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capacitor basics

R

Robin

Jan 1, 1970
0
Sorry Tim, that reply should be tacked onto Jasen Betts' post.

I blame Google beta for that.

Cheers
Robin
 
R

Robin

Jan 1, 1970
0
....yes, so it is. Damn brain, go back to sleep.

Robin
 
R

Roger Dewhurst

Jan 1, 1970
0
Jasen Betts said:
If nothing is touching the other terminal no current will flow through the
one you touch.

Bye.
Jasen

Surely if you touch one terminal of a charged capacitor some of the charge
on one plate will be transferred to your body? Some alarm sensors appear to
work on that principle. For example the #2 pin of a 555 can be held high
via a 1M resistor but will go low enough to trigger the 555 if touched even
though there may be no common ground.

R
 
J

Jasen Betts

Jan 1, 1970
0
No, as originally stated, the voltage will *decrease* with "more
dielectric".

Exacltly what is "more dielectric" supposed to mean?
in the (usual) case where the dielecttric is non-compressible
I'm taking it to mean more distance between the plates.
This is because the dielectric is "stretched" and *that* absorbs energy
hence the voltage must drop a little because energy is conserved.

no.

stretching takes work. (opposing the electrostatic attraction)
so energy is added to the system.

also consider the amount of charge. with no path to the other plate the
charge will stay in the capacitor, so assuming that the "extra dielectric"
displaces the plate the capacitance will fall and the voltage will rise.
See Feynman's Lectures on Physics volume 2 chapter 10 section 3.

I don't have a copy handy.
 
K

Kunal

Jan 1, 1970
0
In that case, you and the lightning are inside the capacitor, made up
of the ground as one plate and the cloud as the other plate, with the
air between being the dielectric.

You can call lightning as a sort of "dielectric breakdown". In fact, if
you apply enough voltage across the capacitor (by enough voltage I mean
enough to cause a dielectric breakdown), you'd probably simulate a
"lightning discharge". In the case of the lightning strike, the
dielectric is air.
 
B

Bill Bowden

Jan 1, 1970
0
Exacltly what is "more dielectric" supposed to mean?
in the (usual) case where the dielecttric is non-compressible
I'm taking it to mean more distance between the plates.

No. The volume is constant but the dielectric does not completely fill
the volume. It's a type of variable capacitor where the dielectric can
be slid in and out between the plates. The voltage rises as the
dielectric is removed, and falls when inserted.

-Bill
 
J

Jasen Betts

Jan 1, 1970
0
Surely if you touch one terminal of a charged capacitor some of the charge
on one plate will be transferred to your body?

some tiny amount possibly, but not a sustained current to significantly
discharge the capacitor, and not again if you touched it again.
Some alarm sensors appear to
work on that principle. For example the #2 pin of a 555 can be held high
via a 1M resistor but will go low enough to trigger the 555 if touched even
though there may be no common ground.

more likely the 555 input sees some fraction of the mains ripple voltage when
touched..

Bye.
Jasen
 
J

Jasen Betts

Jan 1, 1970
0
No. The volume is constant but the dielectric does not completely fill
the volume. It's a type of variable capacitor where the dielectric can
be slid in and out between the plates. The voltage rises as the
dielectric is removed, and falls when inserted.

-Bill

air (or vacuum) can be considered a dielectric...
by replacing it with a solid (or liquid) insulator
you aren't adding dielectric you're exchanging it...


Bye.
Jasen
 
J

John Fields

Jan 1, 1970
0
air (or vacuum) can be considered a dielectric...
by replacing it with a solid (or liquid) insulator
you aren't adding dielectric you're exchanging it...

---
So what?

There's still _more_ dielectric in there than there was before.

The point really is that since:

Q
V = ---
C

For a fixed Q, if C increases, for _any_ reason, V will decrease.
And vice-versa.

For example, consider a capacitor with a lead zirconate titanate
dielectric with a dielectric constant of 7000 charged to 1V. Remove
the dielectric and the voltage across the plates will rise to 7000V!
 
T

Tim Williams

Jan 1, 1970
0
John Fields said:
For example, consider a capacitor with a lead zirconate titanate
dielectric with a dielectric constant of 7000 charged to 1V. Remove
the dielectric and the voltage across the plates will rise to 7000V!

Except it won't: the dielectric retains the charge ala static cling. ;)

Conservations of charge and energy do slightly different things when you
move things around.

