A Pure Scientific Site

On Jul 21, 5:21 pm, Hamid.V.Ans...@gmail.com wrote:
On Jul 21, 7:08 pm, John Larkin



jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Mon, 21 Jul 2008 07:48:35 -0700 (PDT), Max65 <mpor...@tele2.it
wrote:

Yes, the motion of dielectric generates a voltage.
Connect a dielectricless capacitor to a battery.
Now move a dielectric toward the capacitor and
finally between its plates. This MOTION causes
gathering more charge on the capacitor. And this
means flowing of current in the circuit (because
of creating a voltage).

The work you made to move the dielectric generated the voltage not the
dielectric itself.

The dielectric will be pulled into the space between the plates; you
don't have to push it. The mechanical work required to insert the
dielectric is thus negative. And no voltage is generated, since the
plates are stated to be connected to a battery.

New existence of dielectric close to the plates of the capacitor
causes more charge to be gathered onto the plates. This more
chrge has passed through the battery (dut to some additional
voltage source we must attribute to this approach).
No the capacitance increased. That is all.
 
On Jul 21, 4:55 pm, Hamid.V.Ans...@gmail.com wrote:
On Jul 21, 5:29 pm, MooseFET <kensm...@rahul.net> wrote:



On Jul 21, 4:11 am, Hamid.V.Ans...@gmail.com wrote:

On Jul 20, 10:53 pm, John Fields <jfie...@austininstruments.com
wrote:

On Sun, 20 Jul 2008 09:25:50 -0700 (PDT), Hamid.V.Ans...@gmail.com
wrote:

On Jul 20, 6:25 pm, John Fields <jfie...@austininstruments.com> wrote:
On Sat, 19 Jul 2008 19:45:08 -0700 (PDT), Hamid.V.Ans...@gmail.com
wrote:

On Jul 20, 2:44 am, John Fields <jfie...@austininstruments.com> wrote:
On Sat, 19 Jul 2008 03:56:11 -0700 (PDT), Hamid.V.Ans...@gmail.com
wrote:

Only visit:
http://hvansari.googlepages.com  or
http://www.geocities.com/hamid_vasigh

---
Your premise seems to be that if the dielectric constant of material
interposed between parallel conductors is greater than that of the
vacuum, then the energy storage in that system should decrease.  

The opposite, in real life, seems to be true.

JF

Dielectric attracts more charge onto the plates of the capacitor,
so it plays the role of an emf supply.

---
That may be true once the capacitor has been charged and the charging
supply has been disconnected from the capacitor, but while the
capacitor is being charged the dielectric is acting like a spring,

When a dielectricless capacitor is connected to a battery it gathers
some charge.

---
Unless it's broken, there is no such thing as a dielectricless
capacitor.
---

When a dielectric is inserted between the plates of
this capacitor it causes gathering more charge. It seems that
you agree that in this state the dielectrics acts as an emf supply..

---
No, I don't.  It acts as something which can concentrate the electric
field between the plates and allow more electrons to be gathered than
if a dielectric with a lower dielectric constant was in there, much
like the inductance of a coil can be increased by concentrating the
magnetic field it creates by using high-permeability materials to do
so.
---

But why do you think that the situation is different when
instead of a battery we use an alternative votage?

---
I don't. The situation is the same whatever you use for a voltage
source.
---

In the paper it is analysed that here too, dielectric acts as an additional
emf supply.

---
Only if the dielectric is piezoelectric, I believe.

JF- Hide quoted text -

Don't you believe that everything able to flow charge in
a circuit acts as a source of potential?

A resistor doesn't work as a voltage source

What do you mean? Where can a resistor alone cause a
current to flow?
You suggest that we add a dielectric to a capacitor connected to a
voltage source. If we instead add a resistor between the plates, we
also get a current flow. Adding the resistor just as much causes a
current as adding the dielectric.

But approaching a dielectric can.
Complete nonsense. You must have an external voltage source. This
makes the dielectric no better that the resistor.



When you
bring your comb near to a charged electroscope a current
flows between the leaves and the outside plate of the >electroscope.
Yes. When you move a comb with a charge on it, you are causing a
current flow.

Do you think any capacitance has been changed or approaching
of the dielectric has acted as a source of potential?
No, I do not think what I think you are suggesting I think. You
haven't explained what you mean very well but as near as I can tell
your question is based on a false assumption.


Why motion
of wires in magnetic fields in generators can be a voltage
source but motion of dielectric in an electric field cannot?!
A bird can fly. Why can't an elephant unless you attach rockets?
 
