Explanations & Calculations
1)

- If the total capacitance is C,
C1C=21+51+101=54=0.8μF(less than the lowest one)
Q=0.8μF×12V=9.6μC
2)

- The sum of charge on each capacitor is equal to 50. Then
q1+q2+q3V(2μF+5μF+10μF)V=50μC=50μC=2.9V
- If the total capacitance is C,
C=2+5+10=17μF
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