mutual capacitance?

[George Herold]

On Wednesday, February 13, 2019 at 10:14:31 AM UTC-5, George Herold wrote:

On Wednesday, February 13, 2019 at 12:56:49 AM UTC-5, John Larkin wrote:
On Tue, 12 Feb 2019 15:12:11 -0800 (PST), RichD
r_delaney2001@yahoo.com> wrote:

On February 12, John Larkin wrote:
Looking at network theory and the duality theorems,
why is there no mutual capacitance? i.e. electric
flux linkages, symmetric to mutual inductance and B flux.

There are two commonly given values for the capacitance between the
earth and the moon, 160uF and 3 uF. I think one is 2-wire capacitance
and the smaller one is 3-wire.


2-wire, 3-wire?
Explicate please -

https://www.dropbox.com/s/ir45h6qd8gjryl0/2wire_3wire.JPG?dl=0

The 3-wire measurement ignores Cy and Cz.

If the moon moved away from earth, Cem would approach zero, but Ceu
and Cmu wouldn't change.

Most good c-meters will do 3-wire measurement, which allows a small
cap to be measured at the ends of coaxial cables.

OK we should be able to work out the earth-moon capacitance as a
physics problem.
Here it is using method of images.
http://www.iue.tuwien.ac.at/phd/wasshuber/node77.html

(Hmm that is for sphere's of equal radius.)

So the equation from above (equal radius spheres) is,
C_between = 4*pi*e_0 * Radius^2/distance.
(The paper uses (a) for radius.)

The next correction is of order radius^2/ distance^2
(for r<<d)

So for distant spheres the inter capacitance is
about the self capacitance * radius/ distance.

For spheres of unequal radius the answer is going
to be the same, but replace radius^2, with
radius_a * radius_b

4*pi*e_0*r_e*r_m/dist_e-m

Viola, 3.2 nF is about the Earth- Moon C.
I can't seem to make this square with my
previous estimate... but since I end up with
the same number... that's my problem.

George H.

As a first approximation we could guess that the Earth's C to the
universe is decreased by the ratio of the field lines that hit the
moon, to all of them... Which is pi*R_moon ^2/ (4*pi*Dist_E-M^2)
R_earth ~6.4 x10^6 m C_earth ~ 640 uF
R_moon ~1.7 x10^6 m
and Dist_E-M ~3.8x10^8 m.

Putting that all in.. and hopefully making no mistakes I get a drcrease
of 5x10^-6 or C_e-m ~3,200 pF .... 3.2 nF
What was your number for C_e-m?

George H.

Of course this is going to only be true at low frequency...
Speed of light and all.



--

John Larkin Highland Technology, Inc

lunatic fringe electronics

 

[George Herold]

On Wednesday, February 13, 2019 at 4:10:21 PM UTC-5, Phil Hobbs wrote:

On 2/13/19 10:51 AM, jfeng@my-deja.com wrote:
On Wednesday, February 13, 2019 at 7:14:31 AM UTC-8, George Herold wrote:
OK we should be able to work out the earth-moon capacitance as a
physics problem.
This is one of the solved problems in Maxwell's "Treatise on Electricity and Magnetism". Of course, this was before we had mks units, so the answer is given in meters instead of farads (or maybe he was jealous of Michael Faraday).


1 pF = 1.12 cm

Oh dear, I thought 1 cm = 1.12 pF?
the capacitance of a 1 cm sphere.

George h.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC / Hobbs ElectroOptics
Optics, Electro-optics, Photonics, Analog Electronics
Briarcliff Manor NY 10510

http://electrooptical.net
http://hobbs-eo.com

 

[Phil Hobbs]

On 2/13/19 8:25 PM, George Herold wrote:

On Wednesday, February 13, 2019 at 4:10:21 PM UTC-5, Phil Hobbs wrote:
On 2/13/19 10:51 AM, jfeng@my-deja.com wrote:
On Wednesday, February 13, 2019 at 7:14:31 AM UTC-8, George Herold wrote:
OK we should be able to work out the earth-moon capacitance as a
physics problem.
This is one of the solved problems in Maxwell's "Treatise on Electricity and Magnetism". Of course, this was before we had mks units, so the answer is given in meters instead of farads (or maybe he was jealous of Michael Faraday).


1 pF = 1.12 cm
Oh dear, I thought 1 cm = 1.12 pF?
the capacitance of a 1 cm sphere.

Yup, I got it backwards.