Hi John,

In the plate-dielectric case, if you place the dielectric outside the
plates and let it get sucked in, and there's no friction, the
dielectric plate will oscillate back and forth in the plane of the
plates, almost escaping the plates at extreme excursions, in (I think)
a sinusoidal motion [1]. The sum of electric and kinetic energy will
always be 0.5 joules.
I agree, that's what I meant in a previous message since you say that
the "sum of electric and kinetic energy will always be 0.5 joules".
So when the dielecric is moved in the plane of the plates, the kinetic
energy increases reducing the electric one and vice versa when the
dielectric is stopped, am I correct?


Massimo
 
Why motion
of wires in magnetic fields in generators can be a voltage
source but motion of dielectric in an electric field cannot?!
The magnetic field has to be variable to produce energy, and the
energy is produced by the movement into the variable magnetic field.
When you enter the dielectric into the electrostatic field of the two
plates, only during the movement to enter the field you may convert
the movement to charges on the dielectric faces, but when you stop the
dielectric or the whole dielectric material is entered into the field,
you can't have any change of the charges on the dielectric faces.
In each case it's your work converted to electric energy, the
dielectric itself can't never be considered a voltage source.

Massimo
 
On Tue, 22 Jul 2008 03:14:36 -0700 (PDT), Hamid.V.Ansari@gmail.com
wrote:

On Jul 22, 5:49 am, MooseFET <kensm...@rahul.net> wrote:

You suggest that we add a dielectric to a capacitor connected to a
voltage source.  If we instead add a resistor between the plates, we
also get a current flow.  Adding the resistor just as much causes a
current as adding the dielectric.

We want to have capacitor in our circuit. By connecting the two
plates of the capacitor by a resistor we will have no longer any
capacitor.
---
Not true.

The circuit's Q and impedance will have changed, but the value of the
capacitance will have stayed what it was before the resistor was
added.
---

But adding dielectric between the plates won't destroy
the apacitor.
---
Again not true, if you're using "destroy" in the sense of changing the
capacitance, since it will certainly change if the dielectric changes.

Remember:

A
C = K e0 ---
d

where e0 is the dielectric constant of the dielectric.

JF
 
In my former message I wrote "variable magnetic field", I meant a
magnetic field having the flux not the same in any point of the space
where it exists, of course.

Massimo
 
On Tue, 22 Jul 2008 05:33:40 -0500, John Fields
<jfields@austininstruments.com> wrote:

On Tue, 22 Jul 2008 03:14:36 -0700 (PDT), Hamid.V.Ansari@gmail.com
wrote:

On Jul 22, 5:49 am, MooseFET <kensm...@rahul.net> wrote:

You suggest that we add a dielectric to a capacitor connected to a
voltage source.  If we instead add a resistor between the plates, we
also get a current flow.  Adding the resistor just as much causes a
current as adding the dielectric.

We want to have capacitor in our circuit. By connecting the two
plates of the capacitor by a resistor we will have no longer any
capacitor.

---
Not true.

The circuit's Q and impedance will have changed, but the value of the
capacitance will have stayed what it was before the resistor was
added.
---

But adding dielectric between the plates won't destroy
the apacitor.

---
Again not true, if you're using "destroy" in the sense of changing the
capacitance, since it will certainly change if the dielectric changes.

Remember:

A
C = K e0 ---
d

where e0 is the dielectric constant of the dielectric.
---
Aarghhhh!!!

'K' is the dielectric constant of the dielectric.

JF
 
On Jul 22, 5:49 am, MooseFET <kensm...@rahul.net> wrote:
On Jul 21, 4:55 pm, Hamid.V.Ans...@gmail.com wrote:





On Jul 21, 5:29 pm, MooseFET <kensm...@rahul.net> wrote:

On Jul 21, 4:11 am, Hamid.V.Ans...@gmail.com wrote:

On Jul 20, 10:53 pm, John Fields <jfie...@austininstruments.com
wrote:

On Sun, 20 Jul 2008 09:25:50 -0700 (PDT), Hamid.V.Ans...@gmail.com
wrote:

On Jul 20, 6:25 pm, John Fields <jfie...@austininstruments.com> wrote:
On Sat, 19 Jul 2008 19:45:08 -0700 (PDT), Hamid.V.Ans...@gmail..com
wrote:

On Jul 20, 2:44 am, John Fields <jfie...@austininstruments.com> wrote:
On Sat, 19 Jul 2008 03:56:11 -0700 (PDT), Hamid.V.Ans...@gmail.com
wrote:

Only visit:
http://hvansari.googlepages.com  or
http://www.geocities.com/hamid_vasigh

---
Your premise seems to be that if the dielectric constant of material
interposed between parallel conductors is greater than that of the
vacuum, then the energy storage in that system should decrease.  

The opposite, in real life, seems to be true.

JF

Dielectric attracts more charge onto the plates of the capacitor,
so it plays the role of an emf supply.

---
That may be true once the capacitor has been charged and the charging
supply has been disconnected from the capacitor, but while the
capacitor is being charged the dielectric is acting like a spring,

When a dielectricless capacitor is connected to a battery it gathers
some charge.

