Twenty seven charged water droplets each with a diameter of 2 mm and a charge fo `10^(-12)C` coalesce to form a single drop. Calculate the potential of the bigger drop.

Explanation

To solve the problem of finding the potential of a bigger drop formed by the coalescence of 27 smaller charged water droplets, we can follow these steps:

### Step 1: Understand the parameters of the smaller droplets
- Each smaller droplet has a diameter of 2 mm, which gives a radius \( r \):
\[ r = \frac{2 \text{ mm}}{2} = 1 \text{ mm} = 1 \times 10^{-3} \text{ m} \]
- Each droplet has a charge \( q \):
\[ q = 10^{-12} \text{ C} \]

### Step 2: Calculate the total charge of the bigger drop
- When 27 droplets coalesce, the total charge \( q' \) on the bigger drop is:
\[ q' = 27q = 27 \times 10^{-12} \text{ C} = 2.7 \times 10^{-11} \text{ C} \]

### Step 3: Calculate the volume of the smaller droplets
- The volume \( V \) of one smaller droplet is given by the formula for the volume of a sphere:
\[ V = \frac{4}{3} \pi r^3 \]
Substituting \( r = 1 \times 10^{-3} \text{ m} \):
\[ V = \frac{4}{3} \pi (1 \times 10^{-3})^3 = \frac{4}{3} \pi \times 10^{-9} \text{ m}^3 \]
- The total volume of 27 smaller droplets is:
\[ V_{\text{total}} = 27 \times V = 27 \times \frac{4}{3} \pi (1 \times 10^{-3})^3 = 36 \pi \times 10^{-9} \text{ m}^3 \]

### Step 4: Find the radius of the bigger drop
- The volume of the bigger drop \( V' \) must equal the total volume of the smaller droplets:
\[ V' = \frac{4}{3} \pi R^3 \]
Setting \( V' = V_{\text{total}} \):
\[ \frac{4}{3} \pi R^3 = 36 \pi \times 10^{-9} \]
Dividing both sides by \( \pi \) and multiplying by \( \frac{3}{4} \):
\[ R^3 = 27 \times 10^{-9} \implies R = (27 \times 10^{-9})^{1/3} = 3 \times 10^{-3} \text{ m} = 3 \text{ mm} \]

### Step 5: Calculate the potential of the bigger drop
- The potential \( V' \) of a charged sphere is given by:
\[ V' = \frac{k q'}{R} \]
where \( k \) (Coulomb's constant) is approximately \( 9 \times 10^9 \text{ Nm}^2/\text{C}^2 \).
- Substituting \( q' = 2.7 \times 10^{-11} \text{ C} \) and \( R = 3 \times 10^{-3} \text{ m} \):
\[ V' = \frac{9 \times 10^9 \times 2.7 \times 10^{-11}}{3 \times 10^{-3}} \]
Simplifying:
\[ V' = \frac{24.3 \times 10^{-2}}{3} = 8.1 \text{ V} \]

### Final Answer
The potential of the bigger drop is:
\[ \boxed{8.1 \text{ V}} \]