capacitor basics

...yes, so it is. Damn brain, go back to sleep.

Robin

...Looks right here...

Tim

-- Deep Fryer: a very philosophical monk. Website:

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.

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

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

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.

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.

I don't have a copy handy.

--

Bye.
   Jasen

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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!

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

-- Deep Fryer: a very philosophical monk. Website:

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

some tiny amount possibly, but not a sustained current to significantly discharge the capacitor, and not again if you touched it again.

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

Bye. Jasen

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

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

Tim

-- Deep Fryer: a very philosophical monk. Website:

^^^^^ 2.5mJ?

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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. :-)

Wasn't that what caused the big bang?

--
?

Michael A. Terrell
Central Florida