Vacuum has an inherent dielectric constant, known as the permittivity of a vacuum (or of 'free space'.) It also has an inherent inductance, known as the permeability of vacuum. These are values you can look up. They are both very close to what we find in air at atmospheric pressure.

A vacuum capacitor does indeed store charge in its vacuum, odd as that may seem. They typically have very small capacitance values, just as we find in air-dielectric capacitors. The air doesn't store the charge in an air-dielectric capacitor, of course: you can blow the air out from between the plates and the charge will remain.

There are lots easier ways to determine the electrical qualities of air at low pressures than the one suggested. While a balloon experiment would be fun, I've done the same thing with a small vacuum pump, my trusty ignition-coil high-voltage power supply and a suitably-rigged jelly jar.

What you're looking for is something called the Paschen curve for air. One axis of the Paschen curve plot is the voltage and the other is the pressure--though I think that there's a provision for the electrode gap in there somewhere (memory fails at times.) It turns out that this curve is fairly linear near atmospheric pressure. The voltage necessary for arc initiation is lowest at a partial vacuum. It rises for very high vacuums and for very high pressures.

M Kinsler

512 E Mulberry St. Lancaster, Ohio USA 740 687 6368



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Mark Kinsler wrote:

Vacuum has an inherent dielectric constant, known as the permittivity of a vacuum (or of 'free space'.) It also has an inherent inductance, known as the permeability of vacuum. These are values you can look up. They are both very close to what we find in air at atmospheric pressure.

A vacuum capacitor does indeed store charge in its vacuum, odd as that may seem. They typically have very small capacitance values, just as we find in air-dielectric capacitors. The air doesn't store the charge in an air-dielectric capacitor, of course: you can blow the air out from between the plates and the charge will remain.

The air dielectric capacitor is the finest example of why the charge is in the plate, not the dielectric -
and how do the plates know that the dielectric is a vacuum and therefor "the vacuum stores the charge" or that the dielectric is air and since the air can be blown away, the plate needs to store the charge?

It is basic material physics - plates store charge - dielectrics insulate the charges.


This leads to another way to explain things. If a capacitor is charged, one plate has a negative charge and the other plate has a positive charge. If the dielectric stored the charge, how does the dielectric keep the negative charges away from the positive charges?

That IS the purpose of the dielectric - to insulate the charges opposing charges on the plates from one another.


Doug Hale

There are lots easier ways to determine the electrical qualities of air at low pressures than the one suggested. While a balloon experiment would be fun, I've done the same thing with a small vacuum pump, my trusty ignition-coil high-voltage power supply and a suitably-rigged jelly jar.

What you're looking for is something called the Paschen curve for air. One axis of the Paschen curve plot is the voltage and the other is the pressure--though I think that there's a provision for the electrode gap in there somewhere (memory fails at times.) It turns out that this curve is fairly linear near atmospheric pressure. The voltage necessary for arc initiation is lowest at a partial vacuum. It rises for very high vacuums and for very high pressures.

M Kinsler

512 E Mulberry St. Lancaster, Ohio USA 740 687 6368



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开云体育

>> That IS the purpose of the dielectric - to insulate the charges opposing
>> charges on the plates from one another.

?

Well said.

?

Rama

Try this link. I think it is a good explaination though it does not
include the value of the permittivity of free space, 8.85 x 10-12
farad per meter (F/m).


solidstate/dielect.htm

Yes, the dielectric can increase the charge stored in a capacitor.
Actually, the dielectric always increases the charge stored in a
capacitor.



--- In Electronics_101@y..., Doug Hale <doughale@x> wrote:

Mark Kinsler wrote:

Vacuum has an inherent dielectric constant, known as the

permittivity of a

vacuum (or of 'free space'.) It also has an inherent inductance,

known as

the permeability of vacuum. These are values you can look up.

They are

both very close to what we find in air at atmospheric pressure.

A vacuum capacitor does indeed store charge in its vacuum, odd as

that may

seem. They typically have very small capacitance values, just as

we find in

air-dielectric capacitors. The air doesn't store the charge in an
air-dielectric capacitor, of course: you can blow the air out from

between

the plates and the charge will remain.

The air dielectric capacitor is the finest example of why the

charge is

in the plate, not the dielectric -
and how do the plates know that the dielectric is a vacuum and

therefor

"the vacuum stores the charge" or that the dielectric is air and

since

the air can be blown away, the plate needs to store the charge?

It is basic material physics - plates store charge - dielectrics
insulate the charges.


This leads to another way to explain things. If a capacitor is

charged,

one plate has a negative charge and the other plate has a positive
charge. If the dielectric stored the charge, how does the

dielectric

keep the negative charges away from the positive charges?

That IS the purpose of the dielectric - to insulate the charges

opposing

charges on the plates from one another.


Doug Hale

There are lots easier ways to determine the electrical qualities

of air at

low pressures than the one suggested. While a balloon experiment

would be

fun, I've done the same thing with a small vacuum pump, my trusty
ignition-coil high-voltage power supply and a suitably-rigged

jelly jar.

What you're looking for is something called the Paschen curve for

air. One

axis of the Paschen curve plot is the voltage and the other is the
pressure--though I think that there's a provision for the

electrode gap in

there somewhere (memory fails at times.) It turns out that this

curve is

fairly linear near atmospheric pressure. The voltage necessary

for arc

initiation is lowest at a partial vacuum. It rises for very high

vacuums

and for very high pressures.

