In the figure shown, the potential drop across each capacitor is (assume diodes to be ideal)
 

 

The diode connected parallel to the battery is reverse bias. So, current will not pass through it. So, total emf divided among $C_1$  and   $C_2$and in the inverse ratio of their capacitances.
$V=\frac{2}{C}$
$\frac{V_1}{V_2}=\frac{C_2}{C_1}$
$V_1=\frac{V{C}_2}{C_1+C_2},V_2=\frac{V{C}_1}{C_1+C_2}$