mutual capacitance?
RichD
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
Phil Hobbs
Mutual capacitance does exist, e.g. the capacitance between the plates
of a differential (3-plate) variable capacitor. Actually, mutual
capacitance is the usual kind of capacitance we think about. 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.)
Cheers
Phil Hobbs
--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics Electro-optics Photonics Analog Electronics
55 Orchard Rd
Briarcliff Manor NY 10510
845-480-2058
email: hobbs (atsign) electrooptical (period) net
http://electrooptical.net
Darwin123
> But, what about mutual inductance? Why is there no
> mutual capacitance? By symmetry, shouldn't a 'mutual
> capacitor' exist, linking electric flux?
Because the physically relevant quantity of a circuit is the
complex impedance. There is a mutual complex impedance, that includes
the effects of all linear electronic elements. The complex impedance
is analogous to the resistance, but it includes all linear circuit
quantities: the resistance, the capacitances and the inductances.
The complex impedance of an inductor is sqrt(-1)(angular
frequency) (inductance). The complex impedance of a capacitor is 1/
sqrt(-1)(angular frequency)(capacitance). The complex impedance of a
resistor is the real part of the complex impedance.
The complex impedance of a circuit is analogous to the
resistance. There is a total complex impedance of a circuit is
calculated using the well known equations for total resistance taught
in elementary school, only with complex number. The total resistance
of the circuit is simply the real part of the total complex impedance
calculated this way.
One can define a mutual inductance and a mutual capacitance from
the imaginary part of the mutual complex impedance. However, the
mutual inductance will be determined by the mutual capacitance, and
vica versa. All properties important to the circuit as a whole are
contained in the mutual complex impedance.
The analogy that your subconscious is working on is likely to be
the complex impedance. This is more important in problems involving AC
current than for DC current. However, current passing through a
circuit with capacitors and inductors can't be DC (i.e., steady)
anyway. The complex impedance of a circuit can be calculated by
formulas analogous to the formulas for total resistance in DC
current.
So I recommend that you look up complex impedance. I think a lot
of things will come clear when you understand the concept.
Chris Richardson
>
> But, what about mutual inductance? Why is there no mutual capacitance?
> By symmetry, shouldn't a 'mutual capacitor' exist, linking electric
> flux?
Mutual? I thought that your question was about _dual_ elements.
A dual relation arises when current and voltage are exchanged.
For a capacitor the relation between current and voltage is:
I = C * dV/dt
Exchanging I and V, we must include inductance:
V = L * dI/dt
Hence, a capacitor is dual to an inductor.
glen herrmannsfeldt
> 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?
> Mutual capacitance does exist, e.g. the capacitance between the plates
> of a differential (3-plate) variable capacitor.
Yes. But also for the usual frequencies real capacitors are closer
to ideal than real inductors. For coupling between nearby
transmission lines, such as multiple twisted pairs in one cable,
both inductance and capacitance are important.
> Actually, mutual
> capacitance is the usual kind of capacitance we think about. 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.)
-- glen
glen herrmannsfeldt
(snip)
> A dual relation arises when current and voltage are exchanged.
> For a capacitor the relation between current and voltage is:
> I = C * dV/dt
> Exchanging I and V, we must include inductance:
> V = L * dI/dt
> Hence, a capacitor is dual to an inductor.
Yes. What is dual to a transformer? (Coupled inductors)
-- glen
1treePetrifiedForestLane
with every other plate, throw a term in
for the angle between them and sum, or what ever.
Androcles
"glen herrmannsfeldt" <g...@ugcs.caltech.edu> wrote in message
news:itga64$9o4$1...@dont-email.me...
| In sci.physics.electromag Phil Hobbs
<pcdhSpamM...@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?
|
| > Mutual capacitance does exist, e.g. the capacitance between the plates
| > of a differential (3-plate) variable capacitor.
|
| Yes. But also for the usual frequencies real capacitors are closer
| to ideal than real inductors.
That's a very strange statement to make. How would you define
"ideal"? Surely an ideal inductor has zero resistance and zero
diameter wire with as many turns as you like, wound on a bobbin
of no length and no diameter, all turns of the same length, and
that's before considering the core material.
The ideal capacitor has plates with any area you choose and a
zero gap between them, yet do not touch, and that's before
considering the dialectric material between them.
Neither can be ideal.
