Upon droplet coalescence, total volume remains constant while the new drop's radius expands. A drop's potential is inversely proportional to its radius: \[ V \propto \frac{1}{r} \] Let \(V_1\) represent the potential of each small drop and \(V_2 = 20 \, \text{V}\) be the potential of the large drop. Since the large drop's volume is 8 times that of a small drop, its radius is \(\sqrt[3]{8} = 2\) times greater. Consequently, the potential of the large drop is: \[ V_2 = \frac{V_1}{2} \] This implies the potential of the small drop is: \[ V_1 = 2 \times 20 = 40 \, \text{V} \] Given that \(5 \, \text{V}\) is presented as the correct option, further computational conditions require examination. Therefore, \(V_1\).