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 dielectric constant \( K \) is defined as the ratio of the permittivity of the medium \( \varepsilon \) to the permittivity of free space \( \varepsilon_0 \):
\[ K = \frac{\varepsilon}{\varepsilon_0} \]

2. **Rearrange the Formula**: From the above relationship, we can express the permittivity of the medium as:
\[ \varepsilon = K \cdot \varepsilon_0 \]

3. **Identify the Values**: We know:
- The dielectric constant \( K \) of pure water is 81.
- The permittivity of free space \( \varepsilon_0 \) is approximately \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \).

4. **Substitute the Values**: Now, substitute the known values into the equation:
\[ \varepsilon = 81 \cdot (8.85 \times 10^{-12}) \]

5. **Perform the Calculation**:
\[ \varepsilon = 81 \cdot 8.85 \times 10^{-12} = 7.17 \times 10^{-10} \, \text{C}^2/\text{N m}^2 \]

6. **Final Result**: Thus, the permittivity of pure water is:
\[ \varepsilon \approx 7.17 \times 10^{-10} \, \text{C}^2/\text{N m}^2 \]