mutual capacitance?

[Jos Bergervoet]

On 6/19/2011 7:28 AM, Benj wrote:

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

...

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!

Hmm.. How would that change things, Benj?

You did notice that QM is the "science of ignorance", right? We don't
know what the actual forces are that flip a coin,

Really? Then we may have a new "flipping coin paradox! How can a
coin flip if we don't know of any force being applied to it?!

--
Jos

 

[Salmon Egg]

In article <itnfod$16n$3@dont-email.me>,
glen herrmannsfeldt <gah@ugcs.caltech.edu> wrote:

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?

I have a capacitive transformer--sort of-- not far away from where I am
sitting now in what truly is an electronic junk pile. A vendor sent it
to me to try out. I believe it has found some use in battery operated
fluorescent lamp lanterns. I am not sure where else.

It consists of a piezoelectric ceramic bar. It polarized across the bar
for about half its length and along the bar for the rest of its length.
If you apply a voltage across the cross-polarized half at the bar's
mechanical resonance, a high voltage appears at the other end. This acts
as the starting and running source for the lamp. Because of its high
leakage reactance, it is the lamps ballast.

Getting back to your spherical capacitors, the voltage you measure is
because of MUTUAL capacitance. Unless the spheres are extremely far
apart, the mutual capacitance between them is much greater than their
self capacitances. Again. I refer you to Smythe. It is a comprehensive
but difficult book.

--

Sam

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

 

[Jos Bergervoet]

On 6/20/2011 6:29 PM, Salmon Egg wrote:

In article<itnfod$16n$3@dont-email.me>,
glen herrmannsfeldt<gah@ugcs.caltech.edu> wrote:

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?

I have a capacitive transformer--sort of-- not far away from where I am
sitting now in what truly is an electronic junk pile. A vendor sent it
to me to try out. I believe it has found some use in battery operated
fluorescent lamp lanterns. I am not sure where else.

I remember playing with one of those.. They break apart quite easily. As
far as I know they actually convert E-field into mechanical
vibrations and then back to E-field. This of course is different
from a magnetic transformer where the energy transfer is completely
by the EM field alone.

It consists of a piezoelectric ceramic bar. It polarized across the bar
for about half its length and along the bar for the rest of its length.
If you apply a voltage across the cross-polarized half at the bar's
mechanical resonance, a high voltage appears at the other end. This acts
as the starting and running source for the lamp. Because of its high
leakage reactance, it is the lamps ballast.

It does, however, not 'succeed to replacing the functions of the
lamp ballast:
1) High ignition voltage by resonance
2) Limiting the current during operation.
3) Filter out harmonics to prevent EMI.
4) Provide an inductive load to the switch-mode converter (for
loss-less switching).

At least points 3 and 4 are not very well fulfilled! So not much
savings in component count..

Getting back to your spherical capacitors, the voltage you measure is
because of MUTUAL capacitance. Unless the spheres are extremely far
apart, the mutual capacitance between them is much greater than their
self capacitances.

Not at all! If they are 1 diameter apart, then the mutual C is
already about 3 times _lower_ then the self-C. Which means that
with an AC voltage on one sphere, the voltage division would
give about 1/4th of the voltage on a floating neighbor sphere.

Again. I refer you to Smythe. It is a comprehensive
but difficult book.

Yes, Smythe is cool!

--
Jos

 

[Salmon Egg]

In article <4dffae02$0$49041$e4fe514c@news.xs4all.nl>,
Jos Bergervoet <jos.bergervoet@xs4all.nl> wrote:

It does, however, not 'succeed to replacing the functions of the
lamp ballast:
1) High ignition voltage by resonance
2) Limiting the current during operation.
3) Filter out harmonics to prevent EMI.
4) Provide an inductive load to the switch-mode converter (for
loss-less switching).

At least points 3 and 4 are not very well fulfilled! So not much
savings in component count..

Getting back to your spherical capacitors, the voltage you measure is
because of MUTUAL capacitance. Unless the spheres are extremely far
apart, the mutual capacitance between them is much greater than their
self capacitances.

Not at all! If they are 1 diameter apart, then the mutual C is
already about 3 times _lower_ then the self-C. Which means that
with an AC voltage on one sphere, the voltage division would
give about 1/4th of the voltage on a floating neighbor sphere.

Again. I refer you to Smythe. It is a comprehensive
but difficult book.

