A Pure Scientific Site

On Jul 20, 8:36 pm, MooseFET <kensm...@rahul.net> wrote:

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.- Hide quoted text -

- Show quoted text -

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).

 

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?

 

[John Larkin]

On Mon, 21 Jul 2008 03:49:13 -0700 (PDT), Hamid.V.Ansari@gmail.com
wrote:

"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)."

I can tell that you don't own a lot of test equipment.

John

 

[John Larkin]

On Mon, 21 Jul 2008 07:48:35 -0700 (PDT), Max65 <mporzio@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.

John

 

[MooseFET]

On Jul 20, 4:45 pm, John Fields <jfie...@austininstruments.com> wrote:

On Sun, 20 Jul 2008 15:26:37 -0700 (PDT), MooseFET

kensm...@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.

The triboelectric would happen if you experimentally took a hammer to
the capacitor. It happens when the dielectric material is damaged.

I also missed the magneto-hydrodynamic voltage that you could create
in a dielectric if you placed it in the field of a superconducting
magnet and set off an atomic weapon within one meter of it.

 

[MooseFET]

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.

 

[MooseFET]

On Jul 21, 4:02 am, Hamid.V.Ans...@gmail.com wrote:

On Jul 20, 8:36 pm, MooseFET <kensm...@rahul.net> wrote:



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.- Hide quoted text -

- Show quoted text -

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 current only flows only if something else is creating the
voltage. The dielectric increases the capacitance and hence the
current flows. Without the external voltage source, there is no flow
of current. The dielectric is in no way making a voltage. It is only
responding to the externally supplied voltage.

 

[John Fields]

On Mon, 21 Jul 2008 03:49:13 -0700 (PDT), Hamid.V.Ansari@gmail.com
wrote:

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.

---
You seem to have it backwards, since what causes the capacitance of
the capacitor to increase when a dielectric is interposed between its
plates isn't that the dielectric is generating a voltage, but that
it's changing the electric field around the plates when an external
voltage source is connected across the plates, allowing more charge to
be accumulated than would be possible without the presence of the
dielectric.
---

Copacitance meter wrongly attributes this capability
of gathering more charge, to the increase of capacitance
of the capacitor.

---
Not true.

Since the charge accumulated in a capacitor can be described by:


Q = CV


Where Q is the quantity of accumulated charge, in coulombs,

C is the capacitance, in farads, and

V is the voltage across the plates, in volts,

Then we can rearrange to solve for capacitance, like this:


Q It
C = --- = ----
V V


So, if we were to inject a known current into the capacitor and kept
track of the time it took for the voltage across the capacitor to rise
to a particular value, then we'd have Q and V, and a simple division
would get us C, which is roughly what most capacitance meters do.

Just for grins, let's assume that we manage to force a constant
current of 1 ampere into a capacitor and that the time it takes for
the voltage across its terminals to rise to one volt is one second.

Then:


It 1A * 1s
C = --- = --------- = 1F
V 1V


Now, assume that we do the same thing with another capacitor but that
the voltage rises to 1 volt in 2 seconds.

Then we have:


It 1A * 2s
C = --- = --------- = 2F
V 1V


So it's clear that the capacitance meter is measuring quantity of
charge accumulated, which goes with capacitance.
---

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.

---
But you haven't presented a scientific work, all you've done is made a
number of ridiculous assertions which you can't back up with credible
evidence.
---

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)."

---
What you're suggesting is that a resonant circuit with a vacuum
dielectric capacitor will be resonant at a lower frequency than the
identical circuit with some other dielectric between the plates.

That's utterly preposterous and, while you may think your "paper" is
meritorious, I suggest it's more suited for use as toilet tissue.
---

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.

---
Two wrongs don't make a right.
---

He did not
want me to point his name in my paper (maybe because he didn't
want to believe what he saw).

---
Or, more than likely, because he didn't want to be associated with
you.

JF

 

[Max65]

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.
Any voltage or current generators are energy converters indeed (dynamo
are motion to electric, battery are chemical to electric, solar panels
are photon to electric etc.).
A voltage generator without current doesn't exists and vice versa,
they are always energy "generators" indeed (meaning converters, of
course).
If you are right, you should tell me what kind of energy your
generator converts to produce the voltage, otherwise we are talking
about such kind of perpetual machine that produce energy from nothing
(if true, you solved the current international energetic crisis,
congratulations).

