A Pure Scientific Site

[John Fields]

On Sat, 19 Jul 2008 03:56:11 -0700 (PDT), Hamid.V.Ansari@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

 

[MooseFET]

On Jul 19, 4:44 pm, 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.

I had read those same "papers" some time ago. The error was pointed
out back then. It isn't very scientific to leave a known error in
place.

 

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.

 

[clicliclic@freenet.de]

Hamid.V.Ans...@gmail.com schrieb:

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.

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.)

Martin.

 

[Pieter]

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.

 

[John Fields]

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.

---
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,
allowing charge to accumulate until the pressure exerted by the spring
equals the pressure exerted by the charging supply.

All else being equal, the material the spring is made of will
determine the spring rate of the spring and the amount of energy which
can be stored for any given pressure exerted to compress the spring.

The same is true for the dielectric, with its energy storage
capability being determined by the material from which it's made.

It follows, then, that if the capacitance of a capacitor is determined
by the area of the plates, the distance between them, and the relative
static permittivity (dielectric constant) of the material the
dielectric is made from, changes in dielectric constant of the
dielectric will result in changes of capacitance in otherwise
identical capacitors.

In other words,

A
C = K e0 ---
d

Where, for a single parallel plate capacitor:

C is the capacitance in farads,

K is the dielectric constant of the dielectric,

8.85E-12 coulomb˛
e0 is -------------------, the permittivity of free space,
newton-meters˛

A is the area of the surface in contact with the dielectric of
one plate of the capacitor, in square meters, and,

d is the distance between the plates (the thickness of the
dielectric) in meters.

It's easy to see, therefore, that C goes with K, everything else being
equal.
---

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.

---
Then what's "shown" in the paper is wrong, as is the implication that
capacitors have anything to do with scale maps showing the division of
land.

JF

 

[Mark Thorson]

MooseFET wrote:

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

It can, if it becomes electret, 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.

 

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. 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.
But why do you think that the situation is different when
instead of a battery we use an alternative votage? In the paper
it is analysed that here too, dielectric acts as an additional
emf supply.

allowing charge to accumulate until the pressure exerted by the spring
equals the pressure exerted by the charging supply.

All else being equal, the material the spring is made of will
determine the spring rate of the spring and the amount of energy which
can be stored for any given pressure exerted to compress the spring.

The same is true for the dielectric, with its energy storage
capability being determined by the material from which it's made.

It follows, then, that if the capacitance of a capacitor is determined
by the area of the plates, the distance between them, and the relative
static permittivity (dielectric constant) of the material the
dielectric is made from, changes in dielectric constant of the
dielectric will result in changes of capacitance in otherwise
identical capacitors.

In other words,

               A
     C = K e0 ---
               d

Where, for a single parallel plate capacitor:

      C is the capacitance in farads,

      K is the dielectric constant of the dielectric,

              8.85E-12 coulomb˛
      e0 is  -------------------, the permittivity of free space,
               newton-meters˛

      A is the area of the surface in contact with the dielectric of
        one plate of the capacitor, in square meters, and,

      d is the distance between the plates (the thickness of the
        dielectric) in meters.

It's easy to see, therefore, that C goes with K, everything else being
equal.
---

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.

---
Then what's "shown" in the paper is wrong, as is the implication that
capacitors have anything to do with scale maps showing the division of
land.

JF- Hide quoted text -

- Show quoted text -

 

[MooseFET]

On Jul 19, 7:45 pm, 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. 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.

The correct term is "claimed" not "shown". If I say that I am typing
this while sitting on a unicorn, that is a claim, I haven't shown you
that.

You claim that a dielectric attracts more charge. This is not what it
does. It allows a larger charge to be displaced for a given voltage
between the plates. This is the very definition of it increasing the
capacitance. The dielectric is not acting on its own. It is reacting
to the applied voltage gradient and as a result is increasing the
capacitance. If it acted on its own, the motion of the dielectric
would be enough to generate a voltage on the capacitor.

 

[MooseFET]

On Jul 19, 11:06 pm, "cliclic...@freenet.de" <cliclic...@freenet.de>
wrote:

Hamid.V.Ans...@gmail.com schrieb:



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.

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.)

Even for the old single layer ones you need the microscope. The shape
of the plates of the capacitor is often very rough making the surface
area much higher than you would see with the unaided eye.