Yes, Smythe is cool!

Items 1 and 2 are met very well indeed.

My trial gizmo worked very well as a transformer. IIRC i ran at about
30kHz.

Point 3 is not met well by inductive ballasts. My shortwave receiver
just cannot operate near CFLs because of EMI. If inderstand your point
4, it is irrelevant. You just design your electronics for a capacitive
load, and not an inductive load.

--

Sam

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

 

[whit3rd]

On Saturday, June 18, 2011 11:02:17 AM UTC-7, Benj wrote:

On Jun 18, 5:26 am, glen herrmannsfeldt <g....@ugcs.caltech.edu> wrote:

Capacitance is proportional to the radius.

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"? \

The historic development of electricity included years of experimenting
with 'capacitors' that were, in fact, metal shells (suspended
on insulating strings). The capacitance of two balls and the
ability to induce a charge (ground ball2, move it near ball1, remove
the ground, then pull ball2 far from ball1) made it possible to
get a repeatable quantity of charge many times (because
the induced charge on ball2 doesn't at all discharge ball1).

So, 'an isolated sphere' means a spherical conductor far from (many
radii from) the nearest other conductors. The formula for
the capacitance of an isolated sphere is quite valid. Also, useful.

 

[whit3rd]

On Saturday, June 18, 2011 4:33:57 PM UTC-7, Salmon Egg wrote:

I do not know of a common circuit in which self capacitance is an
important feature.

Every dry winter day when you shock yourself on a light switch,
your self capacitance IS an important feature.

 

[Jos Bergervoet]

On 6/20/2011 11:29 PM, Salmon Egg wrote:

In article<4dffae02$0$49041$e4fe514c@news.xs4all.nl>,
Jos Bergervoet<jos.bergervoet@xs4all.nl> wrote:

It does, however, not 'succeed to replacing the functions of the
lamp ballast:
1) High ignition voltage by resonance
2) Limiting the current during operation.
3) Filter out harmonics to prevent EMI.
4) Provide an inductive load to the switch-mode converter (for
loss-less switching).

At least points 3 and 4 are not very well fulfilled! So not much
savings in component count..

Getting back to your spherical capacitors, the voltage you measure is
because of MUTUAL capacitance. Unless the spheres are extremely far
apart, the mutual capacitance between them is much greater than their
self capacitances.

Not at all! If they are 1 diameter apart, then the mutual C is
already about 3 times _lower_ then the self-C. Which means that
with an AC voltage on one sphere, the voltage division would
give about 1/4th of the voltage on a floating neighbor sphere.

Again. I refer you to Smythe. It is a comprehensive
but difficult book.

Yes, Smythe is cool!

Items 1 and 2 are met very well indeed.

My trial gizmo worked very well as a transformer. IIRC i ran at about
30kHz.

Point 3 is not met well by inductive ballasts. My shortwave receiver
just cannot operate near CFLs because of EMI.

That means it is sensitive. It does not disprove that the ballast
coil reduces the harmonics. It actually is essential to meet the
FCC requirements. With your gizmo you will fail, unless you add a
coil after all!

If inderstand your point 4, it is irrelevant.

Well, let's say you understood it as well as you understood
Smythe about the two-sphere capacitances.

You just design your electronics for a capacitive
load, and not an inductive load.

And also you "just" design it to generate a pure sine wave, so
no harmonics need be suppressed! We'll "just" let you do all the
design for us and I'm sure we can remove all components in all
our circuits!

--
Jos

 

[George Herold]

On Jun 21, 12:14 am, whit3rd <whit...@gmail.com> wrote:

On Saturday, June 18, 2011 11:02:17 AM UTC-7, Benj wrote:
On Jun 18, 5:26 am, glen herrmannsfeldt <g....@ugcs.caltech.edu> wrote:

Capacitance is proportional to the radius.
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"? \

The historic development of electricity included years of experimenting
with 'capacitors' that were, in fact, metal shells (suspended
on insulating strings).  The capacitance of two balls and the
ability to induce a charge (ground ball2, move it near ball1, remove
the ground, then pull ball2 far from ball1) made it  possible to
get a repeatable quantity of charge many times (because
the induced charge on ball2 doesn't at all discharge ball1).

So, 'an isolated sphere' means a spherical conductor far from (many
radii from) the nearest other conductors.   The formula for
the capacitance of an isolated sphere is quite valid.   Also, useful.

