A capacitor of capacitance C_1 = 1 mu F can w | vedclass

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(C) In a series combination,the charge $Q$ on each capacitor is the same.First,calculate the maximum charge each capacitor can hold:$(Q_1)_{max} = C_1

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(C) In a series combination,the charge $Q$ on each capacitor is the same.First,calculate the maximum charge each capacitor can hold:$(Q_1)_{max} = C_1 V_1 = 1\ \mu F 	imes 6\ kV = 6\ \mu C$.$(Q_2)_{max} = C_2 V_2 = 3\ \mu F 	imes 4\ kV = 12\ \mu C$.Since the capacitors are in series,the maximum charge the system can hold is limited by the capacitor with the smaller maximum charge,which is $6\ \mu C$.When the system holds a charge of $Q = 6\ \mu C$,the potential difference across $C_1$ is $V_1 = 6\ kV$ and the potential difference across $C_2$ is $V_2 = Q / C_2 = 6\ \mu C / 3\ \mu F = 2\ kV$.The total maximum voltage the series combination can withstand is $V_{total} = V_1 + V_2 = 6\ kV + 2\ kV = 8\ kV$.

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