1. **Problem statement:** We have five capacitors $C_1, C_2, C_3, C_4, C_5$ in a circuit. Capacitors $C_2, C_3, C_4,$ and $C_5$ each have a charge of $6.0\ \mu C$. The potential drop across $C_3$ is $V_3 = 11\ V$. We need to find the charge stored by $C_1$ ($Q_1$) and the potential drop across $C_4$ ($V_4$). 2. **Relevant formulas and rules:** - The charge $Q$ on a capacitor is related to its capacitance $C$ and voltage $V$ by $Q = CV$. - Capacitors in series have the same charge. - Capacitors in parallel have the same voltage. 3. **Assumptions and analysis:** Since $C_2, C_3, C_4, C_5$ all have the same charge $6.0\ \mu C$, they are likely in series or arranged so that these capacitors share the same charge. Given $V_3 = 11\ V$ and $Q_3 = 6.0\ \mu C$, we can find $C_3$: $$C_3 = \frac{Q_3}{V_3} = \frac{6.0\times10^{-6}}{11} = 5.45\times10^{-7}\ F$$ 4. **Finding $Q_1$:** If $C_1$ is in series with the others, it will have the same charge $Q_1 = 6.0\ \mu C$. 5. **Finding $V_4$:** Using $Q_4 = 6.0\ \mu C$ and $C_4 = C_3 = 5.45\times10^{-7}\ F$ (assuming identical capacitors), $$V_4 = \frac{Q_4}{C_4} = \frac{6.0\times10^{-6}}{5.45\times10^{-7}} = 11\ V$$ **Final answers:** $$Q_1 = 6.0\ \mu C$$ $$V_4 = 11\ V$$