Don't you believe that everything able to flow
charge in a circuit acts as a source of potential?
Yes! The work you made moving the dielectric is a source of potential.

Have a nice day.

 

[John Larkin]

On Mon, 21 Jul 2008 08:52:19 -0700 (PDT), Max65 <mporzio@tele2.it>
wrote:

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.

John
Hi John,
I don't agree, even if I never did any test about, I guess that if I
move the dielectric into an electrostatic field and the dielectric
changes someways because of its interaction with that field, I have to
do such kind of work to move it inside the field (probably it's a very
little work, but I've to do it), and maybe that work will be return as
an additional voltage on the plates since the energy I put into the
system has to go somewere when the dielectric stops its movement.
It's like to move one closed loop of wire into a constant magnetic
field, the adittional work made to counteract the magnetic field (that
want to keep steady the wire) is converted to current into the wire
and dissipated into heating.
Don't you agree?
Let me know.
Have a great day

Massimo

It's a little simpler to first consider a pair of parallel plates,
charged to some voltage difference but not connected to a battery or
anything.

Assume the charge is Q = Co * V, where Co is the capacitance between
the plates. Let's use V=1 volt, C=1 farad. Stored energy is 0.5*c*v^2=
0.5 joules.

Now let the dielectric slide in between the plates. Say Er=2 and the
capacitance goes up to Cd = 2 * Co, namely 2 farads. Since charge is
conserved, and C doubled, the voltage drops in half. So we have 0.5
volts, 2 farads.

We have twice the C but half the V, so the electrical energy in the
system has fallen to half of what it was before: 0.25 joules.

Where did the other 0.25 joules of energy disappear to?

It was delivered to your hand as you slid the dielectric in. The
dielectric was *pulled* into the cap. It could have lifted a weight,
and done work.

This is just like the classic puzzle: 2 caps, each 1F, one charged to
1 volt, the other to zero. Connect them in parallel. Q is conserved,
so the voltage drops to 0.5 across both. The energy drops from 0.5 to
0.25 joules. Where did the other 0.25 j go?

John

 

[Max65]

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.

John
Hi John,

I don't agree, even if I never did any test about, I guess that if I
move the dielectric into an electrostatic field and the dielectric
changes someways because of its interaction with that field, I have to
do such kind of work to move it inside the field (probably it's a very
little work, but I've to do it), and maybe that work will be return as
an additional voltage on the plates since the energy I put into the
system has to go somewere when the dielectric stops its movement.
It's like to move one closed loop of wire into a constant magnetic
field, the adittional work made to counteract the magnetic field (that
want to keep steady the wire) is converted to current into the wire
and dissipated into heating.
Don't you agree?
Let me know.
Have a great day

Massimo

 

[John Fields]

On Mon, 21 Jul 2008 10:40:09 -0700, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Mon, 21 Jul 2008 08:52:19 -0700 (PDT), Max65 <mporzio@tele2.it
wrote:

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.

John
Hi John,
I don't agree, even if I never did any test about, I guess that if I
move the dielectric into an electrostatic field and the dielectric
changes someways because of its interaction with that field, I have to
do such kind of work to move it inside the field (probably it's a very
little work, but I've to do it), and maybe that work will be return as
an additional voltage on the plates since the energy I put into the
system has to go somewere when the dielectric stops its movement.
It's like to move one closed loop of wire into a constant magnetic
field, the adittional work made to counteract the magnetic field (that
want to keep steady the wire) is converted to current into the wire
and dissipated into heating.
Don't you agree?
Let me know.
Have a great day

Massimo

It's a little simpler to first consider a pair of parallel plates,
charged to some voltage difference but not connected to a battery or
anything.

Assume the charge is Q = Co * V, where Co is the capacitance between
the plates. Let's use V=1 volt, C=1 farad. Stored energy is 0.5*c*v^2=
0.5 joules.

Now let the dielectric slide in between the plates. Say Er=2 and the
capacitance goes up to Cd = 2 * Co, namely 2 farads. Since charge is
conserved, and C doubled, the voltage drops in half. So we have 0.5
volts, 2 farads.

We have twice the C but half the V, so the electrical energy in the
system has fallen to half of what it was before: 0.25 joules.

Where did the other 0.25 joules of energy disappear to?

It was delivered to your hand as you slid the dielectric in. The
dielectric was *pulled* into the cap. It could have lifted a weight,
and done work.