Thanks for the voice of reason whit3rd.

I wanted to add that anyone who has tried to get short response times
from a high input impedance circuit knows all about self
capacitance.

Short thin traces and no ground plane!

George H.

 

[Darwin123]

On Jun 18, 7:45 pm, Salmon Egg <Salmon...@sbcglobal.net> wrote:

of fields. It took Maxwell to understand that aspect. The true law is
div B = 0. From that show me that lines of force are closed.
The equivalent formula for the Electric field is:

div E = p
where E is the electric field and p is thr electric charge density.
When the charge density is zero (p=0),
div E = 0
Under the condition of zero electric charge density, the electric
field lines are closed.

 

[Szczepan Bialek]

"Benj" <bjacoby@iwaynet.net> napisal w wiadomosci
news:960acb8e-8d6a-4f78-8eee-6fb4a59bbe9e@q30g2000yqb.googlegroups.com...
On Jun 18, 7:24 pm, 1treePetrifiedForestLane <Space...@hotmail.com>
wrote:

don't see how "capacitance to infinity" is not
the sum of capacitances with all other plates
in Universe.

Interesting question! Capacitance between two conducting bodies
depends ONLY on geometry. Now we started with concentric spheres and

allowed the outer sphere to get larger and larger (we won't use the
mathematical term "infinite"). And what we found was that at some very
large diameter the capacitance no longer depends on the diameter of
the outer sphere. But what if we repeat the experiment with the outer
conductor being a hemisphere? Or as you suggest a bunch of segmented
plates around the universe.

"Infinite" is here doubled. It is obvious that in "infinity" (distance) is
the "infinity" number of charges.

Or what if we substitute a flat plate for
the inner sphere? I suggest that for a sphere as the outer conductor

gets far away, the shape of the inner terminal again is the only thing
determining capacitance.

For "doubled" infinity.

I presume that every geometrical shape for
the inner terminal has some kind of "equivalent sphere" that one can

use to calculate self-capacitance.

Maxwell calculated the self capacitance for a long cylinder and for a disc.

But it is not clear to me that if
the outer terminal is say a hemisphere instead of a full sphere that

the capacitance will be the same no matter how far away it is. Same
would go for a "universe" made up of various plates distributed around
in some manner.

If in infinity is infinity number of electrons no matter.

But to calculate the self capaticance you must take the both works. To push
electrons from infinity to the sphere and to compress the electrons on the
sphere.
The both are the stored energy.

Of course in "hydraulic analogy" the second work do not exists. But to
calculate the self capacitance you must use the gas analogy.

S*

 

[Benj]

On Jun 18, 7:24 pm, 1treePetrifiedForestLane <Space...@hotmail.com>
wrote:

don't see how "capacitance to infinity" is not
the sum of capacitances with all other plates
in Universe.

Interesting question! Capacitance between two conducting bodies
depends ONLY on geometry. Now we started with concentric spheres and
allowed the outer sphere to get larger and larger (we won't use the
mathematical term "infinite"). And what we found was that at some very
large diameter the capacitance no longer depends on the diameter of
the outer sphere. But what if we repeat the experiment with the outer
conductor being a hemisphere? Or as you suggest a bunch of segmented
plates around the universe. Or what if we substitute a flat plate for
the inner sphere? I suggest that for a sphere as the outer conductor
gets far away, the shape of the inner terminal again is the only thing
determining capacitance. I presume that every geometrical shape for
the inner terminal has some kind of "equivalent sphere" that one can
use to calculate self-capacitance. But it is not clear to me that if
the outer terminal is say a hemisphere instead of a full sphere that
the capacitance will be the same no matter how far away it is. Same
would go for a "universe" made up of various plates distributed around
in some manner.

 

[Salmon Egg]

In article
<726ffdf4-28b1-45cf-a1bc-13545de0e4bb@u30g2000vby.googlegroups.com>,
Darwin123 <drosen0000@yahoo.com> wrote:

On Jun 18, 7:45 pm, Salmon Egg <Salmon...@sbcglobal.net> wrote:
of fields. It took Maxwell to understand that aspect. The true law is
div B = 0. From that show me that lines of force are closed.
The equivalent formula for the Electric field is:
div E = p
where E is the electric field and p is thr electric charge density.
When the charge density is zero (p=0),
div E = 0
Under the condition of zero electric charge density, the electric
field lines are closed.

