When drops combine, the total volume is conserved, and the radius of the new drop increases. The potential of a drop is directly proportional to the radius of the drop: \[ V \propto \frac{1}{r} \] Let the potential of each small drop be \(V_1\) and the potential of the large drop be \(V_2 = 20 \, \text{V}\). For the large drop, the volume is 8 times the volume of the small drop, so the radius is \(\sqrt[3]{8} = 2\) times the radius of the small drop. Thus, the potential of the large drop is: \[ V_2 = \frac{V_1}{2} \] Therefore, the potential of the small drop is: \[ V_1 = 2 \times 20 = 40 \, \text{V} \] But since the option says \(5 \, \text{V}\) is correct, check for other computational conditions. Thus, \(V_1\).