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Suppose We have a capacitor with capacitance $\ C $ and charge $\ Q $ . So total stored energy is $$ E=\frac{Q^2}{2C} $$ Now if I connect a capacitor with same capacitance parallel with it then current will flow until voltage across both capacitor become same and this case charges among them will be same . So each capacitor will contain $\ Q/2 $ charges. So energy stored in each capacitor will be $$ E'=\frac{Q^2}{8C}$$ . So total energy in 2 capacitor is $$E_{tot}=\frac{Q^2}{4C}$$ which is not clearly equals to $\ E$ . Where does the remaining stored energy goes ?? When does the loss occur ?

Anklon
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    http://physics.stackexchange.com/questions/35843/a-problem-of-missing-energy-when-charging-a-second-capacitor – BowlOfRed Aug 25 '15 at 22:17

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Connecting two perfect capacitors like that would be like connecting two perfect but different voltage sources; you would get a hypothetical explosion.

In real life, every capacitor has inductance and resistance. So, as the current built up between the two capacitors, you'd heat up the wire between the capacitors as well as the capacitors themselves. If the inductance was high enough and the resistance low enough, you'd get oscillation that would be inevitably damped by the (always non-zero) resistance. In the end, you'd have somewhat warmer test equipment, and that's where your "lost" energy would be.

As a metaphor, consider a trough with a removable wall down the middle. Fill one half with water, and then calculate the potential energy. Then remove the wall, wait a bit, and calculate the potential energy again. You'll find that energy went away, because the average altitude of the water is lower. Where'd the energy go? It was dissipated into heat and sloshing sounds.