Martin.

 

[John Fields]

On Sun, 20 Jul 2008 09:25:50 -0700 (PDT), Hamid.V.Ansari@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

 

[Jamie]

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"

 

[John Larkin]

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.

Get a c-meter, a couple of metal plates, and some plastic, and try it.

How do they manage to pack 22 uF into an 0805 capacitor?

John

 

[Jamie]

John Larkin 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.


Get a c-meter, a couple of metal plates, and some plastic, and try it.

How do they manage to pack 22 uF into an 0805 capacitor?

John

Smoke and Mirrors!

http://webpages.charter.net/jamie_5"

 

[clicliclic@freenet.de]

MooseFET schrieb:

On Jul 19, 11:06?pm, "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.)

Even for the old single layer ones you need the microscope. The shape
of the plates of the capacitor is often very rough making the surface
area much higher than you would see with the unaided eye.

Certainly this would be preferable. The plates are something like
sreen-printed blobs of solderable silver paint, but I assumed without
mention that, for ceramic dielectrics, the uncertainty in plate area
couldn't possibly invalidate the general conclusion of the
calculation.

I haste to add that one mustn't pick a 1pF model for this, but
something like a 10nF disk.

(Even if the particular type of ceramic dielectric is known,
especially for class-two ceramics there also remains a large
uncertainty in the dielectric constant.)

Martin.

 

[John Fields]

On Sun, 20 Jul 2008 15:26:37 -0700 (PDT), MooseFET
<kensmith@rahul.net> wrote:

On Jul 20, 11:53 am, John Fields <jfie...@austininstruments.com
wrote:
[....]

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

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

Triboelectric != piezoelectric

...so you missed a case.

---
I don't see it, since triboelectric effects are generated by materials
touching each other and then separating:

http://en.wikipedia.org/wiki/Triboelectric_effect

That certainly isn't the situation in which a capacitor's dielectric
finds itself while in the inside of a capacitor.

One could conceivably make a case for pyroelectricity, though.

JF

 

[MooseFET]

On Jul 20, 2:15 pm, "cliclic...@freenet.de" <cliclic...@freenet.de>
wrote:

MooseFET schrieb:



On Jul 19, 11:06?pm, "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.)

Even for the old single layer ones you need the microscope.  The shape
of the plates of the capacitor is often very rough making the surface
area much higher than you would see with the unaided eye.

Certainly this would be preferable. The plates are something like
sreen-printed blobs of solderable silver paint, but I assumed without
mention that, for ceramic dielectrics, the uncertainty in plate area
couldn't possibly invalidate the general conclusion of the
calculation.

Since the capacitance depends strongly on the area and the distance
between, having a large uncertainty in those values will make the
capacitance value you compute very inaccurate.

I haste to add that one mustn't pick a 1pF model for this, but
something like a 10nF disk.

(Even if the particular type of ceramic dielectric is known,
especially for class-two ceramics there also remains a large
uncertainty in the dielectric constant.)

Martin.

 

[MooseFET]

On Jul 20, 11:53 am, John Fields <jfie...@austininstruments.com>
wrote:
[....]

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

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

Triboelectric != piezoelectric

...so you missed a case.

JF

 

[MooseFET]

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.

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.

 

On Jul 20, 10:06 am, "cliclic...@freenet.de" <cliclic...@freenet.de>
wrote:

Hamid.V.Ans...@gmail.com schrieb:





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.

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.)

Martin.- Hide quoted text -

- Show quoted text -

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

"It is shown that, contrary
to the current belief, capacitance of a capacitor does not at all
depend on the dielectric used in it and depends only on the
configuration of its conductors. ....
It is shown that existence of dielectric in the capacitor of a
circuit causes attraction of more charges onto the capacitor because
of the polarization of the dielectric. Then, in electric circuits we
should consider the capacitor's dielectric as a source of potential
not think wrongly that existence of dielectric changes the
capacitor's capacitance. Difference between these two understandings
are verified completely during some examples, and some experiments
are proposed for testing the theory. For example it is shown that
contrary to what the current theory predicts, resonance frequency of
a circuit of RLC will increase by inserting dielectric into the
capacitor (without any change of the geometry of its conductors)."


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).