Date: Wed, 23 Feb 2011 11:16:08

Author: Jerry DiMarco

Subject: Re: Tin Foil Capacitor

Post:

Thank you for that reference (Ed-3). That demo and Ed-4 have made me
reevaluate what a dielectric does. Texts mention only induced dipoles and
flux in dielectrics. There is no mention of charge leaving the plates and
residing on the surface of the dielectric. But Ed-3 and other references
seem to suggest charge does go onto some dielectrics. If so, then why
doesn't more charge go onto the plates? The dielectric constants of glass
and quartz aren't that different. Why does the demo work with one but not
the other (asking beyond what has already been said)? What would happen if
we used a material with a high dielectric constant?

Jerry
D


At 2/23/2011 09:34 AM, you wrote:
>Paul,
>
>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).
>
>Cliff
>
>-----Original Message-----
>From: tap-l-owner@lists.ncsu.edu [mailto:tap-l-owner@lists.ncsu.edu] On
>Behalf Of Paul Nord
>Sent: Wednesday, February 23, 2011 8:43 AM
>To: tap-l@lists.ncsu.edu
>Cc: Paul Nord
>Subject: Re: [tap-l] Tin Foil Capacitor
>
>Cliff, Jerry,
>
>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
>the dielectric.
>
>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?
>
>Paul



From tap-l-owner@lists.ncsu.edu Wed Feb 23 13:32:57 2011

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