---
Unless it's broken, there is no such thing as a dielectricless
capacitor.
---

When a dielectric is inserted between the plates of
this capacitor it causes gathering more charge. It seems that
you agree that in this state the dielectrics acts as an emf supply.

---
No, I don't.  It acts as something which can concentrate the electric
field between the plates and allow more electrons to be gathered than
if a dielectric with a lower dielectric constant was in there, much
like the inductance of a coil can be increased by concentrating the
magnetic field it creates by using high-permeability materials to do
so.
---

But why do you think that the situation is different when
instead of a battery we use an alternative votage?

---
I don't. The situation is the same whatever you use for a voltage
source.
---

In the paper it is analysed that here too, dielectric acts as an additional
emf supply.

---
Only if the dielectric is piezoelectric, I believe.

JF- Hide quoted text -

Don't you believe that everything able to flow charge in
a circuit acts as a source of potential?

A resistor doesn't work as a voltage source

What do you mean? Where can a resistor alone cause a
current to flow?

You suggest that we add a dielectric to a capacitor connected to a
voltage source.  If we instead add a resistor between the plates, we
also get a current flow.  Adding the resistor just as much causes a
current as adding the dielectric.
We want to have capacitor in our circuit. By connecting the two
plates of the capacitor by a resistor we will have no longer any
capacitor. But adding dielectric between the plates won't destroy
the apacitor.

But approaching a dielectric can.

Complete nonsense.  You must have an external voltage source.  This
makes the dielectric no better that the resistor.

When you
bring your comb near to a charged electroscope a current
flows between the leaves and the outside plate of the >electroscope.

Yes.  When you move a comb with a charge on it, you are causing a
current flow.

Do you think any capacitance has been changed or approaching
of the dielectric has acted as a source of potential?

No, I do not think what I think you are suggesting I think.  You
haven't explained what you mean very well but as near as I can tell
your question is based on a false assumption.

Why motion
of wires in magnetic fields in generators can be a voltage
source but motion of dielectric in an electric field cannot?!

A bird can fly.  Why can't an elephant unless you attach rockets?- Hide quoted text -

- Show quoted text -
 
On Jul 22, 12:12 pm, Max65 <mpor...@tele2.it> wrote:
Why motion
of wires in magnetic fields in generators can be a voltage
source but motion of dielectric in an electric field cannot?!

The magnetic field has to be variable to produce energy, and the
energy is produced by the movement into the variable magnetic field.
When you enter the dielectric into the electrostatic field of the two
plates, only during the movement to enter the field you may convert
the movement to charges on the dielectric faces, but when you stop the
dielectric or the whole dielectric material is entered into the field,
you can't have any change of the charges on the dielectric faces.
In each case it's your work converted to electric energy, the
dielectric itself can't never be considered a voltage source.
What you stated above is exactly what I mean. Anyhow, we
agree that we must attribute a source of potential to the
"motion of dielectric" (when moving toward the plates)
although you attribute it to the first
word ("motion" and the factor causing it) while I have attributed
it to the second word ("dielectric"). If it is the only difference
I accept your choice.
 
although you attribute it to the first
word ("motion" and the factor causing it) while I have attributed
it to the second word ("dielectric"). If it is the only difference
I accept your choice.
Uhmm... By my own point of view, there is a big difference attributing
the electric charge increase (if any, as I already said, I never
investigated about) to the movement or to the dielectric itself.
It's like to say that the dynamo produces energy because of its coil
not because of it rotates by an external mechanical energy. _It's the
external energy incoming into the dynamo system which is converted to
electric energy_. The first hypothesis (where the motion is the cause
of the possible energy conversion) respects the law of conservation of
energy, while the second doesn't.

Massimo
 
On Tue, 22 Jul 2008 06:23:31 -0700 (PDT), MooseFET
<kensmith@rahul.net> wrote:

On Jul 22, 3:33 am, John Fields <jfie...@austininstruments.com> wrote:
On Tue, 22 Jul 2008 03:14:36 -0700 (PDT), Hamid.V.Ans...@gmail.com
wrote:

On Jul 22, 5:49 am, MooseFET <kensm...@rahul.net> wrote:
You suggest that we add a dielectric to a capacitor connected to a
voltage source.  If we instead add a resistor between the plates, we
also get a current flow.  Adding the resistor just as much causes a
current as adding the dielectric.

We want to have capacitor in our circuit. By connecting the two
plates of the capacitor by a resistor  we will have no longer any
capacitor.

---
Not true.

The circuit's Q and impedance will have changed, but the value of the
capacitance will have stayed what it was before the resistor was
added.

