125 drops of water each of radius `2 mm` and carrying charge of `1 nC` are made to form a bigger drop. Find the capacitance and potential of the bigger drop.

Explanation

To solve the problem of finding the capacitance and potential of a bigger drop formed by combining 125 smaller drops of water, we can follow these steps:

### Step 1: Calculate the total charge on the bigger drop
Given:
- Charge on each small drop, \( q = 1 \, \text{nC} = 1 \times 10^{-9} \, \text{C} \)
- Number of drops, \( n = 125 \)

The total charge \( Q \) on the bigger drop is given by:
\[ Q = n \cdot q = 125 \cdot (1 \times 10^{-9}) = 125 \times 10^{-9} \, \text{C} = 1.25 \times 10^{-7} \, \text{C} \]

### Step 2: Calculate the radius of the bigger drop
The volume of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]

The volume of 125 small drops is equal to the volume of the bigger drop. Therefore:
\[ 125 \cdot V_{\text{small}} = V_{\text{big}} \]
\[ 125 \cdot \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3 \]
Where \( r = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \).

Cancelling \( \frac{4}{3} \pi \) from both sides:
\[ 125 \cdot r^3 = R^3 \]
\[ R^3 = 125 \cdot (2 \times 10^{-3})^3 \]
Calculating \( (2 \times 10^{-3})^3 \):
\[ (2 \times 10^{-3})^3 = 8 \times 10^{-9} \, \text{m}^3 \]
Thus,
\[ R^3 = 125 \cdot 8 \times 10^{-9} = 1000 \times 10^{-9} = 1 \times 10^{-6} \, \text{m}^3 \]
Taking the cube root:
\[ R = (1 \times 10^{-6})^{1/3} = 10^{-2} \, \text{m} = 10 \, \text{mm} \]

### Step 3: Calculate the capacitance of the bigger drop
The capacitance \( C \) of a spherical conductor is given by:
\[ C = 4 \pi \epsilon_0 R \]
Where \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \).

Substituting the values:
\[ C = 4 \pi (8.85 \times 10^{-12}) (10 \times 10^{-3}) \]
Calculating:
\[ C = 4 \pi (8.85 \times 10^{-12}) (10^{-2}) = 4 \pi (8.85 \times 10^{-14}) \approx 1.11 \times 10^{-12} \, \text{F} = 1.11 \, \text{pF} \]

### Step 4: Calculate the potential of the bigger drop
The potential \( V \) of the bigger drop can be calculated using the formula:
\[ V = \frac{Q}{C} \]
Substituting the values:
\[ V = \frac{1.25 \times 10^{-7}}{1.11 \times 10^{-12}} \approx 112.61 \, \text{V} \]

### Final Results
- **Capacitance of the bigger drop**: \( C \approx 1.11 \, \text{pF} \)
- **Potential of the bigger drop**: \( V \approx 112.61 \, \text{V} \)