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Re: Capacitor charge, were is it?
Dave,
There is a very simple answer to your apparent paradox....
> There is a classic problem in which two identical capacitors are connected
> with a switch. Before the switch is thrown one capacitor has a certain
> voltage, the other none. If we give numbers, let's say C=1uF and V=1000V on
> the first capacitor. The energy is (cv^2)/2 = 0.5 joules. Now after the
> switch is closed both capacitors will have v/2 = 500V across them. This can
> actually be done. But now the energy in the system is half the original
> value.
> (1uf*500^2)/2 = 0.125 joules/cap times two caps = 0.250 joules. Were is the
> other half?
It is radiated away in effecting the transfer.
BTW, the coil performs the opposite trick via the permeable medium
- transfer charge from a large cap to a small one.
> The classic things I've been taught have me confused with the actual
> workings of nature. If two charged plates (air dielectric) have a certain
> charge Q and then a dielectric is inserted between them with a K>1, the
> voltage should decrease since the charge hasn't changed but the capacitance
> has increased.
It does decrease.
Malcolm