# Work and electric potential energy of a capacitor

## Homework Statement

There's no problem per se, I'm just confused by the definition of work and electric potential energy: work done by the electric field is defined to be the negative of delta_U, the change in electric potential energy. This definition makes sense to me.

However, according to my textbook, it is the other way round for capacitors: the total work to charge a capacitor is defined as Q^2 / (2C), which according to my textbook is the same as the electric potential energy U of the charged capacitor. Where did the minus sign go?

## Homework Equations

W(done by electric field) = - delta_U

Q^2 / (2C) = W = U

Thanks.

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Delphi51
Homework Helper
I don't really see the conflict. In the first case, the electric field does work on something so the field or whatever is maintaining it loses energy, its ΔU is negative, so you need -ΔU to get a positive value for the work done to a charge.

In the second case, a battery does work to charge a capacitor. The battery loses energy and has a negative ΔUb = -QV, while the capacitor gains energy and has a positive ΔUc = Q²/(2C).

Now I'm wondering if ΔUc = -ΔUb. Yikes; I may lose sleep over this!
The factor of 2 must come into it due to the battery with voltage V pouring charge into a capacitor while its voltage rises from zero to V, average V/2. Half the energy is lost . . . in the wire?

ehild
Homework Helper
The battery has got stored chemical energy and this decreases when the buttery supplies charges to the capacitor. Some chemical reactions occur inside the battery, metal ions go into the electrolyte from one of the electrodes and hydrogen evolves on the other one. This process sooner or later makes the battery flat. During the process, also some heat evolves. So the energy loss comes partly because of the change of chemical energy in the battery and also because of the evolved heat.

ehild