# Initial charge in parallel plate capacitor [closed]

Two capacitors $$C_1$$ and $$C_2$$ (where $$C_1 > C_2$$) are charged to the same initial potential difference $$\Delta V_i$$. The charged capacitors are removed from the battery, and their plates are connected with opposite polarity. The switches S1 and S2 are then closed. Find the final potential difference $$\Delta V_f$$ between $$a$$ and $$b$$ after the switches are closed as in figure down below.

The solution finds the initial charge $$Q_i = Q_{1i}+Q_{2i} = C_1 \Delta V_i -C_2 \Delta V_i = (C_1 - C_2) \Delta V_i$$. I don't understand why is there a negative sign for $$Q_{2i}$$. Can someone explain ?

The capacitor plates in figure 26.12(a) are connected in a way that the positive plate of one capacitor is connected by a wire to the negative plate of the other capacitor (It's written in your question, the plates are connected with opposite polarity). Because there is no battery or an external source of voltage, the electrons from the negative plate neutralize a part of the positive charge, so the charges would be balanced, and hence lower, which is why the negative sign is used for $$Q_{2i}$$
• Since the figure is symmetrical, why don't we put a negative sign for $Q_{1i}$ instead of $Q_{2i}$ ? Does the initial charge $Q_i$ have to be strictly positive ? Jun 15, 2021 at 10:11
• You absolutely can. However, note that $C_1$ > $C_2$ necesarily implies $Q_1$ > $Q_2$ because the voltages are the same, so the final charge that remains needs to be a positive one ($Q_1 - Q_2$) Jun 15, 2021 at 10:13