Hi Mario, The link Rubin posted explains the situation quite well. I'm not sure about why there's a threashold point where the voltage starts decreasing though. It's not static electricity being generated - it's just a consequence of the conservation of charge. Richard On 14 December 2014 at 01:59, 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? > > > > Cheers, > > Mario > > > > If the plates are charged and the power source is removed with the same > dielectric the voltage will increase since the capacitance will decrease. > V=3DQ/C. This also means that the energy will increase but it is not for = free > since we need to do work to pull the plates apart (the two plates attract > each > other). > > The voltage will only increase until a certain distance is reached, then = it > will decrease with increased distance. This is because the Electric field > will > decrease (the density of the filed lines decreases) when the distance get= s > too > large compared to the sizes of the plates. > > I found it quite well explained here: > < > http://physics.stackexchange.com/questions/82429/why-does-the-voltage-inc= rease- > when-capacitor-plates-are-separated> > > /Ruben > -- > 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 .