Hi Richard, thanks for Your reply. So the two armatures, now separed, would have developed a static electricity potential of inverse sign? Let's make another virtual experiment then. I charge the capacitor while it has a short-close dielectric. If I touch either terminal, nothing happens. Now I separe the armatures. If I touch either of them with my hand, will a spark (from static electrici= ty) arise? If I touch one terminal (-) electrons will flows from it to me, if I insted touch the other terminal (+) it will rob electrons from my body. Correct? Cheers, Mario At 11.23 2014.12.13, Richard Prosser wrote: >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 wil= l >take energy to pull the plates apart to overcome the attractive forces. So >maybe conservation of charge applies & voltage would increase linearly wit= h >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 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 >> >> -- >> http://www.piclist.com/techref/piclist PIC/SX FAQ & list archive >> View/change your membership options at >> http://mailman.mit.edu/mailman/listinfo/piclist >> >--=20 >http://www.piclist.com/techref/piclist PIC/SX FAQ & list archive >View/change your membership options at >http://mailman.mit.edu/mailman/listinfo/piclist --=20 http://www.piclist.com/techref/piclist PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist .