Exploring the Relationship Between Electric Potential Energy and Capacitors

[abcdefg10645]

all my question is listed in the file below

thanks for your help

 

[Doc Al]

1/2 QV is the energy stored in the capacitor. The other half of the energy was dissipated during charging as heat or radiation.

 

[Meir Achuz]

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.
Then $$U=\int_0^VCV'dV'=(1/2)CV^2=(1/2)QV.$$

 

[Doc Al]

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!