M Kinsler

512 E Mulberry St. Lancaster, Ohio USA 740 687 6368



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Try this link too.

,,sid9_gci211945,00.html

Yes, a vacuum is a dielectric. The dielectric is defined as
a "supporter of electrostatic fields". That means, the charge is in
the field and the field is in the dielectric and...


The charge is in the dielectric. Even if it's a vacuum. Something
for nothing, it just doesn't seem right.

Mr. Manifold (?)
Can you be more specific in stating that dielectric always INCREASES the
charge stored in a capacitor? I think there must be situations where change
of dielectric DECREASES it.

SUNAN

No I can't. That one slipped out from memory and not an attributable
reference. I will try and find support and you can try and find a
way to refute it!

Dang, I'll have to open that physics book yet...

--- In Electronics_101@y..., Sunantoro <SUNANTORO@K...> wrote:

Mr. Manifold (?)
Can you be more specific in stating that dielectric always

INCREASES the

charge stored in a capacitor? I think there must be situations

where change

of dielectric DECREASES it.

SUNAN

-----Original Message-----
From: manifold [SMTP:manifold_1@y...]
Sent: Tuesday, October 30, 2001 7:28 AM
To: Electronics_101@y...
Subject: [Electronics_101] Re: vacuum and charge

Try this link. I think it is a good explaination though it

does not

include the value of the permittivity of free space, 8.85 x 10-12

farad per

meter (F/m).

<
/>

solidstate/dielect.htm
Yes, the dielectric can increase the charge stored in a

capacitor.

Actually, the dielectric always increases the charge stored in a

capacitor.

Ok, here is part of it. This is the part that I remembered about
permitivity and permeability related to the speed of light. Check
out this link:




1
-----------
_________
C = /
/ uo * eo
&#92;/

Nothing goes faster than the speed of light in a vacuum so at least
we now have the relationship between a constant and another
variable. So now I have to answer this question:

o Can we have a a situation where the permittity decreases while the
permeability increases?

still looking...

--- In Electronics_101@y..., "manifold" <manifold_1@y...> wrote:

No I can't. That one slipped out from memory and not an

attributable

reference. I will try and find support and you can try and find a
way to refute it!

Dang, I'll have to open that physics book yet...

--- In Electronics_101@y..., Sunantoro <SUNANTORO@K...> wrote:

Mr. Manifold (?)
Can you be more specific in stating that dielectric always

INCREASES the

charge stored in a capacitor? I think there must be situations

where change

of dielectric DECREASES it.

SUNAN

-----Original Message-----
From: manifold [SMTP:manifold_1@y...]
Sent: Tuesday, October 30, 2001 7:28 AM
To: Electronics_101@y...
Subject: [Electronics_101] Re: vacuum and charge

Try this link. I think it is a good explaination though it

does not

include the value of the permittivity of free space, 8.85 x 10-12

farad per

meter (F/m).

<

/>

solidstate/dielect.htm
Yes, the dielectric can increase the charge stored in a

capacitor.

Actually, the dielectric always increases the charge stored in a

capacitor.

Ah ha, finally the source of confusion.

The charge IS in the plates, the field IS in the dielectric.

The charge and the field ARE NOT the same thing.

An electric field exists between two opposite charges.

Doug Hale


manifold wrote:

Doug,

The charge IS in the plates, the field IS in the dielectric.

Cook seemed to fall short of saying that. First off, the charge
he referred to was not the resulting stored energy but that
what produced the storage in the first place.

The charge and the field ARE NOT the same thing.

An electric field exists between two opposite charges.

OK, I see that. Now, does the potential at the capacitor terminals
constitute the charge. I guess it does, I think I have always thought
of the charge as all those stored electrons. Where do they live, or
do they live at all? Maybe the charge doesn't consist of misplaced
electrons but the field. Countless years of oversimplification of a
concept. But as I said, it didn't undermine my ability to use capacitors.
Not any more than using conventional current or electron flow.

Jim

Lets look at some of the mathamatical relationships:

C = capacitance in Farads
Q = charge in Coulohms
V = voltage in Volts
I = current in amperes
t = time in seconds

C = Q/V
I = Q/t or Q = I t

substituting
C = I t/V or I = C * V/t
which is actually I = C dV/dt ( the little d sort of means the "change in" - it is actually the derivitive - which is the slope of a curve - or "change in")
which is saying that the voltage accross a one farad capacitor will change at the rate of one volt per second at one amp of current flow.

Notice that in the final equation (I = C dV/dt) we didn't worry about electric field strength or charge - just current and voltage.

So was all the other discusions worthless - not hardly - you have a better idea how a capacitor works - it is science instead of magic - I can use science - I cant use magic.

So as an engineer, to me a capacitor is I = C dV/dt. But as a scientist, it is plates, dielectrics, charges and fields.

I can design many interesting things with I=C dV/dt, but it ain't magic - its the way things are.


Doug Hale





C of a capacitor is determined by the area of the plates and the dielectric constant

Jim Purcell wrote:

How about inductors:

The engineering equation for inductors is simular th that of capacitors.

L = inductance in henri
V = voltage in volts
I = current in amperes
t = time in seconds


V = L dI/dt

which means that the current through a one henrie inductor will change at the rate of one ampere per second with a one volt drop across it.

Doug Hale



Doug Hale wrote: