A capacitor of capacitance $C_1 = 1\ \mu F$ can with stand maximum voltage $V_1= 6\ kV$ (kilo-volt) and another capacitor of capacitance $ C_2 = 3\ \mu F$ can withstand maximum voltage $V_2 = 4\ kV$. When the two capacitors are connected in series, the combined system can withstand a maximum voltage of......$kV$
Diffcult
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(c) As $Q = CV, (Q_1)_{max}=10^{-6} × 6 × 10^3 = 6 \ \mu C$
While ($Q_2)_{max}= 3 × 10^{-6} × 4 × 10^3 = 12\ \mu C$
However in series charge is same so maximum charge on $C_2$ will also be $6 \,µC$ (and not $12 \ \mu C$) and potential difference across it $V_2 = 6 / 3 = 2\,kV$ and as in series $V =V_1 + V_2 \,$so $\,V_{max}= 6\,kV + 2\,kV = 8\,kV$
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