Date: Wed, 23 Feb 2011 09:34:25
Author: Cliff Bettis
Subject: Re: Tin Foil Capacitor
Dielectric materials vary widely in their properties (linearity, temperature
dependence, time dependence, breakdown field and probably others I am not
aware of). This is why you find capacitors with a variety of dielectrics
(glass, paper, mica, ceramic, oil and air). So what happens depends on what
kind of dielectric you use. In the standard dissectible capacitor
demonstration, the dielectric is given a polarization that makes it resemble
an electrets. If you use a quartz glass or just air for a dielectric, this
doesn't happen (See Freier and Anderson Ed-3).
From: firstname.lastname@example.org [mailto:email@example.com] On
Behalf Of Paul Nord
Sent: Wednesday, February 23, 2011 8:43 AM
Cc: Paul Nord
Subject: Re: [tap-l] Tin Foil Capacitor
Is the better question: What happens when you pull the capacitor apart?
You're doing work to separate the charges. You're adding an air layer to
Each side of the capacitor by itself has a very small capacitance. To hold
the same amount of charge the voltage would have to become huge.
What happens to the electric field inside the dielectric when the adjoining
plate is removed? Is it pinned there? Is it weaker or stronger because
there is no conducting surface to "terminate" the field lines?
On Feb 23, 2011, at 7:42 AM, Cliff Bettis wrote:
> The behavior of the dissectible capacitor depends on the nature of the
> dielectric. For instance, if the usual demonstration is done with quartz
> glass instead of soda glass, it will retain the charge on the plates.
> -----Original Message-----
> From: firstname.lastname@example.org [mailto:email@example.com] On
> Behalf Of Jerry DiMarco
> Sent: Tuesday, February 22, 2011 6:53 PM
> To: firstname.lastname@example.org
> Subject: Re: [tap-l] Tin Foil Capacitor
> I'm familiar with that demo but there are lots of other demos about
> charge flowing into and out of capacitors, and capacitors powering
> mechanical or electrical devices. Textbook treatment of the subject also
> mention current flowing on to plates, and more current flowing on to
> when a dielectric is inserted. Discussion of dielectrics is about the
> dipoles that are formed due to the E-field between the plates. Only the
> residual charge left over after a capacitor is discharged is attributed to
> dielectric absorption. So where does all the charging current go?
> If you claim the dielectric stores the charge, how does that explain
> a capacitor's ability to act as a filter in a power supply, or as a source
> of power in a remote location? How would you explain the action of an RC
> At 2/22/2011 03:18 PM, you wrote:
>> If you take the capacitor apart and remove the dielectric, you can touch
>> the plates together and will find that there is no (or only a tiny tiny
>> fraction of the) charge on them. Put it back together and it will be
>> On Feb 22, 2011, at 2:31 PM, chuck britton wrote:
>>> Vacuum polarization ! ? ! ? ! ?
>>> At 12:30 PM -0700 2/22/11, Jerry DiMarco wrote:
>>>> Taking one problem at a time, are you saying there is no charge
>> deposited on the plates of a capacitor?
>>>> Jerry D
>>>> At 2/21/2011 03:28 PM, you wrote:
>>>>> But in the case of the rolled foil capacitor there are two foils and
>> two insulators. The "charge" is actually held in the polarization of the
>> dielectric and not on the surface of the conductors.
From email@example.com Wed Feb 23 11:22:53 2011