George Herold
The other day I wanted to determine the corner frequency of an RC
circuit that feeds into an opamp. I couldn't 'break' into the circuit
to inject a little test signal, so I wrapped a bit of plastic coated
wire about the lead going into the opamp. This formed a bit of ?
mutual? capacitance between the wire and lead. (Each has separate
capacitance to ground.) I then sent a square wave into the wire, and
this gave me little charge pulses into the opamp/ RC circuit.
I'm not sure if this maps mathematically into mutual inductance,
George H.
huhie
"RichD" <r_dela...@yahoo.com> wrote in message
news:f8b96838-84f7-4e19...@k15g2000pri.googlegroups.com...
>A network theorem states that every circuit has a
> dual; voltage sources become current sources, etc.
what network theorem is that ?
>
> But, what about mutual inductance? Why is there no
> mutual capacitance? By symmetry, shouldn't a 'mutual
> capacitor' exist, linking electric flux?
obviously you are wrong. Magnetic fields lines are closed loops, no
magnetic monopoles, and E fields bi-polar.
>
> --
> Rich
glen herrmannsfeldt
(snip on duality)
>> But, what about mutual inductance? Why is there no
>> mutual capacitance? By symmetry, shouldn't a 'mutual
>> capacitor' exist, linking electric flux?
> obviously you are wrong. Magnetic fields lines are closed loops, no
> magnetic monopoles, and E fields bi-polar.
Reminds me of an undergrad physics lecture demonstration showing
the equivalence between open and closed end air columns (organ pipes),
and open and shorted coaxial transmission lines.
During the demonstration, the lecturer figured out that the
analogy was backwards. The closed tube (pressure antinode)
coresponds to the open end coax (voltage antinode).
To correct this, the next lecture had the same setup, but with
a current probe on the oscilloscope. Shorted end is a current antinode.
Some might have tried to explain away the difference, and not
bother with the current probe. Now I still remember it over
thirty years later.
-- glen
Salmon Egg
<f8b96838-84f7-4e19...@k15g2000pri.googlegroups.com>,
RichD <r_dela...@yahoo.com> wrote:
A good reference is Smythe's Static and Dynamic Electricity.
Almost all capacitance we deal with is mutual capacitance. Two
conductors are involved, For a conductor a and a conductor b, the self
capacitance is the charge divided potential difference produced on one
conductor as if none of the other conductors were present. Essentially,
it is the capacitance to infinity. That is, self capacitances Caa =
Qa/Vainfinity and Cbb = Qb/Vbinfinity. Using cgs electric units, the
self capacitance of a sphere is its radius in centimeters.
Mutual capacitance Vab = Qab/Vab. It is the charge change produced on a
and b divided by the change of potential difference used to produce the
charge.
--
Sam
Conservatives are against Darwinism but for natural selection.
Liberals are for Darwinism but totally against any selection.
Don Kelly
news:f8b96838-84f7-4e19...@k15g2000pri.googlegroups.com...
A network theorem states that every circuit has a
--
Rich
------------------------
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.
Don Kelly
cross out to reply
Don Kelly
"glen herrmannsfeldt" wrote in message news:itga9f$9o4$2...@dont-email.me...
-- glen
------------------------------------
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
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.
Benj
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. Like
gyrators and some other oddball network elements they do not widely
exist as physical passive elements, but there are certain cases
(certain piezo devices) where they almost exist. They can be made to
exist using active simulation. And who needs a device if you have the
equations? Today, we all believe mathematics is more real than reality
anyway!
Szczepan Bialek
"Phil Hobbs" <pcdhSpamM...@electrooptical.net> napisal w wiadomosci
news:4DFB9514...@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
Itis for 2-cm diameter.
But how much pF has 4-cm diameter sphere?
S*
Szczepan Bialek
"glen herrmannsfeldt" <g...@ugcs.caltech.edu> napisał w wiadomości
news:itggbe$8oj$2...@dont-email.me...
> In sci.physics.electromag huhie <inv...@invalid.invalid> wrote:
>
> (snip on duality)
>>> But, what about mutual inductance? Why is there no
>>> mutual capacitance? By symmetry, shouldn't a 'mutual
>>> capacitor' exist, linking electric flux?
>
>> obviously you are wrong. Magnetic fields lines are closed loops, no
>> magnetic monopoles, and E fields bi-polar.
