If the circumference of a sphere is 2 m, then capacitance of sphere in water would be

Options

A. 2700 pF
B. 2760 pF
C. 2780 pF
D. 2800 pF

Answer

To solve the problem of finding the capacitance of a sphere in water given its circumference, we can follow these steps:

### Step 1: Understand the relationship between circumference and radius
The circumference \( C \) of a sphere is given by the formula:
\[ C = 2 \pi r \]
where \( r \) is the radius of the sphere. Given that the circumference is \( 2 \, \text{m} \), we can rearrange this formula to find the radius.

### Step 2: Calculate the radius
From the circumference formula, we can express the radius as:
\[ r = \frac{C}{2 \pi} \]
Substituting the given circumference:
\[ r = \frac{2 \, \text{m}}{2 \pi} = \frac{1}{\pi} \, \text{m} \approx 0.318 \, \text{m} \]

### Step 3: Use the capacitance formula for a sphere
The capacitance \( C \) of a spherical conductor in a medium is given by:
\[ C = 4 \pi \epsilon r \]
where \( \epsilon \) is the permittivity of the medium. In water, the permittivity can be expressed as:
\[ \epsilon = \epsilon_0 \epsilon_r \]
where \( \epsilon_0 \) is the permittivity of free space (\( 8.85 \times 10^{-12} \, \text{F/m} \)) and \( \epsilon_r \) is the relative permittivity (dielectric constant) of water, which is approximately \( 79 \).

### Step 4: Substitute values into the capacitance formula
Now, substituting the values into the capacitance formula:
\[ C = 4 \pi \epsilon_0 \epsilon_r r \]
Substituting \( \epsilon_0 \), \( \epsilon_r \), and \( r \):
\[ C = 4 \pi (8.85 \times 10^{-12} \, \text{F/m})(79)(\frac{1}{\pi}) \]
This simplifies to:
\[ C = 4 \times 8.85 \times 10^{-12} \times 79 \]

### Step 5: Calculate the capacitance
Calculating the above expression:
\[ C = 4 \times 8.85 \times 10^{-12} \times 79 \approx 2.796 \times 10^{-9} \, \text{F} = 2796 \, \text{pF} \]

### Step 6: Round off the answer
Rounding off \( 2796 \, \text{pF} \) gives approximately \( 2800 \, \text{pF} \).

### Final Answer
The capacitance of the sphere in water is approximately:
\[ C \approx 2800 \, \text{pF} \]

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