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(C) When capacitors are connected in series,the charge $Q$ on each capacitor is the same.For $C_1$,the maximum charge it can hold is $Q_1 = C_1 V_1 =

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(C) When capacitors are connected in series,the charge $Q$ on each capacitor is the same.For $C_1$,the maximum charge it can hold is $Q_1 = C_1 V_1 = (1\,\mu F)(6\,kV) = 6\,\mu C$.For $C_2$,the maximum charge it can hold is $Q_2 = C_2 V_2 = (3\,\mu F)(4\,kV) = 12\,\mu C$.In a series combination,the total charge $Q$ is limited by the capacitor that reaches its maximum charge first. Thus,the maximum charge the series combination can hold is $Q_{max} = \min(Q_1, Q_2) = 6\,\mu C$.The equivalent capacitance of the series combination is $C_{eq} = \frac{C_1 C_2}{C_1 + C_2} = \frac{1 	imes 3}{1 + 3}\,\mu F = 0.75\,\mu F$.The maximum voltage $V_{max}$ that can be applied to the combination is $V_{max} = \frac{Q_{max}}{C_{eq}} = \frac{6\,\mu C}{0.75\,\mu F} = 8\,kV$.

$A$ capacitor $C_1 = 1\,\mu F$ can withstand a maximum voltage of $V_1 = 6\,kV$,and a capacitor $C_2 = 3\,\mu F$ can withstand a maximum voltage of $V_2 = 4\,kV$. If these two capacitors are connected in series,what is the maximum voltage (in $kV$) that can be applied to the combination?

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