Cheers

Phil Hobbs


--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC / Hobbs ElectroOptics
Optics, Electro-optics, Photonics, Analog Electronics
Briarcliff Manor NY 10510

http://electrooptical.net
http://hobbs-eo.com

 

[RichD]

On February 12, Phil Hobbs wrote:

A 1-cm radius isolated sphere has a self-capacitance of 1 cm (Gaussian
units), which is about 1.12 pF.

Well, I have in mind your basic two plate capacitor.
I don't recall self-capacitance -

Self-capacitance is the capacitance of an object with respect to
ground, when it is in a large grounded box. An infinite box, ideally.


Ground isn't necessary. An isolated conductor with a certain amount Q
of free charge on it will have an E field. The voltage V is minus the
line integral of E dot ds from the surface to infinity. The
self-capacitance is Q/V.

Then infinity represents the reference point, at potential zero.

The problem is, in this model, 'infinity' isn't a point,
but a spherical surface.

--
Rich

 

[Phil Hobbs]

On 2/20/19 12:57 AM, RichD wrote:

On February 12, Phil Hobbs wrote:
A 1-cm radius isolated sphere has a self-capacitance of 1 cm
(Gaussian units), which is about 1.12 pF.

Well, I have in mind your basic two plate capacitor. I don't
recall self-capacitance -

Self-capacitance is the capacitance of an object with respect to
ground, when it is in a large grounded box. An infinite box,
ideally.


Ground isn't necessary. An isolated conductor with a certain
amount Q of free charge on it will have an E field. The voltage V
is minus the line integral of E dot ds from the surface to
infinity. The self-capacitance is Q/V.

Then infinity represents the reference point, at potential zero.

The problem is, in this model, 'infinity' isn't a point, but a
spherical surface.

If you've taken lower level undergraduate E&M, you know that
electrostatic fields in source-free regions obey Laplace's equation, i.e.

div grad phi = 0

and that

E = grad phi,

where phi is the (scalar) electric potential. Alternatively, phi is the
line integral of E.

By a vector identity, curl grad phi is identically zero. By Stokes'
theorem, the potential difference along any closed path is the surface
integral of the curl of the potential, so since curl phi = 0, the
potential at any point is independent of how you got there. So the
voltage is well defined everywhere.

The energy density of the field is proportional to the volume integral
of |E|**2. (It's E**2 / 8pi in Gaussian units.)

In order for that to be finite, E has to go to zero at large distances
faster than 1/r. (Actually it's asymptotically 1/r**2, and becomes
purely radial very quickly--all tangential components die off as higher
powers of r.)

Thus there's no voltage difference between points at large distances, so
one point is as good as another.

Because of this, we adopt the simple convention that the potential at
infinity is zero, allowing us to write

energy = 1/2 CV**2, where C is the self-capacitance.

We compute the self-capacitance by doing the volume integral to get the
field energy and equating the two expressions. It's completely
analogous to computing the self-inductance of a solenoid.

No giant spheres required, no 'other plate' need apply.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC / Hobbs ElectroOptics
Optics, Electro-optics, Photonics, Analog Electronics
Briarcliff Manor NY 10510

http://electrooptical.net
http://hobbs-eo.com

 

[Phil Hobbs]

On 2/20/19 9:22 AM, Phil Hobbs wrote:

On 2/20/19 12:57 AM, RichD wrote:
On February 12, Phil Hobbs wrote:
A 1-cm radius isolated sphere has a self-capacitance of 1
cm (Gaussian units), which is about 1.12 pF.

Well, I have in mind your basic two plate capacitor. I don't
recall self-capacitance -

Self-capacitance is the capacitance of an object with respect
to ground, when it is in a large grounded box. An infinite
box, ideally.


Ground isn't necessary. An isolated conductor with a certain
amount Q of free charge on it will have an E field. The voltage
V is minus the line integral of E dot ds from the surface to
infinity. The self-capacitance is Q/V.

Then infinity represents the reference point, at potential zero.

The problem is, in this model, 'infinity' isn't a point, but a
spherical surface.

If you've taken lower level undergraduate E&M, you know that
electrostatic fields in source-free regions obey Laplace's equation,
i.e.

div grad phi = 0

and that

E = grad phi,

where phi is the (scalar) electric potential. Alternatively, phi is
the line integral of E.

By a vector identity, curl grad phi is identically zero. By Stokes'
theorem, the potential difference along any closed path is the
surface integral of the curl of the potential

gradient, so since curl grad phi = 0, the

potential at any point is independent of how you got there. So the
voltage is well defined everywhere.

The energy density of the field is proportional to the volume
integral of |E|**2. (It's E**2 / 8pi in Gaussian units.)

