mutual capacitance?

In sci.physics.electromag Szczepan Bialek <sz.bialek@wp.pl> wrote:
"Phil Hobbs" <pcdhSpamMeSenseless@electrooptical.net> napisal w wiadomosci
news:4DFB9514.2010709@electrooptical.net...

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.).

Itis for 2-cm diameter.
But how much pF has 4-cm diameter sphere?
Click to expand...
Capacitance is proportional to the radius. If not, the unit
would not be cm. To figure it out, find the capacitance between
concentric spheres in the limit that the outer sphere radius
goes to infinity.

For an extra check, find the capacitance of two concentric spheres
of constant spacing (delta r) as the radii increases. For r >> dr,
it is proportional to r squared, as you would expect.

-- glen
 
On Jun 17, 4:34 pm, "huhie" <inva...@invalid.invalid> wrote:
"RichD" <r_delaney2...@yahoo.com> wrote in message

news:f8b96838-84f7-4e19-b352-f470595ebeb8@k15g2000pri.googlegroups.c

obviously you are wrong.  Magnetic fields lines are closed loops, no
magnetic monopoles, and E fields bi-polar.
By bi-polar, you probably mean a point where the field begins and
Click to expand...
ends. Electric circuits have such a point.
If the total charge density is zero, the the electric field has
to be bipolar. Electric field lines begin at a positive charge and end
at a negative charge. In an electric circuit, both electric and
magnetic fields are "bi-polar".
The reason that a circuit is called a circuit is because the
current moves in a closed path. By the generalized Ohms law with
complex impedance, the electric field in the circuit also has to
define the same closed path. There is an effective electromotive
source where the electric field begins and ends. This effective
electromotive source is an effective "bi-pole".
 
On Jun 17, 4:34 pm, "huhie" <inva...@invalid.invalid> wrote:
"RichD" <r_delaney2...@yahoo.com> wrote in message

news:f8b96838-84f7-4e19-b352-f470595ebeb8@k15g2000pri.googlegroups.com...

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

what network theorem is that ?

The Thevenin Equivalent circuit can be converted to Norton Equivalent.
Click to expand...
Theorems in that area seem related to the original question. Those
theorems do not require inductors to have a dual, but voltage sources
become current sources.
 
"glen herrmannsfeldt" <gah@ugcs.caltech.edu> napisał w wiadomości
news:ithr09$4hv$1@dont-email.me...
In sci.physics.electromag Szczepan Bialek <sz.bialek@wp.pl> wrote:

"Phil Hobbs" <pcdhSpamMeSenseless@electrooptical.net> napisal w
wiadomosci
news:4DFB9514.2010709@electrooptical.net...

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.).

Itis for 2-cm diameter.
But how much pF has 4-cm diameter sphere?

Capacitance is proportional to the radius. If not, the unit
would not be cm.
Click to expand...
For a plate capacitor C = S/d and is proportional to surface.

To figure it out, find the capacitance between
concentric spheres in the limit that the outer sphere radius
goes to infinity.
Click to expand...
It is a math joke. E is proportional to the charge density.

For an extra check, find the capacitance of two concentric spheres
of constant spacing (delta r) as the radii increases. For r >> dr,
it is proportional to r squared, as you would expect.
Click to expand...
The self capacitance of a sphere has the two members: proportional to r and
proportional to r squared.
S*
 
On Jun 18, 5:26 am, glen herrmannsfeldt <g...@ugcs.caltech.edu> wrote:

Capacitance is proportional to the radius.  If not, the unit
would not be cm.  To figure it out, find the capacitance between
concentric spheres in the limit that the outer sphere radius
goes to infinity.
Click to expand...
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.

There. Isn't that better now?

Whenever someone uses the word "infinity" they are no longer talking
physics. They are talking mathematics.
 
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.
Click to expand...
The capacitance for concentric spheres has two terms.

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.

-- glen
 
In article <ithr09$4hv$1@dont-email.me>,
glen herrmannsfeldt <gah@ugcs.caltech.edu> wrote:

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

"Phil Hobbs" <pcdhSpamMeSenseless@electrooptical.net> napisal w wiadomosci
news:4DFB9514.2010709@electrooptical.net...

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.).

Itis for 2-cm diameter.
But how much pF has 4-cm diameter sphere?

Capacitance is proportional to the radius. If not, the unit
would not be cm. To figure it out, find the capacitance between
concentric spheres in the limit that the outer sphere radius
goes to infinity.

For an extra check, find the capacitance of two concentric spheres
of constant spacing (delta r) as the radii increases. For r >> dr,
it is proportional to r squared, as you would expect.

-- glen
Click to expand...
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.

--

Sam

Conservatives are against Darwinism but for natural selection.
Liberals are for Darwinism but totally against any selection.
 
In article
<4183b1ff-dbd9-49b9-b95a-ddd615f4767d@f21g2000vbr.googlegroups.com>,
Darwin123 <drosen0000@yahoo.com> wrote:

On Jun 17, 4:34 pm, "huhie" <inva...@invalid.invalid> wrote:
"RichD" <r_delaney2...@yahoo.com> wrote in message

news:f8b96838-84f7-4e19-b352-f470595ebeb8@k15g2000pri.googlegroups.c

obviously you are wrong.  Magnetic fields lines are closed loops, no
magnetic monopoles, and E fields bi-polar.
By bi-polar, you probably mean a point where the field begins and
ends. Electric circuits have such a point.
If the total charge density is zero, the the electric field has
to be bipolar. Electric field lines begin at a positive charge and end
at a negative charge. In an electric circuit, both electric and
magnetic fields are "bi-polar".
The reason that a circuit is called a circuit is because the
current moves in a closed path. By the generalized Ohms law with
complex impedance, the electric field in the circuit also has to
define the same closed path. There is an effective electromotive
source where the electric field begins and ends. This effective
electromotive source is an effective "bi-pole".
Click to expand...
There is no such thing as a line of force. Faraday used lines of force
as a convenience. Smart as he was, he did not understand the mathematics
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.