Huh? The only thing that shows is as much electric flux enters a closed
surface as leaves it. How does that show that an electric flux line
forms a closed loop?

--

Sam

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

 

[Darwin123]

On Jun 22, 12:56 pm, Salmon Egg <Salmon...@sbcglobal.net> wrote:

In article
726ffdf4-28b1-45cf-a1bc-13545de0e...@u30g2000vby.googlegroups.com>,

 Darwin123 <drosen0...@yahoo.com> wrote:
On Jun 18, 7:45 pm, Salmon Egg <Salmon...@sbcglobal.net> wrote:
of fields. It took Maxwell to understand that aspect. The true law is
div B = 0. From that show me that lines of force are closed.
   The equivalent formula for the Electric field is:
div E = p
where E is the electric field and p is thr electric charge density.
When the charge density is zero (p=0),
div E = 0
     Under the condition of zero electric charge density, the electric
field lines are closed.

Huh? The only thing that shows is as much electric flux enters a closed
surface as leaves it. How does that show that an electric flux line
forms a closed loop?
The electric charge density within the closed surface is not zero

in the diagrams that you are thinking of. The electric fields are
being generated by an electric charge. If there is a point charge
inside the closed surface, the charge density at the point is
infinite.
Electric fields can be generated by changing magnetic fields
without any electric charge around. This is called electromotive
inductance. Electric fields generated this way are in a closed loop.
The total charge of a circuit elements is usually zero. So as a
course approximation, the average charge density in a circuit is zero.
Nonzero electric charge accumulates in some parts of the circuit with
capacitance. However, a capacitance is basically a bipolar element.
Some poster in this thread used the word bipolar. Although I have
never heard the word used by an engineer or scientist, it is apparent
what he meant. He was talking about objects where there are equal and
opposite charges separated on the objects. Such an object is bipolar.
A point charge by itself would be monopolar. In this sense, circuit
elements are mostly nonpolar (resistors) or bipolar (capacitors,
batteries). There are no monopolar circuit elements.
Each half of the capacitor has an equal and opposite charge. If
you divided a capacitor in two, each plate has a nonzero electric
charge. There is an electric field in the capacitor that isn't closed.
However, this electric field is generated by what you may call a
bipolar charge.
A circuit can have a total nonzero charge, of course. However,
this nonzero charge doesn't affect the flow of current. The total
nonzero charge may be characterized by the ground potential.
A lot of interesting things are covered by ground potentials.
However, this is not relevant to the question of mutual capacitance.
There are no active monopoles in a circuit. All the electric
moments in a circuit are dipole or higher. Sources of electromotive
power a bipolar. Capacitors can be considered bipolar. However, there
are better ways to analyze the effect of a capacitance in the
circuit.
One of the best ways to figure out what is going on reducing the
entire circuit to a collection of circuit elements with complex
impedance, and currents to Fourier transforms in time. Then the issue
of "charge separation" doesn't come up. You may consider "complex
impedance" a mathematical fiction. Maybe so. However, it is easier to
design electronics with "mathematical fictions" than to sort out the
"charge separation" for each and every element.

 

[1treePetrifiedForestLane]

here is a nice statement from an old textbook:

if little loss of a resonant circuit is
in the capacitator, hte Q of a res. circ. is
practically the Q of the coil unless resistance
is added. It's often convenient to speak
of the Q of an RF coil rather than
its resistance, because the Q of a RF coil is
likely to be more nearly constant over
the useful freq. range of the coil,
than in the effective resistance (sik?)
(The comparitive constancy of Q is due
to the effect of distributed capacitance
in the coil, skin effect in the wires, and
related phenomena.)

 

[John Larkin]

On Fri, 17 Jun 2011 10:49:17 -0700 (PDT), RichD
<r_delaney2001@yahoo.com> 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?

There is dispute about the capacitance between the earth and the moon.
Some people claim about 3 uF, some claim about 160 uF. I think the
first is the "mutual" or 3-terminal capacitance, and the second is the
2-terminal capacitance.


3 uF

earth--------||---------moon
| |
| |
___ ___
___ 710 uF ___ 193 uF
| |
| |
| |
+----------------------+----- universe

John

 

[glen herrmannsfeldt]

In sci.physics.electromag John Larkin <jjlarkin@highnotlandthistechnologypart.com> wrote:

(snip)

There is dispute about the capacitance between the earth and the moon.
Some people claim about 3 uF, some claim about 160 uF. I think the
first is the "mutual" or 3-terminal capacitance, and the second is the
2-terminal capacitance.

