1. **Problem statement:** We have two capacitors $C_1$ and $C_2$ connected in series between points A and B. Given: - Total potential difference $V_{AB} = 18\,V$ - Potential drop across $C_1$ is $V_1 = 6.0\,V$ - Charge on $C_1$ is $Q_1 = 4.0\,\mu C$ We need to find the capacitance $C_2$. 2. **Key concepts and formulas:** - In series, the charge on each capacitor is the same: $Q_1 = Q_2 = Q$. - The total voltage is the sum of voltages across each capacitor: $V_{AB} = V_1 + V_2$. - Capacitance is defined as $C = \frac{Q}{V}$. 3. **Calculate $C_1$:** $$ C_1 = \frac{Q_1}{V_1} = \frac{4.0\,\mu C}{6.0\,V} = \frac{4.0 \times 10^{-6}}{6.0} = 6.67 \times 10^{-7} \,F = 0.667\,\mu F $$ 4. **Find $V_2$:** $$ V_2 = V_{AB} - V_1 = 18 - 6 = 12\,V $$ 5. **Since charge is the same on both capacitors, $Q_2 = Q_1 = 4.0\,\mu C$** 6. **Calculate $C_2$:** $$ C_2 = \frac{Q_2}{V_2} = \frac{4.0\,\mu C}{12\,V} = \frac{4.0 \times 10^{-6}}{12} = 3.33 \times 10^{-7} \,F = 0.333\,\mu F $$ **Final answer:** $$ \boxed{0.333\,\mu C} $$ This means the capacitance of $C_2$ is $0.333\,\mu F$.