Actually the capacitance will have increased for two reasons that have
to do with the resistor having nonzero cross section.
---
Yup! :)

JF
 
On Jul 22, 3:14 am, Hamid.V.Ans...@gmail.com wrote:
On Jul 22, 5:49 am, MooseFET <kensm...@rahul.net> wrote:



On Jul 21, 4:55 pm, Hamid.V.Ans...@gmail.com wrote:

On Jul 21, 5:29 pm, MooseFET <kensm...@rahul.net> wrote:

On Jul 21, 4:11 am, Hamid.V.Ans...@gmail.com wrote:

On Jul 20, 10:53 pm, John Fields <jfie...@austininstruments.com
wrote:

On Sun, 20 Jul 2008 09:25:50 -0700 (PDT), Hamid.V.Ans...@gmail.com
wrote:

On Jul 20, 6:25 pm, John Fields <jfie...@austininstruments.com> wrote:
On Sat, 19 Jul 2008 19:45:08 -0700 (PDT), Hamid.V.Ans...@gmail.com
wrote:

On Jul 20, 2:44 am, John Fields <jfie...@austininstruments.com> wrote:
On Sat, 19 Jul 2008 03:56:11 -0700 (PDT), Hamid.V.Ans...@gmail.com
wrote:

Only visit:
http://hvansari.googlepages.com  or
http://www.geocities.com/hamid_vasigh

---
Your premise seems to be that if the dielectric constant of material
interposed between parallel conductors is greater than that of the
vacuum, then the energy storage in that system should decrease.  

The opposite, in real life, seems to be true.

JF

Dielectric attracts more charge onto the plates of the capacitor,
so it plays the role of an emf supply.

---
That may be true once the capacitor has been charged and the charging
supply has been disconnected from the capacitor, but while the
capacitor is being charged the dielectric is acting like a spring,

When a dielectricless capacitor is connected to a battery it gathers
some charge.

---
Unless it's broken, there is no such thing as a dielectricless
capacitor.
---

When a dielectric is inserted between the plates of
this capacitor it causes gathering more charge. It seems that
you agree that in this state the dielectrics acts as an emf supply.

---
No, I don't.  It acts as something which can concentrate the electric
field between the plates and allow more electrons to be gathered than
if a dielectric with a lower dielectric constant was in there, much
like the inductance of a coil can be increased by concentrating the
magnetic field it creates by using high-permeability materials to do
so.
---

But why do you think that the situation is different when
instead of a battery we use an alternative votage?

---
I don't. The situation is the same whatever you use for a voltage
source.
---

In the paper it is analysed that here too, dielectric acts as an additional
emf supply.

---
Only if the dielectric is piezoelectric, I believe.

JF- Hide quoted text -

Don't you believe that everything able to flow charge in
a circuit acts as a source of potential?

A resistor doesn't work as a voltage source

What do you mean? Where can a resistor alone cause a
current to flow?

You suggest that we add a dielectric to a capacitor connected to a
voltage source.  If we instead add a resistor between the plates, we
also get a current flow.  Adding the resistor just as much causes a
current as adding the dielectric.

We want to have capacitor in our circuit. By connecting the two
plates of the capacitor by a resistor  we will have no longer any
capacitor. But adding dielectric between the plates won't destroy
the apacitor.
You seem to understand very little. When you connect the resistor
between the plates of the capacitor we most certainly still have the
capacitor. We now also has a DC current flowing between the plates
but the capacitance is still there.

[....]
A bird can fly.  Why can't an elephant unless you attach rockets?
You seem to think that attaching rockets onto the elephant will make
it no longer an elephant.
 
On Jul 22, 3:33 am, John Fields <jfie...@austininstruments.com> wrote:
On Tue, 22 Jul 2008 03:14:36 -0700 (PDT), Hamid.V.Ans...@gmail.com
wrote:

On Jul 22, 5:49 am, MooseFET <kensm...@rahul.net> wrote:
You suggest that we add a dielectric to a capacitor connected to a
voltage source.  If we instead add a resistor between the plates, we
also get a current flow.  Adding the resistor just as much causes a
current as adding the dielectric.

We want to have capacitor in our circuit. By connecting the two
plates of the capacitor by a resistor  we will have no longer any
capacitor.

---
Not true.

The circuit's Q and impedance will have changed, but the value of the
capacitance will have stayed what it was before the resistor was
added.
Actually the capacitance will have increased for two reasons that have
to do with the resistor having nonzero cross section.
 
On Sun, 20 Jul 2008 15:13:00 -0400, Jamie
<jamie_ka1lpa_not_valid_after_ka1lpa_@charter.net> wrote:

Pieter wrote:

On Sat, 19 Jul 2008 19:45:08 -0700 (PDT), Hamid.V.Ansari@gmail.com
wrote:


On Jul 20, 2:44 am, John Fields <jfie...@austininstruments.com> wrote:

On Sat, 19 Jul 2008 03:56:11 -0700 (PDT), Hamid.V.Ans...@gmail.com
wrote:


Only visit:
http://hvansari.googlepages.com or
http://www.geocities.com/hamid_vasigh

---
Your premise seems to be that if the dielectric constant of material
interposed between parallel conductors is greater than that of the
vacuum, then the energy storage in that system should decrease.

