To solve this question, we need to analyze the change in potential when small charged droplets coalesce to form a larger drop.
First, let's understand the relevant concept here:
- The potential \(V\) of a spherical drop is given by the formula: \(V = \frac{kQ}{R}\), where \(k\) is the electrostatic constant, \(Q\) is the charge, and \(R\) is the radius of the drop.
- When multiple droplets coalesce, the total volume remains the same. Let the radius of each small drop be \(r\) and the radius of the large drop be \(R'\).
Next, let's compute the radius of the new larger drop:
- The volume of each small drop \(V_d = \frac{4}{3} \pi r^3\).
- For \(n\) such drops, the total initial volume is \(n \times \frac{4}{3} \pi r^3 = \frac{4}{3} \pi nr^3\).
- Let \(R'\) be the radius of the larger drop formed by coalescence. Thus, the volume of the larger drop is \(\frac{4}{3} \pi R'^3\).
- Equating the total volume before and after coalescence: \(\frac{4}{3} \pi nr^3 = \frac{4}{3} \pi R'^3\).
- Solving for \(R'\): \(R'^3 = nr^3 \Rightarrow R' = n^{1/3}r\).
Next, we calculate the potential of the larger droplet:
- The total charge of \(n\) droplets, each with charge \(Q\), is \(nQ\).
- The potential of the large drop, using \(V' = \frac{k \cdot nQ}{R'}\): \(V' = \frac{k \cdot nQ}{n^{1/3}r} = n^{2/3} \cdot \frac{kQ}{r} = n^{2/3}V\).
Thus, when \(n\) such droplets coalesce into one, the potential of the larger drop is \(n^{2/3}V\).
Therefore, the correct answer is: \(n^{2/3}V\)