In order for that to be finite, E has to go to zero at large
distances faster than 1/r. (Actually it's asymptotically 1/r**2, and
becomes purely radial very quickly--all tangential components die off
as higher powers of r.)

Thus there's no voltage difference between points at large distances,
so one point is as good as another.

Because of this, we adopt the simple convention that the potential
at infinity is zero, allowing us to write

energy = 1/2 CV**2, where C is the self-capacitance.

We compute the self-capacitance by doing the volume integral to get
the field energy and equating the two expressions. It's completely
analogous to computing the self-inductance of a solenoid.

No giant spheres required, no 'other plate' need apply.

Cheers

Phil Hobbs

Coffee hadn't kicked in.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC / Hobbs ElectroOptics
Optics, Electro-optics, Photonics, Analog Electronics
Briarcliff Manor NY 10510

http://electrooptical.net
http://hobbs-eo.com

 

[John Larkin]

On Tue, 19 Feb 2019 21:57:05 -0800 (PST), RichD
<r_delaney2001@yahoo.com> wrote:

On February 12, Phil Hobbs wrote:
A 1-cm radius isolated sphere has a self-capacitance of 1 cm (Gaussian
units), which is about 1.12 pF.

Well, I have in mind your basic two plate capacitor.
I don't recall self-capacitance -

Self-capacitance is the capacitance of an object with respect to
ground, when it is in a large grounded box. An infinite box, ideally.


Ground isn't necessary. An isolated conductor with a certain amount Q
of free charge on it will have an E field. The voltage V is minus the
line integral of E dot ds from the surface to infinity. The
self-capacitance is Q/V.

Then infinity represents the reference point, at potential zero.

The problem is, in this model, 'infinity' isn't a point,
but a spherical surface.

A couple of light-years away is probably good enough.

Need long test leads.


--

John Larkin Highland Technology, Inc

lunatic fringe electronics

 

[George Herold]

On Wednesday, February 20, 2019 at 10:28:42 PM UTC-5, John Larkin wrote:

On Tue, 19 Feb 2019 21:57:05 -0800 (PST), RichD
r_delaney2001@yahoo.com> wrote:

On February 12, Phil Hobbs wrote:
A 1-cm radius isolated sphere has a self-capacitance of 1 cm (Gaussian
units), which is about 1.12 pF.

Well, I have in mind your basic two plate capacitor.
I don't recall self-capacitance -

Self-capacitance is the capacitance of an object with respect to
ground, when it is in a large grounded box. An infinite box, ideally.


Ground isn't necessary. An isolated conductor with a certain amount Q
of free charge on it will have an E field. The voltage V is minus the
line integral of E dot ds from the surface to infinity. The
self-capacitance is Q/V.

Then infinity represents the reference point, at potential zero.

The problem is, in this model, 'infinity' isn't a point,
but a spherical surface.

A couple of light-years away is probably good enough.

Need long test leads.

The moon will be far enough away.
A conducting sphere at the moon orbit would change the Earth's C
by less than 1%.
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capsph.html

George H.

--

John Larkin Highland Technology, Inc

lunatic fringe electronics

 

[John Larkin]

On Thu, 21 Feb 2019 05:51:46 -0800 (PST), George Herold
<gherold@teachspin.com> wrote:

On Wednesday, February 20, 2019 at 10:28:42 PM UTC-5, John Larkin wrote:
On Tue, 19 Feb 2019 21:57:05 -0800 (PST), RichD
r_delaney2001@yahoo.com> wrote:

On February 12, Phil Hobbs wrote:
A 1-cm radius isolated sphere has a self-capacitance of 1 cm (Gaussian
units), which is about 1.12 pF.

Well, I have in mind your basic two plate capacitor.
I don't recall self-capacitance -

Self-capacitance is the capacitance of an object with respect to
ground, when it is in a large grounded box. An infinite box, ideally.


Ground isn't necessary. An isolated conductor with a certain amount Q
of free charge on it will have an E field. The voltage V is minus the
line integral of E dot ds from the surface to infinity. The
self-capacitance is Q/V.

Then infinity represents the reference point, at potential zero.

The problem is, in this model, 'infinity' isn't a point,
but a spherical surface.

A couple of light-years away is probably good enough.

Need long test leads.
The moon will be far enough away.
A conducting sphere at the moon orbit would change the Earth's C
by less than 1%.
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capsph.html

George H.

How would you make a decent e-field gradient sensor? Some sort of
vibrating reed thing? Spinning ball with electrodes?

I bet there's e-fields everywhere. Insects? Plants? Doorknobs?
Footprints on the floor? Bicycles? It would be cool if we could see
e-fields.






