Energy stored in the battery of a circuit (single electron box)

Consider the following, a closed loop circuit with two different capacitors in series plus a battery that provides a voltage $V$ (in series with the capacitors). You may add a resistance if that helps.

In any situation (equilibrium or not), the energy of the system is written as:

$$U=\frac{q_1^2}{2C_1}+\frac{q_2^2}{2C_2}-q_2V$$

where $q_1$ and $q_2$ are the charges in each capacitor, and $C_1$ and $C_2$ the respective capacitances.

The first to terms in the formula for $U$ are the energy stored in each capacitor. What I want to know, is about the last term $-q_2V$, why is the negative sign there? If it helps at all the capacitor "2" is at the positive terminal of the battery (why only 2?).

I just want to understand this last term not solve the problem, as I am analyzing a single electron box with Coulomb blockade so the charges in each capacitor are not trivial.

Edit:

Trying to give a picture $$|\big{|}\bigg{|}----|\:|----|\;|----\bigg{|}|----\bigg{|}\big{|}{|}$$

Edit2:

Apparently this energy is also called a free energy.

Edit3: For those no reading the title, here is reference link to the kind of systems I am considering: G. Fiori, Single electron box, 2005

Consider the following, a closed loop circuit with two different capacitors in series plus a battery that provides a voltage $V$

I'm not sure what it is that you're modelling here but, assuming ideal circuit elements for the circuit described (and the picture you've given)

$$q_1 = q_2 = Q$$

and so your first equation is just

$$U = \frac{Q^2}{2}\left( \frac{1}{C_1} + \frac{1}{C_2} - 2\frac{V}{Q}\right)$$

But

$$\frac{V}{Q} = \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$$

and so

$$U = \frac{Q^2}{2}\left(\frac{1}{C_1} + \frac{1}{C_2}\right)(1 - 2) < 0$$

which looks suspicious. I recommend adding some additional information about that which you this circuit is supposed to model.

• I'm not in an ideal case where all charges are equal. I just want to understand why the associate this free energy to that system. – Mauricio Jun 15 '18 at 18:53