>
> Reminds me of an undergrad physics lecture demonstration showing
> the equivalence between open and closed end air columns (organ pipes),
> and open and shorted coaxial transmission lines.
>
> During the demonstration, the lecturer figured out that the
> analogy was backwards. The closed tube (pressure antinode)
> coresponds to the open end coax (voltage antinode).
It is not backwards. Pressure and voltage are the same.
>
> To correct this, the next lecture had the same setup, but with
> a current probe on the oscilloscope. Shorted end is a current antinode.
Shorted ends are like a loop dipole.
>
> Some might have tried to explain away the difference, and not
> bother with the current probe. Now I still remember it over
> thirty years later.
So is time to understand it. Electron gas is like air. The one arm of a
dipole is like Kund's tube.
S*
Szczepan Bialek
"Salmon Egg" <Salm...@sbcglobal.net> napisa� w wiadomo�ci
news:SalmonEgg-6618A...@news60.forteinc.com...
> In article
> <f8b96838-84f7-4e19...@k15g2000pri.googlegroups.com>,
> That is, self capacitances Caa =
> Qa/Vainfinity and Cbb = Qb/Vbinfinity. Using cgs electric units, the
> self capacitance of a sphere is its radius in centimeters.
Benj wrote: "Today, we all believe mathematics is more real than reality
anyway!"
In reality the "self capacitance of a sphere " is its surface and radius
dependent.
S*
.
glen herrmannsfeldt
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
k...@att.bizzzzzzzzzzzz
>
>"glen herrmannsfeldt" wrote in message news:itga9f$9o4$2...@dont-email.me...
>
>In sci.physics.electromag Chris Richardson <ro...@localhost.localdomain>
>wrote:
>
>(snip)
>> A dual relation arises when current and voltage are exchanged.
>
>> For a capacitor the relation between current and voltage is:
>
>> I = C * dV/dt
>
>> Exchanging I and V, we must include inductance:
>
>> V = L * dI/dt
>
>> Hence, a capacitor is dual to an inductor.
>
>Yes. What is dual to a transformer? (Coupled inductors)
What is the dual of the electron? (Magnetic monopole?)
Darwin123
> "RichD" <r_delaney2...@yahoo.com> wrote in message
>
> news:f8b96838-84f7-4e19...@k15g2000pri.googlegroups.c
>
> obviously you are wrong. Magnetic fields lines are closed loops, no
> magnetic monopoles, and E fields bi-polar.
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".
Globemaker
>
> news:f8b96838-84f7-4e19...@k15g2000pri.googlegroups.com...
>
> >A network theorem states that every circuit has a
> > dual; voltage sources become current sources, etc.
>
> what network theorem is that ?
>
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.
Szczepan Bialek
"glen herrmannsfeldt" <g...@ugcs.caltech.edu> napisa� w wiadomo�ci
news:ithr09$4hv$1...@dont-email.me...
> In sci.physics.electromag Szczepan Bialek <sz.b...@wp.pl> wrote:
>>
>> "Phil Hobbs" <pcdhSpamM...@electrooptical.net> napisal w
>> wiadomosci
>> news:4DFB9514...@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.
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.
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.
The self capacitance of a sphere has the two members: proportional to r and
proportional to r squared.
S*
Benj
> 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.
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.
glen herrmannsfeldt
(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.
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
1treePetrifiedForestLane
the sum of capacitances with all other plates
in Universe.
Salmon Egg
glen herrmannsfeldt <g...@ugcs.caltech.edu> wrote:
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.
Salmon Egg
<4183b1ff-dbd9-49b9...@f21g2000vbr.googlegroups.com>,
Darwin123 <drose...@yahoo.com> wrote:
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.
Jamie
The reason a SHORT is called a SHORT is because it takes a short
cut to common..
THere.!
Jamie
RichD
<pcdhSpamMeSensel...@electrooptical.net> wrote:
> RichD wrote:
> > A network theorem states that every circuit has a
> > dual; voltage sources become current sources, etc.
>
> > 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
RichD
> Yes. What is dual to a transformer? (Coupled inductors)
Now you're getting warm...
--
Rich
RichD
> > Hence, a capacitor is dual to an inductor.
>
> Yes. What is dual to a transformer? (Coupled inductors)
>
> 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
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.
Can't picture this one -
--
Rich
RichD
> > >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.
Yes.
The network duality theorem.
Crack your old textbook.
> Those theorems do not require inductors to have a
> dual,
Incorrect.