The opposite, in real life, seems to be true.

JF

Dielectric attracts more charge onto the plates of the capacitor,
so it plays the role of an emf supply. On the other hand it is
shown in the paper that capacitance of a capacitor depends
only on the configuration of its plats not also on the existence
of any dielectric.


I suggest you go to manufacterers of electrolytic capacitors and tell
them they can build them without the dielectric. You will make them
very happy.

Would you be so kind to publish their comments here? That will make us
happy too.

P.
Yes, that would be interesting since I use to be a consulting EE for an
old time capacitor company which recently closed down due to foreign
competition and lack of military orders.

http://webpages.charter.net/jamie_5"
Yes, everybody wants the cheap chinese crap nowadays. Lifetime of 1000
hours or less.

P.
 
On Sun, 20 Jul 2008 15:31:34 -0700 (PDT), MooseFET
<kensmith@rahul.net> wrote:

On Jul 20, 9:51 am, Mark Thorson <nos...@sonic.net> wrote:
MooseFET wrote:

You claim that a dielectric attracts more charge.
This is not what it does.

It can, if it becomes electret,

If you put wheels on a sled you have a wagon.
And if you remove the wheels of a car, you do not have a sled.

which often happens
with use to high-voltage capacitors.  When I was
in college, someone warned me that all high-voltage
above a certain size were required to have a shorting
strap across their terminals when not installed in
equipment.  If you leave the terminals open, charge
can accumulate over time, and you sometimes can draw
a big spark by shorting them together.

Nearly any sort of capacitor that can be used at high voltage should
be stored with its legs shorted. The plastic ones rated for 100V can
give you a nasty zap.
Pieter
 
Yup, as long as the dielectric plate is sucked in by the electric
force. If you place it fully in there by hand, the electric field does
work on your hand going in, and that dumps the 0.25 j into your hand
mechanically, so there's no oscillation and only 0.25 j remains in the
system.

Neat stuff.
You should wear that kind of "special gloves" to avoid this issue, or
try to put a bulb in your mouth and see if light on!
Maybe you'll lose your hair and then (wearing a black tunic) you could
play the uncle Fester in an Addams family movie.

That's a scientific plan!

;-)
 
Pieter wrote:
On Sun, 20 Jul 2008 15:31:34 -0700 (PDT), MooseFET
kensmith@rahul.net> wrote:

On Jul 20, 9:51 am, Mark Thorson <nos...@sonic.net> wrote:
MooseFET wrote:

You claim that a dielectric attracts more charge.
This is not what it does.

It can, if it becomes electret,

If you put wheels on a sled you have a wagon.

And if you remove the wheels of a car, you do not have a sled.
But if you put skis on the wheel-less car,
you do have a sled.
 
On Jul 22, 4:33 pm, Max65 <mpor...@tele2.it> wrote:
although you attribute it to the first
word ("motion" and the factor causing it) while I have attributed
it to the second word ("dielectric"). If it is the only difference
I accept your choice.

Uhmm... By my own point of view, there is a big difference attributing
the electric charge increase (if any, as I already said, I never
investigated about) to the movement or to the dielectric itself.
It's like to say that the dynamo produces energy because of its coil
not because of it rotates by an external mechanical energy. _It's the
external energy incoming into the dynamo system which is converted to
electric energy_. The first hypothesis (where the motion is the cause
of the possible energy conversion) respects the law of conservation of
energy, while the second doesn't.

Massimo
You are quite right. But anyway you know that traditionally we name
a dynamo in a circuit as a source of potential, ie although a dynamo
is itself an object not energy, we have in mind its energy although
not explicitly stated.
 
Hamid.V.Ans...@gmail.com schrieb:
On Jul 21, 9:50?pm, "cliclic...@freenet.de" <cliclic...@freenet.de
wrote:
Hamid.V.Ans...@gmail.com schrieb:

On Jul 20, 10:06?am, "cliclic...@freenet.de" <cliclic...@freenet.de
wrote:
Hamid.V.Ans...@gmail.com schrieb:

Dielectric attracts more charge onto the plates of the capacitor,
so it plays the role of an emf supply. On the other hand it is
shown in the paper that capacitance of a capacitor depends
only on the configuration of its plats not also on the existence
of any dielectric.

In any case, you should try to verify this revolutionary theory before
you publish it!