--

Sam

Conservatives are against Darwinism but for natural selection.
Liberals are for Darwinism but totally against any selection.
 
Darwin123 wrote:
On Jun 17, 4:34 pm, "huhie" <inva...@invalid.invalid> wrote:

"RichD" <r_delaney2...@yahoo.com> wrote in message

news:f8b96838-84f7-4e19-b352-f470595ebeb8@k15g2000pri.googlegroups.c

obviously you are wrong. Magnetic fields lines are closed loops, no
magnetic monopoles, and E fields bi-polar.

By bi-polar, you probably mean a point where the field begins and
ends. Electric circuits have such a point.
If the total charge density is zero, the the electric field has
to be bipolar. Electric field lines begin at a positive charge and end
at a negative charge. In an electric circuit, both electric and
magnetic fields are "bi-polar".
The reason that a circuit is called a circuit is because the
current moves in a closed path. By the generalized Ohms law with
complex impedance, the electric field in the circuit also has to
define the same closed path. There is an effective electromotive
source where the electric field begins and ends. This effective
electromotive source is an effective "bi-pole".
I might as well get in on this band wagon.
Click to expand...

The reason a SHORT is called a SHORT is because it takes a short
cut to common..

THere.!

Jamie
 
don't see how "capacitance to infinity" is not
the sum of capacitances with all other plates
in Universe.
 
"RichD" <r_delaney2001@yahoo.com> wrote in message
news:349a14df-9d87-431e-a55f-64945c44062f@h12g2000pro.googlegroups.com...
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 -
Click to expand...
Capacitance is simply defined as the constant of proportionality between
charge and voltage.

C = Q/V

Best,

Fred Diether
 
"glen herrmannsfeldt" <gah@ugcs.caltech.edu> wrote in message
news:itjq2a$5ks$3@dont-email.me...
In sci.physics.electromag RichD <r_delaney2001@yahoo.com> wrote:

(snip)
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 -

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.
Click to expand...
C = Q/V

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

Best,

Fred Diether
 
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.  
Click to expand...
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.)
Click to expand...
I don't get this one - self capacitance? Capacitance
requires 2 surfaces, plus dielectric, unless things
have changed -

--
Rich
 
On Jun 17, glen herrmannsfeldt <g...@ugcs.caltech.edu> wrote:

Yes.  What is dual to a transformer?  (Coupled inductors)
Click to expand...
Now you're getting warm...

--
Rich
 
On Jun 17, "Don Kelly" <d...@shawcross.ca> wrote:
Hence, a capacitor is dual to an inductor.

Yes.  What is dual to a transformer?  (Coupled inductors)

v1=L11(di1/dt) +L12 (di2/dt)
V2=L21(di1/dt) +L22(di2/dt)
inductive coupling

vs

I1 =C11(dv1/dt) +C12(dv2/dt)
I2=C21(dv1/dt) +C22(dv2/dt)
Capacitive coupling
Click to expand...
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? It's the dual of a transformer, yes/no?
Electric flux linkage, vs. magnetic flux linkage?


It exists-grab a fence wire parallel to and under a transmission line-
get the benefit of C21(dv1/dt) where I2 is the current through your
body- a real world problem.
Click to expand...
Can't picture this one -

--
Rich
 
On Jun 18, Globemaker <alanfolms...@cabanova.com> wrote:
A network theorem states that every circuit has a
dual; voltage sources become current sources, etc.

what network theorem is that ?

The Thevenin Equivalent circuit can be converted to Norton
Equivalent. Theorems in that area seem related to the original
question.
Click to expand...
Yes.
The network duality theorem.
Crack your old textbook.

Those theorems do not require inductors to have a
dual,
Click to expand...
Incorrect.

but voltage sources become current sources.
Click to expand...
--
Rich
 
On Jun 17, "Don Kelly" <d...@shawcross.ca> 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?

Mutual capacitance does exist.
One practical case is a multiconductor power transmission lines
where there is capacitance coupling between conductors and
to images of conductors.
One can form a Potential coefficient matrix P which based
on V=PQ. The form of the terms in this matrix are analogous
to the inductance matrix and involve terms of the form ln Dij/Hij  
where D is the distance between conductors i, j and H is the
distance from conductor i to the image of j  The inverse of P
is a capacitance matrix where the Cii terms are "self"
capacitances and the Cij terms are the "mutual" capacitances.
  In the inductor case one looks at self and mutual impedances
and in the capacitor case, the dual is self and mutual admittances.
Click to expand...
Interesting. That sounds about what I'm looking
for, describing the network in matrix form. It seems
to imply a capacitive transformer.

But it must be fairly obscure, as I've never seen this
formulation. And I don't follow the 'image' thing.


--
Rich
 
On Jun 17, Benj <bjac...@iwaynet.net> 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?

Depends what you mean by "exist".  While as others have pointed out
there is mutual capacitance, it isn't really a true dual to a
transformer. The true dual is the capacitive transformer which retains
all the features of a magnetic transformer. Generally speaking the
device doesn't exist except in certain special circumstances, but it's
widely used as a theoretical aid to network calculations.
Click to expand...
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.

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

--
Rich
 
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.
Click to expand...
How about the van de Graaf generator? Doesn't seem so far off.

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

(snip)
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 -
Click to expand...
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.

If you don't like that, consider the energy in the electric
field around a sphere as a function of the charge on the
sphere. Integrate the energy over all space, and compare
to C*V*V/2 = Q*Q/C/2.

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

-- glen
 
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