> but voltage sources become current sources.
--
Rich
RichD
> 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.
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
RichD
> > 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.
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!
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
glen herrmannsfeldt
(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
glen herrmannsfeldt
(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.
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
glen herrmannsfeldt
(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
FrediFizzx
news:349a14df-9d87-431e...@h12g2000pro.googlegroups.com...
Capacitance is simply defined as the constant of proportionality between
charge and voltage.
C = Q/V
Best,
Fred Diether
FrediFizzx
news:itjq2a$5ks$3...@dont-email.me...
> In sci.physics.electromag RichD <r_dela...@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.
C = Q/V
Must be a typo on your part since you got the first one right. :-)
Best,
Fred Diether
glen herrmannsfeldt
(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 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
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? :-)
Salmon Egg
<5ae7c5e6-b960-48da...@d1g2000yqm.googlegroups.com>,
Benj <bja...@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.
Salmon Egg
glen herrmannsfeldt <g...@ugcs.caltech.edu> 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.
Benj
> 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.
glen herrmannsfeldt
(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
Szczepan Bialek
> In sci.physics.electromag Szczepan Bialek <sz.b...@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*
Szczepan Bialek
"Salmon Egg" <Salm...@sbcglobal.net> napisa� w wiadomo�ci
news:SalmonEgg-56B34...@news60.forteinc.com...
> In article
> <5ae7c5e6-b960-48da...@d1g2000yqm.googlegroups.com>,
> Benj <bja...@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" <bja...@iwaynet.net> napisal w wiadomosci
news:a0de8577-a323-4563...@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" <g...@ugcs.caltech.edu> napisa� w wiadomo�ci
news:itkden$17d$1...@dont-email.me...
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*
Salmon Egg
<bfadfb32-87d0-4369...@e35g2000yqc.googlegroups.com>,
Benj <bja...@iwaynet.net> wrote:
The only proof here is that a little knowledge can be a dangerous thing.
Salmon Egg
There was a server error so I apolgize if this is a repeat/
Salmon Egg
glen herrmannsfeldt <g...@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.
Jamie
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
(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
> In sci.physics.electromag Salmon Egg <Salm...@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
> 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
>> 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
Jos Bergervoet
...
>> 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
glen herrmannsfeldt <g...@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.
Jos Bergervoet
> In article<itnfod$16n$3...@dont-email.me>,
> glen herrmannsfeldt<g...@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
Jos Bergervoet <jos.ber...@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.
Jos Bergervoet
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
Darwin123
> 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.
Benj
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.
Szczepan Bialek
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*
Salmon Egg
<726ffdf4-28b1-45cf...@u30g2000vby.googlegroups.com>,
Darwin123 <drose...@yahoo.com> wrote:
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?
Darwin123
> 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?
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
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
<r_dela...@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
(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
George Herold
<jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
> On Fri, 17 Jun 2011 10:49:17 -0700 (PDT), RichD
>
Cool! It's only the 3 uF that counts for coupling electrostatic
signals between the two.
George H.
John Larkin
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
Szczepan Bialek
"John Larkin" <jjla...@highNOTlandTHIStechnologyPART.com> napisal w
wiadomosci news:r82507p5h5dnfsovk...@4ax.com...
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" <Spac...@hotmail.com> napisal w wiadomosci
news:1a6d4781-cf36-4d7c...@r21g2000pri.googlegroups.com...
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*
John Larkin
Explain please.
>
>In space are ions and electrons. Each body is negatively charged. It is the
>plasma physics.
It's claimed in many places that earth is electrically neutral. I
don't entirely buy that; I'd like to see it measured.
But there's no fundamental reason why a sphere this big, well
insulated in vacuum, couldn't be charged to a gigavolt.
John
Androcles
"John Larkin" <jjla...@highNOTlandTHIStechnologyPART.com> wrote in message
news:32f607t2hgavou0pq...@4ax.com...
Is that a positive or a negative gigavolt?
"Give me but one firm spot on which to stand, and I will move the earth." --
Archimedes.
Voltage, like motion, is relative.
"Give me but one firm cathode on which to rub, and I will charge the
earth." -- Van der Graaf.
Androcles
"Androcles" <Headm...@Hogwarts.physics.June.2011> wrote in message
news:gBHMp.7776$m22....@newsfe05.ams2...