Consider the basic physics example: take one 100uF capacitor (C1) charged to
10V and another (C2) at 0V. C1 has Q = VC = 1mC on it, while C2 has none.
Conservation of charge states that, after connecting them, the charge
remains equal, so we now have a composite 200uF capacitor with 1mC on it. V
= Q/C = 5V. The energy in C1 is E = 1/2CV^2 = 5mJ, but afterwards, total
energy = 0.5 * 200u * 5^2 = 250mJ. (These capacitors were joined with
mathematically perfect wires, yet something got away! ;-)

The point is, the charge remains on the dielectric, with or without the
plates. (Remember, the plates just spread the charge over the surface --
they mean absolutely nothing more mathematically and have are not capacitors
themselves! In air or vacuum (K=1), they serve to spread charge to build
the electric field, which holds the energy.) Conversely, an uncharged
dielectric consumes power as, adding it to a K=1 capacitor, it removes
energy as it is charged.

The overall effect is, yes, voltage decreases as dielectric is slid into the
gap, but no, it is not a reversible process; the charges remain in the
dielectric. The mechanics change here, breakinn down the assumptions made
in certain equations. If instead the dielectric were squeezed elastically,
so that the capacitor's area goes up as distance goes down (or vice-versa,
compressing from the edges) this would be a reversible process, with
mechanical energy exerted and released respectively.

By the way, a dielectric "wants" to move into a charged gap. This ought to
make plenty of sense since it only wants to remove energy from the system!
Lots of things happen, in effect, because they are the route to the lowest
final energy. As shown above, halving the charge between two equal
capacitors will half the energy in the system, pretty favorable I'd say.

Tim
 
B

Bill Bowden

Jan 1, 1970
0
Tim said:
Consider the basic physics example: take one 100uF capacitor (C1) charged to
10V and another (C2) at 0V. C1 has Q = VC = 1mC on it, while C2 has none.
Conservation of charge states that, after connecting them, the charge
remains equal, so we now have a composite 200uF capacitor with 1mC on it. V
= Q/C = 5V. The energy in C1 is E = 1/2CV^2 = 5mJ, but afterwards, total
energy = 0.5 * 200u * 5^2 = 250mJ. (These capacitors were joined with
mathematically perfect wires, yet something got away! ;-)

If there are no I^2R losses, and there are no radiation losses, the
system of two parallel capacitors should retain all the original
energy. So, I would guess the charge would just oscillate back and
forth between the two caps like a tank circuit. What do you suppose the
frequency would be?

-Bill
 
T

Tim Williams

Jan 1, 1970
0
Bill Bowden said:
... So, I would guess the charge would just oscillate back and
forth between the two caps like a tank circuit. What do you suppose the
frequency would be?

Hmm, with perfect wires? I think this is an "only God can divide by zero"
problem. ;-)

Tim
 
J

John Fields

Jan 1, 1970
0
Except it won't: the dielectric retains the charge ala static cling. ;)

Conservations of charge and energy do slightly different things when you
move things around.

Consider the basic physics example: take one 100uF capacitor (C1) charged to
10V and another (C2) at 0V. C1 has Q = VC = 1mC on it, while C2 has none.
Conservation of charge states that, after connecting them, the charge
remains equal, so we now have a composite 200uF capacitor with 1mC on it. V
= Q/C = 5V. The energy in C1 is E = 1/2CV^2 = 5mJ, but afterwards, total
energy = 0.5 * 200u * 5^2 = 250mJ. (These capacitors were joined with ^^^^^
2.5mJ?

mathematically perfect wires, yet something got away! ;-)

The point is, the charge remains on the dielectric, with or without the
plates. (Remember, the plates just spread the charge over the surface --
they mean absolutely nothing more mathematically and have are not capacitors
themselves! In air or vacuum (K=1), they serve to spread charge to build
the electric field, which holds the energy.) Conversely, an uncharged
dielectric consumes power as, adding it to a K=1 capacitor, it removes
energy as it is charged.

The overall effect is, yes, voltage decreases as dielectric is slid into the
gap, but no, it is not a reversible process; the charges remain in the
dielectric. The mechanics change here, breakinn down the assumptions made
in certain equations. If instead the dielectric were squeezed elastically,
so that the capacitor's area goes up as distance goes down (or vice-versa,
compressing from the edges) this would be a reversible process, with
mechanical energy exerted and released respectively.

By the way, a dielectric "wants" to move into a charged gap. This ought to
make plenty of sense since it only wants to remove energy from the system!
Lots of things happen, in effect, because they are the route to the lowest
final energy. As shown above, halving the charge between two equal
capacitors will half the energy in the system, pretty favorable I'd say.

---
Very nice, thanks!

Not that I disbelieve you, but I'm going to build a capacitor over
the weekend that I can charge up and then pull the dielectric out of
to see what happens.

If I can get the leakage down far enough and build a probe with a
high enough impedance, I'll post back with what I find. :)
 
M

Michael A. Terrell

Jan 1, 1970
0
Tim said:
I think this is an "only God can divide by zero" problem. ;-)

Tim


Wasn't that what caused the big bang?
 
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