You might for example take one of those old AM radio tuning capacitors
with air dielectric, connect it to a capacitance meter, and observe
what happens to the reading when you immerse the capacitor in
destillated water, or in turpentine, etc. You will then discover that
your theory is contradicted by experiment, so there must be something
very wrong with it.

Calculating the capacitance of an actual ceramic capacitor according
to your theory, and comparing the result with the specification or a
meter reading might also help you see the problem. I suggest you check
the capacitance of an old single-layer disk capacitor first, and break
it open then. (For modern multilayer models you would need to study
the cross-section under a microscope.)

It is certain that dielectric has role in gathering more charge onto
the plates of a capacitopr. But this is not because of increase in
capacitance of the capacitor but is because of the role of the
dielectric as a source of potential. Copacitance meter wrongly
attributes
this capability of gathering more charge, to the increase of
capacitance
of the capacitor. In practice this may ceate no big problem, but
this is a delicate problem and as a scientific work we must be
able to show the difference.

A capacitance meter measures the capacitance of the object connected
to it by probing the dependence of the voltage across the object on
the charge pumped into (or sucked out of) it, and the result must of
course be attributed to this object - what you write is plain
nonsense.

Of course there is no problem in practice: if the capacitance of all
non-vacuum capacitors in all the electronic equipment on the world
capacitors would not depend on the presence (i.e. the dielectric
constant) of their dielectrics, virtually all equipment would fail.
And apparently you know that. Good. Very good. Why not admit it right
away?

Since there are no measurable or detectable consequences, the
"delicate problem" exists only in your imagination.

In the paper it is shown that such a
difference can be seen when we try to compare resonace frequency
of a dielectricless capacitor with the same capacitor having
dielectric.

Presumably you mean the resonance frequency of a tank circuit
consisting of your test capacitor and a fixed inductance.

Let me quote some part of the abstract of the paper:

" ... "

Some years ago a dear friend from the same overseas area of you
examined this proposed experiment of resonance frequency and
surprisingly reported what I have inserted in the paper. He did not
want me to point his name in my paper (maybe because he didn't
want to believe what he saw).

(Do you perhaps mean "asserted" where you write "inserted"? I haven't
looked at the paper.)

This is his report which I have inserted in my paper:
| Oh, yes, indeed the resonant frequencies do change as
| drastically as you suggest if you put a dielectric with high
| dielectric constant between the parallel plates of a capacitor.
| I've put an example at the end of this posting.
|
| Example of capacitor with high-K dielectric...
| You can buy "disc ceramic" capacitors with about 1.0nF capacitance.
| These are nominally 1cm diameter, with nominally 0.5mm plate
| separation, with dielectric only between the conductive plates.
| The dielectric has a very high dielectric constant. If you resonant
| such a capacitor with, say, a 5mH inductor, you will find its
| resonant frequency will be about 70kHz. You can replace that
| capacitor with one with the same plate size and spacing but air
| dielectric, resulting in roughly 0.5pF capacitance. Then you will
| find that the measured resonant frequency depends on the self-
| resonance of the inductor, because you will be very hard-pressed to
| make a 5mH inductor with self-capacitance as low as 0.5pF. If you
| choose an inductor of, say, 1uH, properly constructed, then you
| might reasonably see the effects of 0.5pF, but now you will be
| dealing with much more awkward (especially if you have limited
| access to good test equipment) resonant frequencies in the hundreds
| of MHz. You will indeed find that the resonant frequency of that
| inductor with the nominal 1.0nF ceramic-dielectric capacitor will
| be on the order of 5MHz. The Q in each case should be high enough
| (with a well-constructed inductor) to give an easily measured
| resonant frequency. I _could_ do the experiment to specifically
| demonstrate the _dramatic_ shift in resonance, and even use other
| dielectrics less extreme, but I feel no need to: as I told you
| before, I _routinely_ design resonant circuits and filters, even
| taking into account the effects of stray capacitance and inductance
| and the resistances of things like circuit board traces where
| appropriate, and within my understanding of the tolerances of the
| parts and the effects of the strays, I'm never surprised. I am
| CERTAINLY never surprised by a resonance shifting higher as I
| increase capacitance, so long as I'm within the practical range of
| the parts I'm using.
|
| Note on 1uH coil: If you make a coil with #18AWG wire, which is
| about 1.0mm diameter, and make that coil with uniformly spaced
| turns, about 2.6cm diameter turns, spaced out 2.5cm total coil
| length, it will have an inductance about 1.0uH, and its first
| parallel self-resonance at about 190MHz. That implies about 0.7pF
| effective self-capacitance. Adding an external 0.5pF capacitance
| would drop the resonant frequency to about 145MHz.


