The capacitors on the right all have the same surface area, $A$. The separations between the two plates of the capacitors with capacitance $C$ are $d$. Now, a new capacitor plate, of total charge $Q_0$ and surface area $A$, is inserted at a distance $x$ from the left plate of the $\frac{C}{2}$ capacitor. After the system has equilibrated, what is the final charge on the left plate of the capacitor (labeled as B on the diagram) that had an original capacitance $\frac{C}{2}$ ? (Chinese Physics Olympiad)
(GRAPH CAN'T COPY)