1. **State the problem:** Four capacitors $C_1, C_2, C_3,$ and $C_4$ are connected in series between terminals A and B. 2. **Given data:** - Potential drops: $V_1 = 1$ V, $V_2 = 2$ V, $V_3 = 3$ V, $V_4 = 4$ V - Charge on $C_3$: $Q_3 = 8 \mu C$ 3. **Find:** - Potential difference $V_{AB}$ between terminals A and B - Charge stored by capacitor $C_1$, $Q_1$ 4. **Key concept:** In series connection, the charge $Q$ on each capacitor is the same: $$Q_1 = Q_2 = Q_3 = Q_4 = Q$$ 5. **Calculate total potential difference:** The total voltage across the series is the sum of individual voltages: $$V_{AB} = V_1 + V_2 + V_3 + V_4 = 1 + 2 + 3 + 4 = 10 \text{ V}$$ 6. **Charge on $C_1$:** Since the capacitors are in series, the charge on each capacitor is equal: $$Q_1 = Q_3 = 8 \mu C$$ **Final answers:** - Potential difference between A and B: $$V_{AB} = 10 \text{ V}$$ - Charge stored by capacitor $C_1$: $$Q_1 = 8 \mu C$$