N drops of mercury of equal radii and possessing equal charges combine to form a big spherical drop. Then the capacitance of the bigger drop compared to each individual drop is

Options

A. N times
B. `N^(2//3)` times
C. `N^(1//3)` times
D. `N^(5//3)` times

Answer

To solve the problem of finding the capacitance of a larger spherical drop formed by combining N smaller drops of mercury, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**: We have N small drops of mercury, each with a radius \( r \) and equal charge. When combined, they form a larger spherical drop. We need to find the capacitance of the larger drop compared to that of each individual drop.

2. **Volume Conservation**: The volume of the larger drop must equal the total volume of the N smaller drops. The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
Therefore, the volume of N small drops is:
\[ V_{\text{small}} = N \times \frac{4}{3} \pi r^3 \]
The volume of the larger drop with radius \( R \) is:
\[ V_{\text{large}} = \frac{4}{3} \pi R^3 \]

3. **Set Volumes Equal**: Since the volumes are equal, we can set them equal to each other:
\[ \frac{4}{3} \pi R^3 = N \times \frac{4}{3} \pi r^3 \]
We can cancel \( \frac{4}{3} \pi \) from both sides:
\[ R^3 = N r^3 \]

4. **Solve for R**: Taking the cube root of both sides gives us:
\[ R = r N^{1/3} \]

5. **Capacitance of a Sphere**: The capacitance \( C \) of a spherical conductor is given by:
\[ C = 4 \pi \epsilon_0 R \]
For the small drop, the capacitance \( C \) is:
\[ C = 4 \pi \epsilon_0 r \]
For the larger drop, the capacitance \( C' \) is:
\[ C' = 4 \pi \epsilon_0 R = 4 \pi \epsilon_0 (r N^{1/3}) = 4 \pi \epsilon_0 r N^{1/3} \]

6. **Ratio of Capacitances**: Now, we find the ratio of the capacitance of the larger drop to that of the smaller drop:
\[ \frac{C'}{C} = \frac{4 \pi \epsilon_0 r N^{1/3}}{4 \pi \epsilon_0 r} = N^{1/3} \]

### Final Answer:
The capacitance of the larger drop compared to each individual drop is:
\[ \frac{C'}{C} = N^{1/3} \]