Step 1: Understanding the Concept:
When multiple small charged drops merge into one large drop, two physical quantities are conserved: total volume and total charge.
The electric potential of a spherical drop depends on both its charge and its radius. Step 2: Key Formula or Approach:
1. Electric Potential of a sphere: $V = \frac{kq}{r}$.
2. Conservation of Charge: $Q_{\text{big}} = n \times q_{\text{small}}$.
3. Conservation of Volume: $V_{\text{big\_drop}} = n \times V_{\text{small\_drop}} \implies \frac{4}{3}\pi R^3 = n \times \frac{4}{3}\pi r^3 \implies R = n^{1/3}r$. Step 3: Detailed Explanation:
Let a small drop have radius $r$ and charge $q$. Its potential is $V = \frac{kq}{r}$.
When $n$ such drops coalesce, the big drop has a total charge:
\[ Q = nq \]
And its new radius $R$ is found from volume conservation:
\[ R = n^{1/3} r \]
The potential $V'$ of the new big drop is:
\[ V' = \frac{kQ}{R} \]
Substitute the expressions for $Q$ and $R$:
\[ V' = \frac{k(nq)}{n^{1/3}r} \]
Separate the terms relating to $n$ and the original small drop's potential:
\[ V' = \left( \frac{n}{n^{1/3}} \right) \left( \frac{kq}{r} \right) \]
Using exponent rules ($n^1 / n^{1/3} = n^{1 - 1/3} = n^{2/3}$):
\[ V' = n^{2/3} \times V \] Step 4: Final Answer:
The potential of the big drop is $\text{n}^{2/3} \cdot \text{V}$.