The equivalent capacitance of the combination between A and B in the given figure is 4 μF.

(i) Calculate the capacitance of the capacitor C.
(ii) Calculate charge on each capacitor if a 12V battery is connected across terminals A and B.
(iii) What will be the potential drop across each capacitor?

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The capacitors given are arranged in a series combination. 
i) Given, equivalent capacitance of two combination is 4 μF. 
Therefore, 
$\frac{1}{{\mathrm{C}}_{\mathrm{AB}}}=\frac{1}{C_A}+\frac{1}{C_B}$
$$\begin{gathered} \frac{1}{{\mathrm{C}}_{\mathrm{AB}}}=\frac{1}{C_A}+\frac{1}{C_B} \\ \frac{1}{4}=\frac{1}{20}+\frac{1}{C_B} \\ \therefore C_B= 5 \mu F \end{gathered}$$ 
 
ii) Charge on capacitor is given by C=qV
Capacitors in series have the same charge .
Equivalent capacitance of the system is 4 μF.
$\therefore$ charge on each capacitor, q =CV
                                           $$\begin{gathered} 4\times 10{ }^{-6} \times 12 \\ =48 \times {10}^{-6} \mathrm{C} \end{gathered}$$  

iii) Potential drop across the capacitor 

$$\begin{gathered} {\mathrm{V}}_{\mathrm{AB} }= {\mathrm{V}}_{\mathrm{a}} + {\mathrm{V}}_{\mathrm{B}} \\ = \raisebox{1ex}{$\mathrm{q}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{C}}_{\mathrm{a}} + {\displaystyle \raisebox{1ex}{$\mathrm{q}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{C}}_{\mathrm{b}}$}\right.}$}\right. \\ =\frac{48\times {10}^{-6}}{20}+\frac{48\times {10}^{-6}}{5} \\ =2.4 \times {10}^{-6} + 9.6 \times {10}^{-6} \\ =12 \times {10}^{-6} \mathrm{V} \end{gathered}$$