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
I blame Google beta for that.
Cheers
Robin
Robin said:Sorry Tim, that reply should be tacked onto Jasen Betts' post.
Jasen Betts said:If nothing is touching the other terminal no current will flow through the
one you touch.
Bye.
Jasen
No, as originally stated, the voltage will *decrease* with "more
dielectric".
This is because the dielectric is "stretched" and *that* absorbs energy
hence the voltage must drop a little because energy is conserved.
See Feynman's Lectures on Physics volume 2 chapter 10 section 3.
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.
in the (usual) case where the dielecttric is non-compressible
I'm taking it to mean more distance between the plates.
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.
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...
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!
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! ;-)
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?
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.
Tim said:I think this is an "only God can divide by zero" problem. ;-)
Tim