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)xx24=16V,V_(2)=(1)/(3)E=(1)/(3)xx24=8V`