--

John Larkin Highland Technology, Inc
picosecond timing precision measurement

jlarkin att highlandtechnology dott com
http://www.highlandtechnology.com

 

[Jasen Betts]

On 2019-02-22, John Larkin <jjlarkin@highland_snip_technology.com> wrote:

On Thu, 21 Feb 2019 05:51:46 -0800 (PST), George Herold
gherold@teachspin.com> wrote:

On Wednesday, February 20, 2019 at 10:28:42 PM UTC-5, John Larkin wrote:
On Tue, 19 Feb 2019 21:57:05 -0800 (PST), RichD
r_delaney2001@yahoo.com> wrote:

On February 12, Phil Hobbs wrote:
A 1-cm radius isolated sphere has a self-capacitance of 1 cm (Gaussian
units), which is about 1.12 pF.

Well, I have in mind your basic two plate capacitor.
I don't recall self-capacitance -

Self-capacitance is the capacitance of an object with respect to
ground, when it is in a large grounded box. An infinite box, ideally.


Ground isn't necessary. An isolated conductor with a certain amount Q
of free charge on it will have an E field. The voltage V is minus the
line integral of E dot ds from the surface to infinity. The
self-capacitance is Q/V.

Then infinity represents the reference point, at potential zero.

The problem is, in this model, 'infinity' isn't a point,
but a spherical surface.

A couple of light-years away is probably good enough.

Need long test leads.
The moon will be far enough away.
A conducting sphere at the moon orbit would change the Earth's C
by less than 1%.
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capsph.html

George H.


How would you make a decent e-field gradient sensor? Some sort of
vibrating reed thing? Spinning ball with electrodes?

I bet there's e-fields everywhere. Insects? Plants? Doorknobs?
Footprints on the floor? Bicycles? It would be cool if we could see
e-fields.

use a field mill

--
When I tried casting out nines I made a hash of it.

 

[Phil Hobbs]

On 2/22/19 3:03 PM, John Larkin wrote:

On Thu, 21 Feb 2019 05:51:46 -0800 (PST), George Herold
gherold@teachspin.com> wrote:

On Wednesday, February 20, 2019 at 10:28:42 PM UTC-5, John Larkin wrote:
On Tue, 19 Feb 2019 21:57:05 -0800 (PST), RichD
r_delaney2001@yahoo.com> wrote:

On February 12, Phil Hobbs wrote:
A 1-cm radius isolated sphere has a self-capacitance of 1 cm (Gaussian
units), which is about 1.12 pF.

Well, I have in mind your basic two plate capacitor.
I don't recall self-capacitance -

Self-capacitance is the capacitance of an object with respect to
ground, when it is in a large grounded box. An infinite box, ideally.


Ground isn't necessary. An isolated conductor with a certain amount Q
of free charge on it will have an E field. The voltage V is minus the
line integral of E dot ds from the surface to infinity. The
self-capacitance is Q/V.

Then infinity represents the reference point, at potential zero.

The problem is, in this model, 'infinity' isn't a point,
but a spherical surface.

A couple of light-years away is probably good enough.

Need long test leads.
The moon will be far enough away.
A conducting sphere at the moon orbit would change the Earth's C
by less than 1%.
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capsph.html

George H.


How would you make a decent e-field gradient sensor? Some sort of
vibrating reed thing? Spinning ball with electrodes?

Vibrating reed electrometers are commonly used to measure electrostatic
voltages.

Cheers

Phil Hobbs



--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC / Hobbs ElectroOptics
Optics, Electro-optics, Photonics, Analog Electronics
Briarcliff Manor NY 10510

http://electrooptical.net
http://hobbs-eo.com

 

[George Herold]

On Friday, February 22, 2019 at 3:04:04 PM UTC-5, John Larkin wrote:

On Thu, 21 Feb 2019 05:51:46 -0800 (PST), George Herold
gherold@teachspin.com> wrote:

On Wednesday, February 20, 2019 at 10:28:42 PM UTC-5, John Larkin wrote:
On Tue, 19 Feb 2019 21:57:05 -0800 (PST), RichD
r_delaney2001@yahoo.com> wrote:

On February 12, Phil Hobbs wrote:
A 1-cm radius isolated sphere has a self-capacitance of 1 cm (Gaussian
units), which is about 1.12 pF.

Well, I have in mind your basic two plate capacitor.
I don't recall self-capacitance -

Self-capacitance is the capacitance of an object with respect to
ground, when it is in a large grounded box. An infinite box, ideally.