This is just like the classic puzzle: 2 caps, each 1F, one charged to
1 volt, the other to zero. Connect them in parallel. Q is conserved,
so the voltage drops to 0.5 across both. The energy drops from 0.5 to
0.25 joules. Where did the other 0.25 j go?

---
W

JF

 

[clicliclic@freenet.de]

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.)

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.)

Martin.

 

[John Larkin]

On Mon, 21 Jul 2008 13:05:25 -0700 (PDT), Max65 <mporzio@tele2.it>
wrote:

Hi John,

This is just like the classic puzzle: 2 caps, each 1F, one charged to
1 volt, the other to zero. Connect them in parallel. Q is conserved,
so the voltage drops to 0.5 across both. The energy drops from 0.5 to
0.25 joules. Where did the other 0.25 j go?

I've an idea about the two caps puzzle.
Assuming no resistance between the two caps, the only solution is an
electromagnetic emmission due to the two voltage steps in zero
seconds.
If you analyze the resulting shape in the frequency domain (instead of
the time domain), you get a zero time pulse with infinite armonics.
That should be the place where the other 0.25j went.
In the true life (we know) that energy spread between the
electromagnetic pulse and the parassitic components of the capacitor
leads.


Massimo

The analogies are nice. In the paralleled-cap case, with no losses,
the two caps oscillate with the connection inductance, exchanging
charge back and forth forever, and the full 0.5 joules stays within
the system.

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.

In both cases, in the real world, half the energy is eventually lost,
and 0.25 joules remain.

John

[1] that would be an interesting differential equation.

 

[Max65]

Hi John,

This is just like the classic puzzle: 2 caps, each 1F, one charged to
1 volt, the other to zero. Connect them in parallel. Q is conserved,
so the voltage drops to 0.5 across both. The energy drops from 0.5 to
0.25 joules. Where did the other 0.25 j go?

I've an idea about the two caps puzzle.
Assuming no resistance between the two caps, the only solution is an
electromagnetic emmission due to the two voltage steps in zero
seconds.
If you analyze the resulting shape in the frequency domain (instead of
the time domain), you get a zero time pulse with infinite armonics.
That should be the place where the other 0.25j went.
In the true life (we know) that energy spread between the
electromagnetic pulse and the parassitic components of the capacitor
leads.


Massimo

 

[John Fields]

On Mon, 21 Jul 2008 17:12:49 -0700 (PDT), Hamid.V.Ansari@gmail.com
wrote:

On Jul 21, 6:48 pm, 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.
Any voltage or current generators are energy converters indeed (dynamo
are motion to electric, battery are chemical to electric, solar panels
are photon to electric etc.).
A voltage generator without current doesn't exists and vice versa,
they are always energy "generators" indeed (meaning converters, of
course).
If you are right, you should tell me what kind of energy your
generator converts to produce the voltage

As you said in dynamo or generator, motion, caused by a human
or flow of water or ..., is converted to electric current in a
magnetic
field. Here too, a similar motion is converted to electric current in
an electric field.

---
But there _is_ no motion in a capacitor where the electric field
"lines" are cut by the dielectric which is analogous to current being
generated by a conductor cutting magnetic field lines.

JF

 

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? But approaching a dielectric can. When you
bring your comb near to a charged electroscope a current
flows between the leaves and the outside plate of the electroscope.
Do you think any capacitance has been changed or approaching
of the dielectric has acted as a source of potential? Why motion
of wires in magnetic fields in generators can be a voltage
source but motion of dielectric in an electric field cannot?!

 

On Jul 21, 6:48 pm, 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.
Any voltage or current generators are energy converters indeed (dynamo
are motion to electric, battery are chemical to electric, solar panels
are photon to electric etc.).
A voltage generator without current doesn't exists and vice versa,
they are always energy "generators" indeed (meaning converters, of
course).
If you are right, you should tell me what kind of energy your
generator converts to produce the voltage

As you said in dynamo or generator, motion, caused by a human
or flow of water or ..., is converted to electric current in a
magnetic
field. Here too, a similar motion is converted to electric current in
an electric field.

, otherwise we are talking
about such kind of perpetual machine that produce energy from nothing
(if true, you solved the current international energetic crisis,
congratulations).

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

Yes! The work you made moving the dielectric is a source of potential.

Have a nice day.

 

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).

 

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.)

Martin.- Hide quoted text -

- Show quoted text -