The series capacitor rule doesn't apply if the connection
between the two is grounded (for example).

3 uF

earth--------||---------moon
| |
| |
___ ___
___ 710 uF ___ 193 uF
| |
| |
| |
+----------------------+----- universe

Given the circuit above, (assume it is smaller scale) one can
measure the 3uF capacitance with the other two in place.

Apply an AC voltage across the 710uF, measure the AC current
across the 193uF. (That is, shunt current through a low
resistance ammeter.)

-- glen

 

[John Larkin]

On Wed, 22 Jun 2011 20:08:49 -0700 (PDT), George Herold
<gherold@teachspin.com> wrote:

On Jun 22, 8:46 pm, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Fri, 17 Jun 2011 10:49:17 -0700 (PDT), RichD

r_delaney2...@yahoo.com> 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?

There is dispute about the capacitance between the earth and the moon.
Some people claim about 3 uF, some claim about 160 uF. I think the
first is the "mutual" or 3-terminal capacitance, and the second is the
2-terminal capacitance.

                3 uF

    earth--------||---------moon
      |                      |
      |                      |
     ___                    ___
     ___  710 uF            ___  193 uF
      |                      |
      |                      |
      |                      |
      +----------------------+----- universe

John

Cool! It's only the 3 uF that counts for coupling electrostatic
signals between the two.


George H.

Right. That one will taper off with distance. The 2-terminal
capacitance wouldn't change much if the moon moved out past Jupiter.
You'd just need longer test leads.

John

 

[George Herold]

On Jun 22, 8:46 pm, John Larkin
<jjlar...@highNOTlandTHIStechnologyPART.com> wrote:

On Fri, 17 Jun 2011 10:49:17 -0700 (PDT), RichD

r_delaney2...@yahoo.com> 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?

There is dispute about the capacitance between the earth and the moon.
Some people claim about 3 uF, some claim about 160 uF. I think the
first is the "mutual" or 3-terminal capacitance, and the second is the
2-terminal capacitance.

                3 uF

    earth--------||---------moon
      |                      |
      |                      |
     ___                    ___
     ___  710 uF            ___  193 uF
      |                      |
      |                      |
      |                      |
      +----------------------+----- universe

John

Cool! It's only the 3 uF that counts for coupling electrostatic
signals between the two.


George H.

 

[Szczepan Bialek]

"John Larkin" <jjlarkin@highNOTlandTHIStechnologyPART.com> napisal w
wiadomosci news:r82507p5h5dnfsovkht64d3eejlkr94ua0@4ax.com...

On Fri, 17 Jun 2011 10:49:17 -0700 (PDT), RichD
r_delaney2001@yahoo.com> 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?


There is dispute about the capacitance between the earth and the moon.
Some people claim about 3 uF, some claim about 160 uF. I think the
first is the "mutual" or 3-terminal capacitance, and the second is the
2-terminal capacitance.


3 uF

earth--------||---------moon
| |
| |
___ ___
___ 710 uF ___ 193 uF
| |
| |
| |
+----------------------+----- universe

John

The 710uF is calculated for the hydraulic analogy. For this value the
Earth's potential would be 10^9V. In reality no such voltage.

In space are ions and electrons. Each body is negatively charged. It is the
plasma physics.

S*

 

[Szczepan Bialek]

"1treePetrifiedForestLane" <Space998@hotmail.com> napisal w wiadomosci
news:1a6d4781-cf36-4d7c-8a97-db1780eafb68@r21g2000pri.googlegroups.com...

here is a nice statement from an old textbook:

if little loss of a resonant circuit is
in the capacitator, hte Q of a res. circ. is
practically the Q of the coil unless resistance
is added. It's often convenient to speak
of the Q of an RF coil rather than
its resistance, because the Q of a RF coil is
likely to be more nearly constant over
the useful freq. range of the coil,
than in the effective resistance (sik?)
(The comparitive constancy of Q is due
to the effect of distributed capacitance
in the coil, skin effect in the wires, and
related phenomena.)

I like it. How old is the book?

The plate capacitor and coil are both capacitors. The both have the large
surfaces to collect the electrons.
The important difference is the rate of the electron flow. From the plates
the electrons can flow quickly, from the coils very slowly.
S*