> 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 fr= ee=20 since we need to do work to pull the plates apart (the two plates attract e= ach=20 other). The voltage will only increase until a certain distance is reached, then it= =20 will decrease with increased distance. This is because the Electric field w= ill=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 /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 .