At 13.59 2014.12.13, Ruben J=F6nsson wrote: >> 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? >>=20 >> Cheers, >> Mario >>=20 > >If the plates are charged and the power source is removed with the same=20 >dielectric the voltage will increase since the capacitance will decrease.= =20 >V=3DQ/C. This also means that the energy will increase but it is not for f= ree=20 >since we need to do work to pull the plates apart (the two plates attract = each=20 >other). > >The voltage will only increase until a certain distance is reached, then i= t=20 >will decrease with increased distance. This is because the Electric field = will=20 >decrease (the density of the filed lines decreases) when the distance gets= too=20 >large compared to the sizes of the plates. > >I found it quite well explained here:=20 >-in= crease- >when-capacitor-plates-are-separated> Very interesting, thank you all. Cheers, Mario >/Ruben --=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 .