John Larkin
<Headm...@Hogwarts.physics.June.2011> wrote:
Assume the universe is neutral, which it probably is. A satellite in
elliptical orbit around Earth, or one doing a flyby, could measure
field gradients and measure Earth's potential against the universe.
John
John Larkin
<Headm...@Hogwarts.physics.June.2011> wrote:
Bizarre. Vacuum is close to a perfect insulator.
John
Androcles
"John Larkin" <jjla...@highNOTlandTHIStechnologyPART.com> wrote in message
Androcles
"John Larkin" <jjla...@highNOTlandTHIStechnologyPART.com> wrote in message
No, I don't make assumptions that I can't support.
| A satellite in
| elliptical orbit around Earth, or one doing a flyby, could measure
| field gradients and measure Earth's potential against the universe.
|
Bizarre, you'll be claiming it can weigh itself when it's weightless next.
What's the potential of the universe?
Szczepan Bialek
"John Larkin" <jjla...@highNOTlandTHIStechnologyPART.com> napisal w
> On Thu, 23 Jun 2011 09:30:28 +0200, "Szczepan Bialek"
> <sz.b...@wp.pl> wrote:
>
>>
>> "John Larkin" <jjla...@highNOTlandTHIStechnologyPART.com> napisal w
>>wiadomosci news:r82507p5h5dnfsovk...@4ax.com...
>>> On Fri, 17 Jun 2011 10:49:17 -0700 (PDT), RichD
>>> <r_dela...@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.
>
> Explain please.
In electrostatics and EM equations the electricity is as the incompressible
massles fluid.
Electron gas is sometimes like the water (hydraulic analogy). But in many
cases not.
>
>
>>
>>In space are ions and electrons. Each body is negatively charged. It is
>>the
>>plasma physics.
>
> It's claimed in many places that earth is electrically neutral. I
> don't entirely buy that; I'd like to see it measured.
Electric field of the Earth is measured for more than 100 years. It is above
100V/m.
>
> But there's no fundamental reason why a sphere this big, well
> insulated in vacuum, couldn't be charged to a gigavolt.
Of course could be charged. But Your hair would be stand up.
Of course the space is a conductor not an insulator.
S*
John Larkin
<Headm...@Hogwarts.physics.June.2011> wrote:
You don't have to support assumptions, you just assume them.
>
>| A satellite in
>| elliptical orbit around Earth, or one doing a flyby, could measure
>| field gradients and measure Earth's potential against the universe.
>|
>Bizarre, you'll be claiming it can weigh itself when it's weightless next.
>What's the potential of the universe?
>
Define it to be zero. It would still be interesting to measure any
field gradient around Earth and anything else a satellite can sweep
past. We do routinely measure magnetic fields.
I guess there could be a net charge imbalance in, say, our galaxy.
Those hyper-nova black hole things have immense magnetic fields, and
fling out jets, so I suppose we could have been bathed with charged
particles at some point.
John
John Larkin
<sz.b...@wp.pl> wrote:
That's just surface field. It doesn't say anything about the planet's
net charge.
>>
>> But there's no fundamental reason why a sphere this big, well
>> insulated in vacuum, couldn't be charged to a gigavolt.
>
>Of course could be charged. But Your hair would be stand up.
>Of course the space is a conductor not an insulator.
That's the argument for earth not having net charge, namely that the
solar wind is conductive.
We have gravitational maps of earth, and magnetic maps, and
temperature maps. I wonder if any satellite has made an electrical
potential map.
John
Androcles
"John Larkin" <jjla...@highNOTlandTHIStechnologyPART.com> wrote in message
Don't argue, just assume it.
|
| >
| >| A satellite in
| >| elliptical orbit around Earth, or one doing a flyby, could measure
| >| field gradients and measure Earth's potential against the universe.
| >|
| >Bizarre, you'll be claiming it can weigh itself when it's weightless
next.
| >What's the potential of the universe?
| >
|
| Define it to be zero. It would still be interesting to measure any
| field gradient around Earth and anything else a satellite can sweep
| past. We do routinely measure magnetic fields.
|
There already is a non-zero gravitational field from Earth, Moon
and Sun, and that is the weakest force acting on the satellite.
| I guess
I'm not interested in your bizarre guesses; these are science
newsgroups, not sci-fi newsgroups. Be a good little boy and
write a letter to Santa, he hasn't heard from you in six months
and he gets lonely. Tell him you want your very own satellite
to measure gravitational, electrostatic and magnetic fields and
I'll assume you'll find one under the Xmas tree come December,
I guess.
John Larkin
<Headm...@Hogwarts.physics.June.2011> wrote:
Oh, I get it, science should never consider things that aren't already
established science.
Be a good little boy and
>write a letter to Santa, he hasn't heard from you in six months
>and he gets lonely. Tell him you want your very own satellite
>to measure gravitational, electrostatic and magnetic fields and
>I'll assume you'll find one under the Xmas tree come December,
>I guess.
>
Is your name really Androcles? Are you really the headmaster at
Hogwarts?
John
Androcles
"John Larkin" <jjla...@highNOTlandTHIStechnologyPART.com> wrote in message
Science is the observation, investigation and explanation of
natural phenomena, in that order. And no, you don't "get it".
Nor does anyone that guesses or makes unfounded assumptions.
Science should never consider things that are fiction.
|
| Be a good little boy and
| >write a letter to Santa, he hasn't heard from you in six months
| >and he gets lonely. Tell him you want your very own satellite
| >to measure gravitational, electrostatic and magnetic fields and
| >I'll assume you'll find one under the Xmas tree come December,
| >I guess.
| >
|
| Is your name really Androcles? Are you really the headmaster at
| Hogwarts?
|
A name is what one is known by. Ten years ago I used a different
pseudonym and the response was "It's Androcles!". So yes, I really
am Androcles. Did you really choose "John Larkin" for a name or
did someone else arbitrarily choose it for you and you just went along
with it?
John Larkin
<Headm...@Hogwarts.physics.June.2011> wrote:
What I suggested is that the electric field gradients around the earth
would be interesting. What's wrong with being interested in stuff like
this? Measuring this would certainly not be science fiction.
When I was a kid, I used to measure the voltage on open-wire antennas,
with a super high impedance voltmeter, and watch as clouds passed
over, and lightning changed the charges. It was cool. I did VLF
atmospheric whistler stuff, too. Some very strange electric stuff goes
on at planetary scales.
People have recently, accidentally, discovered antimatter beams and
gamma ray bursts shooting into space from thunderstorms. Suggesting
that would have got you laughed at not so long ago.
Undiscovered things remain.
>|
>| Be a good little boy and
>| >write a letter to Santa, he hasn't heard from you in six months
>| >and he gets lonely. Tell him you want your very own satellite
>| >to measure gravitational, electrostatic and magnetic fields and
>| >I'll assume you'll find one under the Xmas tree come December,
>| >I guess.
>| >
>|
>| Is your name really Androcles? Are you really the headmaster at
>| Hogwarts?
>|
>A name is what one is known by. Ten years ago I used a different
>pseudonym and the response was "It's Androcles!". So yes, I really
>am Androcles. Did you really choose "John Larkin" for a name or
>did someone else arbitrarily choose it for you and you just went along
>with it?
It's my legal name. Works fine.
So, what's the Hogwarts thing?
John
Androcles
"John Larkin" <jjla...@highNOTlandTHIStechnologyPART.com> wrote in message
What you suggested was "there's no fundamental reason why a
sphere [the Earth] this big, well insulated in vacuum, couldn't be
charged to a gigavolt." Measuring this would certainly be sci-fi,
because if I took a CRT TV to the Moon, took away the glass
and left only a light skeleton to support the shadowmask,
phosphors, anode, cathode, grid, heater, and deflection coils it
would still function.
| When I was a kid, I used to measure the voltage on open-wire antennas,
| with a super high impedance voltmeter, and watch as clouds passed
| over, and lightning changed the charges. It was cool. I did VLF
| atmospheric whistler stuff, too. Some very strange electric stuff goes
| on at planetary scales.
|
| People have recently, accidentally, discovered antimatter beams and
| gamma ray bursts shooting into space from thunderstorms. Suggesting
| that would have got you laughed at not so long ago.
Yes, and it has me laughing now. Are you sure you don't mean anti-photon
gigavolt anti-torpedoes?
| Undiscovered things remain.
