Hi Mario,
Conservation of energy would predict an increase in voltage.  (E=3D1/2 CV^2
so V =3D sqrt(2E/C) ).   It may actually increase more than this as it will
take energy to pull the plates apart to overcome the attractive forces. So
maybe conservation of charge applies & voltage would increase linearly with
the decrease in capacitance (Q=3DCV so V=3DQ/C).

I read a science fiction story once that used this sort of thing as a high
voltage generator. A liquid dielectric was used & was drained out of the
charged capacitor to increase the voltage. Always wanted to try it.

Richard P

On 13 December 2014 at 23:03, Electron <electron2k4@infinito.it> wrote:

>
> Hello,
> suppose we have a capacitor made of 2 armatures and a dielectric.
> Like two rigid sheets of metal with a sheet of paper in between.
>
> We begin with the capacitor assembled, i.e. with the two armatures
> closely together (and the dielectric in the middle).
>
> Now we also charge the capacitor.
>
> We measure tension between the two terminals, but we leave the
> capacitor charged.
>
> Now we separe the two armatures, placing them apart, very distant
> from each other.
>
> My question is: will there still be a voltage accross the two
> terminals?
>
> I think I know the answer (NO) but I'd like to fully understand
> at the intuitive level why. Ok it's the electric field between
> the armatures that causes a difference of potential in the
> terminals.. but why?
> I mean, when I charged the capacitor, I effectively moved charges
> from an armature and transferred them into the other. So why doesn't
> the difference in potential remain?
>
> Cheers,
> Mario
>
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