Whast is the absolute permittivity of mica if its relative permittivity is 5 `[epsilon_(0)=8.85 xx10^(-12)F/m]`

Options

A. `2.5 xx10^(-3) N/m^(2)`
B. `4xx10^(-4) N/ m^(2)`
C. `5xx10^(-4) N/m^(2)`
D. `8xx10^(-4) N/m^(2)`

Explanation

To find the absolute permittivity of mica given its relative permittivity, we can use the relationship between absolute permittivity (ε) and relative permittivity (ε_r). The formula is:

\[ \epsilon = \epsilon_0 \times \epsilon_r \]

Where:
- \( \epsilon \) is the absolute permittivity,
- \( \epsilon_0 \) is the absolute permittivity of free space (vacuum), given as \( 8.85 \times 10^{-12} \, \text{F/m} \),
- \( \epsilon_r \) is the relative permittivity, given as 5 for mica.

### Step-by-step Solution:

1. **Identify the values**:
- \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \)
- \( \epsilon_r = 5 \)

2. **Substitute the values into the formula**:
\[ \epsilon = \epsilon_0 \times \epsilon_r = (8.85 \times 10^{-12} \, \text{F/m}) \times 5 \]

3. **Perform the multiplication**:
\[ \epsilon = 8.85 \times 10^{-12} \times 5 = 44.25 \times 10^{-12} \, \text{F/m} \]

4. **Convert to standard scientific notation**:
\[ \epsilon = 4.425 \times 10^{-11} \, \text{F/m} \]

5. **Final Result**:
The absolute permittivity of mica is:
\[ \epsilon = 4.425 \times 10^{-11} \, \text{F/m} \]