are for Darwinism but totally against any selection.

 

[glen herrmannsfeldt]

In sci.physics.electromag FrediFizzx <fredifizzx@hotmail.com> wrote:

(snip, I wrote)

Usual, but not required. A capacitor stores energy in the
electric field, which you can do with a single electrode,
usually spherical.

Q=CV, or C=V/Q. If you charge a sphere, its voltage change.

C = Q/V

Must be a typo on your part since you got the first one right.

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

-- glen

 

[Benj]

On Jun 18, 10:49 pm, RichD <r_delaney2...@yahoo.com> wrote:

On Jun 17, Benj <bjac...@iwaynet.net> wrote:

A network theorem states that every circuit has a
dual; voltage sources become current sources, etc.

Generally speaking the
device doesn't exist except in certain special circumstances, but it's
widely used as a theoretical aid to network calculations.

Why couldn't it exist?  Interleave plates and
dielectric to link electric flux, analogous to
magnetically coupled coils.

That's really the thrust of the question.

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.

 And who needs a device if you have the
equations? Today, we all believe mathematics is more real than
reality anyway!

Well, if the universe is really a big quantum
computer, then all we need is information theory
and Schrodinger's wave function of  'potentiality',
and no need for terra firma -

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.

 

[Benj]

On Jun 18, 11:22 pm, glen herrmannsfeldt <g...@ugcs.caltech.edu>
wrote:

Next, you can figure out the inductance per unit length for
a long straight wire.

-- glen

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 = 0)

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

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

Right?

 

[Szczepan Bialek]

"glen herrmannsfeldt" <gah@ugcs.caltech.edu> napisał w wiadomości
news:itiuap$1ji$2@dont-email.me...

In sci.physics.electromag Szczepan Bialek <sz.bialek@wp.pl> wrote:

(snip)

The self capacitance of a sphere has the two members:
proportional to r and proportional to r squared.

The capacitance for concentric spheres has two terms.

The next spheres change the self capacitance of the first.

In the limit as the spacing goes to zero, the R squared
term is left, as you would expect. (Parallel plate capacitor
with spacing delta R and area 4 pi R squared.)

In the limit as the outer sphere goes to infinity, the R
term is left.

You can also do it integrating E squared over all space.

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*

 

[Benj]

On Jun 19, 1:54 am, Salmon Egg <Salmon...@sbcglobal.net> wrote::

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.

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.

 

[Szczepan Bialek]

"Salmon Egg" <SalmonEgg@sbcglobal.net> napisał w wiadomości
news:SalmonEgg-56B345.22321118062011@news60.forteinc.com...

In article
5ae7c5e6-b960-48da-96d2-0fd1511a9921@d1g2000yqm.googlegroups.com>,
Benj <bjacoby@iwaynet.net> wrote:

Lets get back to reality, shall we? The whole concept of the
"capacitance" of an isolated sphere is bogus! Please. Just HOW does
one connect their voltage source to "infinity"? Just HOW big is that
"infinite" outer sphere? Well, actually it can be ANY size so long as
it's big enough! So in reality what we are saying is that IF we have
two concentric spheres and we measure the capacitance between them, so
long as the outer sphere is much larger than the inner sphere the
capacitance of the PAIR depends ONLY on the radius of the smaller
inner sphere.

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.

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

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

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.

S*

 

[Szczepan Bialek]

"Benj" <bjacoby@iwaynet.net> napisal w wiadomosci
news:a0de8577-a323-4563-b700-9b0db89c4b3b@e35g2000yqc.googlegroups.com...

. 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.

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*

 

[Szczepan Bialek]

"glen herrmannsfeldt" <gah@ugcs.caltech.edu> napisał w wiadomości
news:itkden$17d$1@dont-email.me...

In sci.physics.electromag Benj <bjacoby@iwaynet.net> wrote:

(snip on van de Graaf and self capacitance)
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.

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


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.

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

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*

 

[glen herrmannsfeldt]

In sci.physics.electromag Benj <bjacoby@iwaynet.net> wrote:

(snip on van de Graaf and self capacitance)

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.

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

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.

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

-- glen

 

[Salmon Egg]

In article
<bfadfb32-87d0-4369-b146-f44710d6de2c@e35g2000yqc.googlegroups.com>,
Benj <bjacoby@iwaynet.net> wrote:

On Jun 18, 11:22 pm, glen herrmannsfeldt <g...@ugcs.caltech.edu
wrote:

Next, you can figure out the inductance per unit length for
a long straight wire.

-- glen

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 = 0)

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

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

Right?

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

--

Sam

Conservatives are against Darwinism but for natural selection.
Liberals are for Darwinism but totally against any selection.

 

[Salmon Egg]

In article
<bfadfb32-87d0-4369-b146-f44710d6de2c@e35g2000yqc.googlegroups.com>,
Benj <bjacoby@iwaynet.net> wrote:

On Jun 18, 11:22 pm, glen herrmannsfeldt <g...@ugcs.caltech.edu
wrote:

Next, you can figure out the inductance per unit length for
a long straight wire.

-- glen

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 = 0)

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

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

Right?

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.

--

Sam

Conservatives are against Darwinism but for natural selection.
Liberals are for Darwinism but totally against any selection.

 

[Salmon Egg]

In article <itkden$17d$1@dont-email.me>,
glen herrmannsfeldt <gah@ugcs.caltech.edu> wrote:

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

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.

--

Sam

Conservatives are against Darwinism but for natural selection.
Liberals are for Darwinism but totally against any selection.

 

[Jamie]

RichD wrote:

On Jun 17, 10:55 am, Phil Hobbs
pcdhSpamMeSensel...@electrooptical.net> wrote:

RichD wrote:

A network theorem states that every circuit has a
dual; voltage sources become current sources, etc.

But, what about mutual inductance? Why is there no
mutual capacitance? By symmetry, shouldn't a 'mutual
capacitor' exist, linking electric flux?

--
Rich

Mutual capacitance does exist, e.g. the capacitance
between the plates of a differential (3-plate) variable capacitor.


That's similar to what I have in mind, but not quite -


There's also self-capacitance, e.g. the self capacitance of a
1-cm diameter sphere in free space is 1.12 pF. (The cgs unit
of capacitance is the centimetre.)


I don't get this one - self capacitance? Capacitance
requires 2 surfaces, plus dielectric, unless things
have changed -

--
Rich
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

 

[glen herrmannsfeldt]

In sci.physics.electromag Salmon Egg <SalmonEgg@sbcglobal.net> wrote:

(snip, I wrote)

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

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.

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.

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

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

 

[Szczepan Bialek]

"glen herrmannsfeldt" <gah@ugcs.caltech.edu> napisał w wiadomości
news:itljhi$buc$2@dont-email.me...

In sci.physics.electromag Salmon Egg <SalmonEgg@sbcglobal.net> wrote:

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

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.

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.

Is there that the Earth has 10^9V?

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

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.

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

 

[Benj]

On Jun 19, 1:32 am, Salmon Egg <Salmon...@sbcglobal.net> wrote:

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.

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?

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'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!

 

[glen herrmannsfeldt]

In sci.physics.electromag Benj <bjacoby@iwaynet.net> wrote:

On Jun 19, 1:32 am, Salmon Egg <Salmon...@sbcglobal.net> wrote:

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

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'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!

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