Answer:
0.84μF
Step-by-step explanation:
Charge is same through both the capacitors since they are in series. Total voltage is the sum of the voltages of the individual capacitors.. So voltage across the 2nd capacitor is 120- 90 =30 V.
Charge across first capacitor is Q = C₁V₁ = 90 x0.28 = 25.2μC
Therefore capacitance of 2nd capacitor =
C₂ = Q÷V₂ = 25.2÷30 = 0.84 μF