Certainly, but that's no excuse for inventing them as you are guessing.
|
| >|
| >| Be a good little boy and
| >| >write a letter to Santa, he hasn't heard from you in six months
| >| >and he gets lonely. Tell him you want your very own satellite
| >| >to measure gravitational, electrostatic and magnetic fields and
| >| >I'll assume you'll find one under the Xmas tree come December,
| >| >I guess.
| >| >
| >|
| >| Is your name really Androcles? Are you really the headmaster at
| >| Hogwarts?
| >|
| >A name is what one is known by. Ten years ago I used a different
| >pseudonym and the response was "It's Androcles!". So yes, I really
| >am Androcles. Did you really choose "John Larkin" for a name or
| >did someone else arbitrarily choose it for you and you just went along
| >with it?
|
| It's my legal name. Works fine.
That isn't what I asked you. Did you really choose "John Larkin"
for a name?
Androcles is my real name. It works fine too, I answer to it and I chose it.
If my name was John Larkin I wouldn't want that as a name and wouldn't
reply, so it wouldn't be a real name.
Are you legally related to a lawyer-in-law?
| So, what's the Hogwarts thing?
It's not Hogwarts, it's Hogwarts.physics. I'm the headmaster because
own it and I say I am. I'm also the CEO, the President and the
Managing Director. I can play many roles and have as many names
as I want, I'm a free man. Are you a free person or are you bound
by convention?
John Larkin
<Headm...@Hogwarts.physics.June.2011> wrote:
Imagine a satellite rotating at some moderate rate. Extend two wires
out, to form a dipole antenna. If there's an electric field gradient,
a sine wave will be induced into the antanna, synchronous to the
rotation. Detect this synchronously to the rotation rate. Gradients in
the nanovolts/meter range should be measurable. Other interesting
signals, like sloshing solar wind potentials, would no doubt be noted.
Maybe some of this has been done.
Vibrating probe electric field sensors work this way.
>
>
>
>| When I was a kid, I used to measure the voltage on open-wire antennas,
>| with a super high impedance voltmeter, and watch as clouds passed
>| over, and lightning changed the charges. It was cool. I did VLF
>| atmospheric whistler stuff, too. Some very strange electric stuff goes
>| on at planetary scales.
>|
>| People have recently, accidentally, discovered antimatter beams and
>| gamma ray bursts shooting into space from thunderstorms. Suggesting
>| that would have got you laughed at not so long ago.
>
>
>Yes, and it has me laughing now. Are you sure you don't mean anti-photon
>gigavolt anti-torpedoes?
OK, laugh at this:
http://www.nasa.gov/mission_pages/GLAST/news/fermi-thunderstorms.html
John
John Larkin
Oops, it's been done:
http://www-ssc.igpp.ucla.edu/personnel/russell/ESS265/POLAR_EFI_Harvey.pdf
http://lasp.colorado.edu/~ergun/PDF/fast_ess/fa_ess.pdf
apparently a lot.
John
Jos Bergervoet
> On Thu, 23 Jun 2011 18:42:23 -0700, John Larkin
>> Imagine a satellite rotating at some moderate rate. Extend two wires
>> out, to form a dipole antenna. If there's an electric field gradient,
>> a sine wave will be induced into the antanna, synchronous to the
>> rotation. Detect this synchronously to the rotation rate. Gradients in
>> the nanovolts/meter range should be measurable. Other interesting
>> signals, like sloshing solar wind potentials, would no doubt be noted.
>> Maybe some of this has been done.
>>
>> Vibrating probe electric field sensors work this way.
>>
>
> Oops, it's been done:
>
> http://www-ssc.igpp.ucla.edu/personnel/russell/ESS265/POLAR_EFI_Harvey.pdf
>
> http://lasp.colorado.edu/~ergun/PDF/fast_ess/fa_ess.pdf
>
> apparently a lot.
You can do it yourself on earth by swinging the end of a cable
around in a vertical circle. If the cable is connected to an
oscilloscope, it will show a low-frequency sine wave. Do it outside,
inside buildings there is hardly any static E-field. And you can
also just swing around a 1:10 voltage probe at the end of its
cable with a piece of extra wire fixed into its signal clamp.
(The piece of extra wire is the actual antenna, the probe just
conveniently connects to the oscilloscope).
It will show that there is E-field outside..
--
Jos
Szczepan Bialek
"John Larkin" <jjla...@highNOTlandTHIStechnologyPART.com> napisal w
Yes. But it say that on the Earth surface is the excess of electrons.
>
>>>
>>> But there's no fundamental reason why a sphere this big, well
>>> insulated in vacuum, couldn't be charged to a gigavolt.