Unconfirmed rumors are worthless in science. Your frame of mind is
that of pseudoscience. Since it is comparatively easy to check your
claim about the independence of capacitance (as defined by the
dependence of voltage across a suitable two-terminal circuit on the
charge pumped in or out) on the presence of a dielectric, you should
do so yourself. Use a tank circuit if you consider this important. (In
fact, thousands of people - hobbyists, students, engineers - must be
probing this dependence year by year, and no deviation from textbook
behavior has surfaced. Go figure.)
There is nothing "surprising" at all in the letter you quote; your
correspondent is obviously an experienced engineer. Using his words to
support your assertion that the "resonance frequency of a circuit of
RLC will increase by inserting dielectric into the capacitor (without
any change of the geometry of its conductors)" would be a grave and
despicable misrepresentation. In fact he emphasizes ("I am CERTAINLY
never surprised") that he never saw deviations from textbook behavior.
Small wonder that he didn't want his name to be associated with your
imaginations; his marks of emphasis indicate that he was already
getting annoyed.

A large part of his remarks concerns the effect of stray or parasitic
capacitances and resistances: when checking capacitors by putting them
in a tank circuit, you must make sure that the capacitor under test
constitutes the dominant capacitance in the circuit, and that the
resistance and capacitance of the inductor, the resistance,
capacitance and inductance of the wiring, and the resistance and
inductance of the capacitor are all negligible.

(To some extent, the quoted passages are ambiguous because the
preceding exchange between you and him is left out and "change" or
"shift" can be read as either "increase" and "rise" or "decrease" and
"fall", etc.; the words must obviously be taken as being in accordance
with textbook behavior. And of course his "am never surprised by xxx"
must be read as "never observe xxx".)

I too am getting annoyed with your pseudo-scientific frame of mind
(read: systematic dishonesty) and am dropping off at this point.

Martin.
 
On Jul 23, 5:21 am, "cliclic...@freenet.de" <cliclic...@freenet.de>
wrote:
Hamid.V.Ans...@gmail.com schrieb:





On Jul 21, 9:50?pm, "cliclic...@freenet.de" <cliclic...@freenet.de
wrote:
Hamid.V.Ans...@gmail.com schrieb:

On Jul 20, 10:06?am, "cliclic...@freenet.de" <cliclic...@freenet.de
wrote:
Hamid.V.Ans...@gmail.com schrieb:

Dielectric attracts more charge onto the plates of the capacitor,
so it plays the role of an emf supply. On the other hand it is
shown in the paper that capacitance of a capacitor depends
only on the configuration of its plats not also on the existence
of any dielectric.

In any case, you should try to verify this revolutionary theory before
you publish it!

You might for example take one of those old AM radio tuning capacitors
with air dielectric, connect it to a capacitance meter, and observe
what happens to the reading when you immerse the capacitor in
destillated water, or in turpentine, etc. You will then discover that
your theory is contradicted by experiment, so there must be something
very wrong with it.

Calculating the capacitance of an actual ceramic capacitor according
to your theory, and comparing the result with the specification or a
meter reading might also help you see the problem. I suggest you check
the capacitance of an old single-layer disk capacitor first, and break
it open then. (For modern multilayer models you would need to study
the cross-section under a microscope.)

It is certain that dielectric has role in gathering more charge onto
the plates of a capacitopr. But this is not because of increase in
capacitance of the capacitor but is because of the role of the
dielectric as a source of potential. Copacitance meter wrongly
attributes
this capability of gathering more charge, to the increase of
capacitance
of the capacitor. In practice this may ceate no big problem, but
this is a delicate problem and as a scientific work we must be
able to show the difference.

A capacitance meter measures the capacitance of the object connected
to it by probing the dependence of the voltage across the object on
the charge pumped into (or sucked out of) it, and the result must of
course be attributed to this object - what you write is plain
nonsense.

Of course there is no problem in practice: if the capacitance of all
non-vacuum capacitors in all the electronic equipment on the world
capacitors would not depend on the presence (i.e. the dielectric
constant) of their dielectrics, virtually all equipment would fail.
And apparently you know that. Good. Very good. Why not admit it right
away?

Since there are no measurable or detectable consequences, the
"delicate problem" exists only in your imagination.

In the paper it is shown that such a
difference can be seen when we try to compare resonace frequency
of a dielectricless capacitor with the same capacitor having
dielectric.

Presumably you mean the resonance frequency of a tank circuit
consisting of your test capacitor and a fixed inductance.

Let me quote some part of the abstract of the paper:

" ... "

Some years ago a dear friend from the same overseas area of you
examined this proposed experiment of resonance frequency and
surprisingly reported what I have inserted in the paper. He did not
want me to point his name in my paper (maybe because he didn't
want to believe what he saw).

(Do you perhaps mean "asserted" where you write "inserted"? I haven't
looked at the paper.)

