Two identical charged spherical drops each of capacitance `C` merge to from a single drop. The resultant capacitance is

Options

A. equal to `2C`
B. greater than `2C`
C. less than `2c` but greater than `C`
D. less than `C`

Explanation

To find the resultant capacitance when two identical charged spherical drops, each with capacitance \( C \), merge to form a single drop, we can follow these steps:

### Step 1: Understand the Concept of Capacitance
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 and \( \epsilon_0 \) is the permittivity of free space.

### Step 2: Calculate the Volume of the Drops
The volume \( V \) of a single spherical drop is given by:
\[ V = \frac{4}{3} \pi r^3 \]
For two identical drops, the total volume \( V_{total} \) is:
\[ V_{total} = 2 \times \frac{4}{3} \pi r^3 = \frac{8}{3} \pi r^3 \]

### Step 3: Find the Radius of the Merged Drop
Let the radius of the new larger drop be \( R \). The volume of the larger drop can be expressed as:
\[ V_{new} = \frac{4}{3} \pi R^3 \]
Since the total volume is conserved, we equate the two volumes:
\[ \frac{8}{3} \pi r^3 = \frac{4}{3} \pi R^3 \]
By simplifying, we can cancel \( \frac{4}{3} \pi \) from both sides:
\[ 2r^3 = R^3 \]
Taking the cube root of both sides gives:
\[ R = 2^{1/3} r \]

### Step 4: Calculate the Capacitance of the New Drop
Now, we can find the capacitance \( C_{new} \) of the new drop using the formula for capacitance:
\[ C_{new} = 4\pi \epsilon_0 R \]
Substituting \( R = 2^{1/3} r \):
\[ C_{new} = 4\pi \epsilon_0 (2^{1/3} r) \]
Since \( C = 4\pi \epsilon_0 r \), we can express \( C_{new} \) in terms of \( C \):
\[ C_{new} = 2^{1/3} \times C \]

### Step 5: Conclusion
Thus, the resultant capacitance when two identical charged spherical drops merge is:
\[ C_{new} = 2^{1/3} C \]