On Sat, 19 Jul 2008 03:56:11 -0700 (PDT), Hamid.V.Ans... at (no spam) 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.

On Sat, 19 Jul 2008 03:56:11 -0700 (PDT), Hamid.V.Ans... at (no spam) 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

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

On Jul 20, 2:44?am, John Fields <jfie... at (no spam) austininstruments.com> wrote:
On Sat, 19 Jul 2008 03:56:11 -0700 (PDT), Hamid.V.Ans... at (no spam) 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.

On Jul 20, 2:44 am, John Fields <jfie... at (no spam) austininstruments.com> wrote:
On Sat, 19 Jul 2008 03:56:11 -0700 (PDT), Hamid.V.Ans... at (no spam) 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.

On Sat, 19 Jul 2008 19:45:08 -0700 (PDT), Hamid.V.Ans... at (no spam) gmail.com
wrote:





On Jul 20, 2:44 am, John Fields <jfie... at (no spam) austininstruments.com> wrote:
On Sat, 19 Jul 2008 03:56:11 -0700 (PDT), Hamid.V.Ans... at (no spam) 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 coulomb²
      e0 is  -------------------, the permittivity of free space,
               newton-meters²

      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 -

On Jul 20, 2:44 am, John Fields <jfie... at (no spam) austininstruments.com> wrote:

On Sat, 19 Jul 2008 03:56:11 -0700 (PDT), Hamid.V.Ans... at (no spam) 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.

Hamid.V.Ans... at (no spam) gmail.com schrieb:



On Jul 20, 2:44?am, John Fields <jfie... at (no spam) austininstruments.com> wrote:
On Sat, 19 Jul 2008 03:56:11 -0700 (PDT), Hamid.V.Ans... at (no spam) 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.

On Jul 20, 2:44 am, John Fields <jfie... at (no spam) austininstruments.com> wrote:
On Sat, 19 Jul 2008 03:56:11 -0700 (PDT), Hamid.V.Ans... at (no spam) 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.

On Jul 19, 11:06?pm, "cliclic... at (no spam) freenet.de" <cliclic... at (no spam) freenet.de
wrote:
Hamid.V.Ans... at (no spam) 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.

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

MooseFET schrieb:



On Jul 19, 11:06?pm, "cliclic... at (no spam) freenet.de" <cliclic... at (no spam) freenet.de
wrote:
Hamid.V.Ans... at (no spam) 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.

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

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.