What is the permittivity, if the electric constant of water is 80?

Options

A. `708xx10^(-12)F//m`
B. `700xx10^(-12)F//m`
C. `708xx10^(-14)F//m`
D. `708.32xx10^(-2)F//m`

Explanation

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

### Step-by-Step Solution:

1. **Understand the Relationship**: The dielectric constant (K) is related to permittivity (ε) and the permittivity of free space (ε₀) by the formula:
\[ K = \frac{\epsilon}{\epsilon_0} \]
Rearranging this gives us:
\[ \epsilon = K \cdot \epsilon_0 \]

2. **Identify Known Values**: From the problem, we know:
- Dielectric constant of water (K) = 80
- Permittivity of free space (ε₀) = \(8.85 \times 10^{-12} \, \text{F/m}\) (farads per meter)

3. **Substitute the Values**: Now, substitute the known values into the rearranged formula:
\[ \epsilon = 80 \cdot (8.85 \times 10^{-12}) \]

4. **Perform the Calculation**:
- First, calculate \(80 \cdot 8.85\):
\[ 80 \cdot 8.85 = 708 \]
- Now, multiply by \(10^{-12}\):
\[ \epsilon = 708 \times 10^{-12} \, \text{F/m} \]

5. **Final Result**: Thus, the permittivity of water is:
\[ \epsilon = 7.08 \times 10^{-10} \, \text{F/m} \]