all my question is listed in the file below
thanks for your help
1/2 QV is the energy stored in the capacitor. The other half of the energy was dissipated during charging as heat or radiation.
Q and V for a capacitor are proportional to each other by Q=CV so the simple E=QV does not apply. The change in energy is dU=Vdq=CVdV.
Clem is, of course, correct. Perhaps I misinterpreted your question (or just answered a different question!) about charging a capacitor with a battery. The energy imparted to the charge by the battery is QV, but the energy stored in the capacitor only half that. Sorry about that!
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