# Electric potential energy

1. Nov 29, 2008

### abcdefg10645

all my question is listed in the file below

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2. Nov 29, 2008

### Staff: Mentor

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

3. Nov 30, 2008

### clem

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

Last edited: Nov 30, 2008
4. Nov 30, 2008

### Staff: Mentor

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!