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There is no paradox involved in what happens to the stored energy when connecting two capacitors together of the same capacity, where one is initially charged and the other is not charged. Anyone who understands RC charging of a capacitor understands where half the energy goes.

Yes, please explain where the electrons go...

Hi mr2. I have pondered this as well. The total charge from one capacitor when divided between another capacitor of the same value of capacitance as the first capacitor doesn't actually disappear anywhere. The total charge divided between the two same valued capacitors remains the same. It is only the total energy that drops by half. I think it is a valid question to ask where half the energy went when the charge from one capacitor is divided over to a second capacitor of the same capacitance value. If I pour water into an empty cylinder, the water pours in the same all the way to the top, but charge is not like water in that charge wants to repel like charge with a relatively strong force. So, the more charge we cram into a capacitor, the more energy it takes to cram that charge into the capacitor, as the stronger the repelling force of the charge that is already crammed in there becomes the more the capacitor is charged. When we discharge this crammed in charge to another capacitor of the same capacitance value, the charge on the first capacitor then divides evenly between the two capacitors. There is now half the charge on each capacitor, giving a total charge the same as before the discharge, but the total energy of the two capacitors is only half the energy that was originally on the first capacitor before we discharged it to the second capacitor. Where does half the original energy go? Presumably it is dissipated in resistive losses as heat in the wire connecting the two capacitors together and in the capacitor plates, although if you think about it the total energy should drop by about the same for different wire resistance values and for different capacitor ESR values, should it not? This then makes me wonder where did half of the energy really all go? I have done experiments similar to what I believe you are describing where I can end up with more energy from such a discharge than just half the original energy before the discharge, using three capacitors and a transformer primary between two main capacitors, so it is still a mystery to me where the energy really all does go when doing such a capacitor discharge. Where does half the energy really go? Can it all be accounted for as heat gain due to resistive losses? If I place a transformer primary winding between the two capacitors, I have a higher total wire resistance, but I can collect some of the normal energy losses on a third capacitor on the transformer secondary (through a rectifier diode) and end up with somewhat more than one half of the original energy after the discharge. There is more wire resistance between the two capacitors, but I end up with more than half the energy after the discharge. Something doesn't quite seem to add up there. What might I be overlooking? At any rate this does seem to support the notion that the energy that is normally lost in such a capacitor discharge can be captured at least to some extent with the right setup... It seems it would be hard to capture most of it however. All the best...

To really understand this problem, you have to understand what the capacitor is actually doing.The bucket analogy works, if you think of it as filling the bucket through a hole at the bottom of the bucket.the more water you put in, the more the potential energy resists further filling of the bucket.Follow the curve, and you see how this affects charging time.

Indeed resistive losses, no matter how small the resistance between the caps.See this document clearly illustrating the effect.

My patent:https://patentscope.wipo.int/search/en/detail.jsf?docId=WO2011005103

The truth is far simpler. It is almost all just resistive loss. Charge any capacitor from a voltage source through a resistor and the resistor dissipates the same amount of energy as ends up stored in the capacitor. The resistor value just contributes to how long the charging process takes.Hooking two capacitors together does radiate a little bit. The lion's share of energy loss goes directly to heat dissipated in the wiring and ESR of the two capacitors.

resistance doesn't account for but a fraction of the losses in the system.we're talking about a small conducting wire, which can be substituted with a superconductor @ 0 Ohms. E = 1/2 stillThis effect was explained in several physics papers 25-30 yrs ago. College professors still like to use it to get their students to think.If you look at both the kinetic and potential energies of the system, everything is accounted for.We are only using 1/2 of the potential energy.