>>
>>Of course could be charged. But Your hair would be stand up.
>>Of course the space is a conductor not an insulator.
>
>
> That's the argument for earth not having net charge, namely that the
> solar wind is conductive.
Solar wind consists of ions, electrons and dust. Ions and electrons are
plasma. In plasma all bodies have the excess of electrons.
> We have gravitational maps of earth, and magnetic maps, and
> temperature maps. I wonder if any satellite has made an electrical
> potential map.
Your "surface field" is weather dependent. You know that: "When I was a kid,
I used to measure the voltage on open-wire antennas, with a super high
impedance voltmeter, and watch as clouds passed over, and lightning changed
the charges". John.
S*
>
Androcles
"John Larkin" <jjla...@highNOTlandTHIStechnologyPART.com> wrote in message
You are not even aware of Faraday's law, curl E = -dB/dt, or
how a generator works. Your hypothetical rotating dipole would
detect a magnetic field, not an electrostatic one. An interesting
compass dipole should detect that just by no doubt pointing.
Imagine that.
| >
| >| When I was a kid, I used to measure the voltage on open-wire antennas,
| >| with a super high impedance voltmeter, and watch as clouds passed
| >| over, and lightning changed the charges. It was cool. I did VLF
| >| atmospheric whistler stuff, too. Some very strange electric stuff goes
| >| on at planetary scales.
| >|
| >| People have recently, accidentally, discovered antimatter beams and
| >| gamma ray bursts shooting into space from thunderstorms. Suggesting
| >| that would have got you laughed at not so long ago.
| >
| >
| >Yes, and it has me laughing now. Are you sure you don't mean anti-photon
| >gigavolt anti-torpedoes?
|
|
| OK, laugh at this:
|
| http://www.nasa.gov/mission_pages/GLAST/news/fermi-thunderstorms.html
|
Bwhahahahahahaha!
"You MAY BE witnessing anti-matter in the making."
Bwhahahahahahaha!
Bwhahahahahahaha!
Bwhahahahahahaha!
Androcles
"John Larkin" <jjla...@highNOTlandTHIStechnologyPART.com> wrote in message
Someone detected the Earth's magnetic field? You should tell a
sailor like Columbus about it, he might find that useful knowing
which way is North.
He could launch a satellite to tell if his compass is working
properly, apparently a lot.
John Larkin
<Headm...@Hogwarts.physics.June.2011> wrote:
We're talking about electric fields, not magnetic fields. They are
different. Look it up.
The preferred detector seems to be a couple of conductive spheres, out
on the ends of some extended booms. That makes sense, to move the most
pickup capacitance as far out as possible. Better than a simple
dipole.
John
John Larkin
<Headm...@Hogwarts.physics.June.2011> wrote:
Totally wrong. Zero loop area.
An interesting
>compass dipole should detect that just by no doubt pointing.
>Imagine that.
>
>| >
>| >| When I was a kid, I used to measure the voltage on open-wire antennas,
>| >| with a super high impedance voltmeter, and watch as clouds passed
>| >| over, and lightning changed the charges. It was cool. I did VLF
>| >| atmospheric whistler stuff, too. Some very strange electric stuff goes
>| >| on at planetary scales.
>| >|
>| >| People have recently, accidentally, discovered antimatter beams and
>| >| gamma ray bursts shooting into space from thunderstorms. Suggesting
>| >| that would have got you laughed at not so long ago.
>| >
>| >
>| >Yes, and it has me laughing now. Are you sure you don't mean anti-photon
>| >gigavolt anti-torpedoes?
>|
>|
>| OK, laugh at this:
>|
>| http://www.nasa.gov/mission_pages/GLAST/news/fermi-thunderstorms.html
>|
>Bwhahahahahahaha!
>"You MAY BE witnessing anti-matter in the making."
>Bwhahahahahahaha!
>Bwhahahahahahaha!
>Bwhahahahahahaha!
>Are you sure you don't mean anti-photon gigavolt anti-torpedoes?
>
>
>
I should believe you, and not NASA?
No thanks.
John
Androcles
"John Larkin" <jjla...@highNOTlandTHIStechnologyPART.com> wrote in message
We are talking about moving a conductor that has a sinusoidal
voltage induced in it. That's what happens in a generator. LOOK IT UP!
| The preferred detector seems to
I've had enough of your "seems to" and "maybe." Look up some basic
physics before you preach nonsense.