hey ,
?
here we go....thanks for the proof....
?
When you remove the dielectric slab then as per the
formula
?
C=keA/d the capacitance would decrease and hence
the voltage would increase...
Since the charge is constant...Q=CV
?
Try that...!!
?
So in short the charge stays on the plates...(I did
this 2 years back..)
Regards :-),
?
--himanshu sharma
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----- Original Message -----
Sent: Tuesday, October 30, 2001 9:54
AM
Subject: Re: [Electronics_101] Capacitor
- Charge- Energy
Sunantoro,
> When we charge a capacitor, we
actually pull-out the electrons from one
> plate and at the same time
provide more electrons on the other plate.
All this thinking is making
my head hurt (figure of speech only). I haven't t\
thought this much about
'tonix in ages. I retired last May but it was even
before that when I last
brain stormed about it.
OK, here's an experiment. Take a conductor and
charge it. I can do charge
some dielectric materials, but not conductors.
The if the charge is really
on the plates it would be there when the
dielectric is removed. Now if
you mean that you can measure the potential
of the charge at the plates,
I can buy that.
Jim
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