Re: electricity

Tom Potter wrote:
"PD" <TheDraperFamily@xxxxxxxxx> wrote in message
news:1193758998.597220.9700@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
On Oct 30, 6:48 am, "Tom Potter" <tdp1...@xxxxxxxxx> wrote:
"Sam Wormley" <sworml...@xxxxxxxxx> wrote in message

news:nCoVi.176801$Fc.166135@xxxxxxxxxxxx

G=EMC^2 Glazier wrote:
Sam My thinking tells me the capacitor does not store electrons.
You should spend some time in a library, Glazier, while you
still can... zip up your pants, get to the library, and learn
something about capacitors and their history. Go on.... git!
Sammy, if you visited the library with Glazier,
you'd discover that the charge on a capacitor
is stored in the dielectric,

and if you remove the dielectric,
the charge goes with it.
Oh, good heavens, Potter. Either you don't know when not to say
anything when you don't know what you're talking about, or you don't
even know when you don't know what you're talking about.

There will be a residual charge
left on a capacitor after the dielectric is removed,
due to the dielectric properties of space
( distance * permittivity of space ),

as space is something you can't move around.
(Contrary to what GTR says.)

Excuse me folks.
"PD" is right.

I was thinking in terms of
charge = voltage * capacitance = voltage * permittivity * distance

and if you lowered the capacitance by taking away the dielectric,
that this would lower the charge,
but note that voltage and distance also come into play,
and it takes energy to remove the dielectric.

The system (Capacitor plates plus the dielectric)
were CHARGED by some electromotive force,
and when you move the dielectric,
you don't affect the charge,
but you do affect the voltage and energy,
by the work you do on the system.

Sorry if I mislead anyone.

PD has a much better handle on this,
so I'll shut up and go to sleep.


Somebody is faking "Tom Potter" or he decided to kiss ass!
-Sam

.

Relevant Pages

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