This is his report which I have inserted in my paper:
|        Oh, yes, indeed the resonant frequencies do change as
| drastically as you suggest if you put a dielectric with high
| dielectric constant between the parallel plates of a capacitor.
| I've put an example at the end of this posting.
|
| Example of capacitor with high-K dielectric...
| You can buy "disc ceramic" capacitors with about 1.0nF capacitance.
| These are nominally 1cm diameter, with nominally 0.5mm plate
| separation, with dielectric only between the conductive plates.
| The dielectric has a very high dielectric constant. If you resonant
| such a capacitor with, say, a 5mH inductor, you will find its
| resonant frequency will be about 70kHz.  You can replace that
| capacitor with one with the same plate size and spacing but air
| dielectric, resulting in roughly 0.5pF capacitance. Then you will
| find that the measured resonant frequency depends on the self-
| resonance of the inductor, because you will be very hard-pressed to
| make a 5mH inductor with self-capacitance as low as 0.5pF. If you
| choose an inductor of, say, 1uH, properly constructed, then you
| might reasonably see the effects of 0.5pF, but now you will be
| dealing with much more awkward (especially if you have limited
| access to good test equipment) resonant frequencies in the hundreds
| of MHz. You will indeed find that the resonant frequency of that
| inductor with the nominal 1.0nF ceramic-dielectric capacitor will
| be on the order of 5MHz. The Q in each case should be high enough
| (with a well-constructed inductor) to give an easily measured
| resonant frequency. I _could_ do the experiment to specifically
| demonstrate the _dramatic_ shift in resonance, and even use other
| dielectrics less extreme, but I feel no need to:  as I told you
| before, I _routinely_ design resonant circuits and filters, even
| taking into account the effects of stray capacitance and inductance
| and the resistances of things like circuit board traces where
| appropriate, and within my understanding of the tolerances of the
| parts and the effects of the strays, I'm never surprised. I am
| CERTAINLY never surprised by a resonance shifting higher as I
| increase capacitance, so long as I'm within the practical range of
| the parts I'm using.
|
| Note on 1uH coil:  If you make a coil with #18AWG wire, which is
| about 1.0mm diameter, and make that coil with uniformly spaced
| turns, about 2.6cm diameter turns, spaced out 2.5cm total coil
| length, it will have an inductance about 1.0uH, and its first
| parallel self-resonance at about 190MHz. That implies about 0.7pF
| effective self-capacitance. Adding an external 0.5pF capacitance
| would drop the resonant frequency to about 145MHz.

Unconfirmed rumors are worthless in science. Your frame of mind is
that of pseudoscience. Since it is comparatively easy to check your
claim about the independence of capacitance (as defined by the
dependence of voltage across a suitable two-terminal circuit on the
charge pumped in or out) on the presence of a dielectric, you should
do so yourself. Use a tank circuit if you consider this important. (In
fact, thousands of people - hobbyists, students, engineers - must be
probing this dependence year by year, and no deviation from textbook
behavior has surfaced. Go figure.)

There is nothing "surprising" at all in the letter you quote; your
correspondent is obviously an experienced engineer. Using his words to
support your assertion that the "resonance frequency of a circuit of
RLC will increase by inserting dielectric into the capacitor (without
any change of the geometry of its conductors)" would be a grave and
despicable misrepresentation. In fact he emphasizes ("I am CERTAINLY
never surprised") that he never saw deviations from textbook behavior.
Small wonder that he didn't want his name to be associated with your
imaginations; his marks of emphasis indicate that he was already
getting annoyed.

A large part of his remarks concerns the effect of stray or parasitic
capacitances and resistances: when checking capacitors by putting them
in a tank circuit, you must make sure that the capacitor under test
constitutes the dominant capacitance in the circuit, and that the
resistance and capacitance of the inductor, the resistance,
capacitance and inductance of the wiring, and the resistance and
inductance of the capacitor are all negligible.

(To some extent, the quoted passages are ambiguous because the
preceding exchange between you and him is left out and "change" or
"shift" can be read as either "increase" and "rise" or "decrease" and
"fall", etc.; the words must obviously be taken as being in accordance
with textbook behavior. And of course his "am never surprised by xxx"
must be read as "never observe xxx".)

I too am getting annoyed with your pseudo-scientific frame of mind
(read: systematic dishonesty) and am dropping off at this point.
Before declaring his report he asked me by email for paying him the
expences of some experiments performed by him which he claimed are
in support of the prediction of my paper. I was not able to act as he
asked me,
and continue discussion about it in usenet. Finally this dear friend
decided
to publish his report in usent which I repeated it now. That is all,
without
anythig else. He insisted that I shoud perform necessary experiments
myself and analyse the results myself. Unfortunately neither then nor
even
now, contrary to my desire, my life has not defined
based on experimental or even theoretical scientific researches.
If interested in the subject you can perform similar experiments
at least to prove decisively that this prediction is wrong.
 

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