In the circuit shown, the potential drop across each capacitor is (assuming the two diodes are ideal)

Explanation

The diode `D_1` is reverse biased (open circuit), but the diode `D_2` is forward biased (short circuit).

` therefore ` the potential of the battery divides across the two capacitors in the inverse ratio of their capacities.
` i.e., (V_1)/(V_2) =(C_2 )/(C_1) = 8/4 = 2/1`
`V_1 = 2/3 E =2/3 xx 24 = 16 V `
` V_2 = 1/3 E=1/3 xx 24 = 8V `