Ground isn't necessary. An isolated conductor with a certain amount Q
of free charge on it will have an E field. The voltage V is minus the
line integral of E dot ds from the surface to infinity. The
self-capacitance is Q/V.

Then infinity represents the reference point, at potential zero.

The problem is, in this model, 'infinity' isn't a point,
but a spherical surface.

A couple of light-years away is probably good enough.

Need long test leads.
The moon will be far enough away.
A conducting sphere at the moon orbit would change the Earth's C
by less than 1%.
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capsph.html

George H.


How would you make a decent e-field gradient sensor? Some sort of
vibrating reed thing? Spinning ball with electrodes?

No (good) idea... An electric dipole in a gradient feels a force.
I think an electric quadrupole in a E field gradient would feel a
torque. (though it's a bit hard for me to 'see'.)

I never heard of vibrating reeds...
This is a nice lecture/ chapter.
http://www.feynmanlectures.caltech.edu/II_09.html

George H.

I bet there's e-fields everywhere. Insects? Plants? Doorknobs?
Footprints on the floor? Bicycles? It would be cool if we could see
e-fields.







--

John Larkin Highland Technology, Inc
picosecond timing precision measurement

jlarkin att highlandtechnology dott com
http://www.highlandtechnology.com

 

[RichD]

On February 11, Jasen Betts wrote:

Looking at network theory and the duality theorems,
why is there no mutual capacitance? i.e. electric
flux linkages, symmetric to mutual inductance and B flux.

you'll see something like mutual capacitance in piezo-electric
transformers.

Sounds interesting.
The flux from one capacitor, flows through another?


--
Rich

 

[Jasen Betts]

On 2019-02-28, RichD <r_delaney2001@yahoo.com> wrote:

On February 11, Jasen Betts wrote:
Looking at network theory and the duality theorems,
why is there no mutual capacitance? i.e. electric
flux linkages, symmetric to mutual inductance and B flux.

you'll see something like mutual capacitance in piezo-electric
transformers.


Sounds interesting.
The flux from one capacitor, flows through another?

charging one capacitor puts a voltage on the other capacitor according
to the ratio of the piezo transformer. but it's piezoelectric, it's not
really a capacitance effect.

--
Rich

--
When I tried casting out nines I made a hash of it.

 

[whit3rd]

On Friday, February 22, 2019 at 12:04:04 PM UTC-8, John Larkin wrote:

On Thu, 21 Feb 2019 05:51:46 -0800 (PST), George Herold

How would you make a decent e-field gradient sensor? Some sort of
vibrating reed thing? Spinning ball with electrodes?

A dielectric fiber will polarize and align with E-field.. From field lines (because
of Laplace's law) you can map gradients in uncharged spaces

Input data will look ike this:

<https://www.hvfx.co.uk/images/hair-raising-2.jpg>

 

[John Larkin]

On Sun, 3 Mar 2019 18:09:13 -0800 (PST), whit3rd <whit3rd@gmail.com>
wrote:

On Friday, February 22, 2019 at 12:04:04 PM UTC-8, John Larkin wrote:
On Thu, 21 Feb 2019 05:51:46 -0800 (PST), George Herold

How would you make a decent e-field gradient sensor? Some sort of
vibrating reed thing? Spinning ball with electrodes?

A dielectric fiber will polarize and align with E-field.. From field lines (because
of Laplace's law) you can map gradients in uncharged spaces

Input data will look ike this:

https://www.hvfx.co.uk/images/hair-raising-2.jpg

Can you calibrate that?


--

John Larkin Highland Technology, Inc

lunatic fringe electronics

 

[George Herold]

On Monday, March 4, 2019 at 11:42:52 PM UTC-5, John Larkin wrote:

On Sun, 3 Mar 2019 18:09:13 -0800 (PST), whit3rd <whit3rd@gmail.com
wrote:

On Friday, February 22, 2019 at 12:04:04 PM UTC-8, John Larkin wrote:
On Thu, 21 Feb 2019 05:51:46 -0800 (PST), George Herold

How would you make a decent e-field gradient sensor? Some sort of
vibrating reed thing? Spinning ball with electrodes?

A dielectric fiber will polarize and align with E-field.. From field lines (because
of Laplace's law) you can map gradients in uncharged spaces

Input data will look ike this:

https://www.hvfx.co.uk/images/hair-raising-2.jpg

Can you calibrate that?

Hmm I think that's ~1.5 long haired blonde,
about 1kV/m in MKS units. :^)

George H.

--

John Larkin Highland Technology, Inc

lunatic fringe electronics