A spherical drop of capacitance `12 muF` is broken into eight drops of equal radius. What is the capacitance of each small drop in `muF`?

Explanation

To solve the problem, we need to find the capacitance of each small drop after a spherical drop of capacitance `12 µF` is divided into eight smaller drops of equal radius.

### Step-by-Step Solution:

1. **Understand the relationship between capacitance and radius**:
The capacitance \( C \) of a spherical conductor is given by the formula:
\[ C = 4 \pi \epsilon_0 R \]
where \( R \) is the radius of the sphere.

2. **Given capacitance of the large drop**:
We know that the capacitance of the large drop is:
\[ C = 12 \, \mu F \]

3. **Volume conservation**:
When the large drop is broken into 8 smaller drops, the total volume remains constant. The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi R^3 \]
Let \( R \) be the radius of the large drop and \( r \) be the radius of each small drop. The volume of the large drop is:
\[ V_{large} = \frac{4}{3} \pi R^3 \]
The volume of one small drop is:
\[ V_{small} = \frac{4}{3} \pi r^3 \]
Since there are 8 small drops, the total volume of the small drops is:
\[ V_{total} = 8 \times V_{small} = 8 \times \frac{4}{3} \pi r^3 \]

4. **Setting volumes equal**:
Since the volume is conserved:
\[ \frac{4}{3} \pi R^3 = 8 \times \frac{4}{3} \pi r^3 \]
We can cancel \( \frac{4}{3} \pi \) from both sides:
\[ R^3 = 8r^3 \]

5. **Solving for the radius of the small drops**:
Taking the cube root of both sides gives:
\[ R = 2r \]

6. **Finding the capacitance of the small drop**:
The capacitance of each small drop can be calculated using the radius \( r \):
\[ C_{small} = 4 \pi \epsilon_0 r \]
Since \( R = 2r \), we can substitute \( r = \frac{R}{2} \):
\[ C_{small} = 4 \pi \epsilon_0 \left(\frac{R}{2}\right) = 2 \cdot 4 \pi \epsilon_0 R \]
Now, substituting \( 4 \pi \epsilon_0 R = 12 \, \mu F \):
\[ C_{small} = 2 \cdot 12 \, \mu F = 6 \, \mu F \]

### Final Answer:
The capacitance of each small drop is \( 6 \, \mu F \).

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