Dielectric constant of pure water is 81. Its permittivity will be

Options

A. `7.16 xx 10^(-10)` MKS units
B. `8.86 xx 10^(-12)` MKS units
C. `1.02 xx 10^(13)` MKS units
D. Cannot be calculated

Answer

To find the permittivity of pure water given its dielectric constant, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Relationship**: The permittivity of a medium (\( \epsilon \)) is related to the permittivity of free space (\( \epsilon_0 \)) and the dielectric constant (\( \epsilon_r \)) of the medium by the formula:
\[ \epsilon = \epsilon_r \cdot \epsilon_0 \]

2. **Identify Given Values**:
- Dielectric constant of pure water (\( \epsilon_r \)) = 81
- Permittivity of free space (\( \epsilon_0 \)) = \( 8.86 \times 10^{-12} \, \text{F/m} \) (in mkS units)

3. **Substitute the Values into the Formula**:
\[ \epsilon = 81 \cdot (8.86 \times 10^{-12} \, \text{F/m}) \]

4. **Perform the Multiplication**:
\[ \epsilon = 81 \cdot 8.86 \times 10^{-12} = 7.16 \times 10^{-10} \, \text{F/m} \]

5. **Final Result**: The permittivity of pure water is:
\[ \epsilon = 7.16 \times 10^{-10} \, \text{F/m} \]