Electro magnetic waves travel in a medium with speed of `2xx10^(8)m//sec`. The relative permeability of the medium is `1` find relative permittivity.

Options

A. `2.25`
B. `4.5`
C. `0.5`
D. 4

Answer

To find the relative permittivity (\( \epsilon_r \)) of a medium in which electromagnetic waves travel at a speed of \( 2 \times 10^8 \, \text{m/s} \) and given that the relative permeability (\( \mu_r \)) is 1, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between speed, permeability, and permittivity**:
The speed of electromagnetic waves in a medium (\( V \)) is given by the formula:
\[ V = \frac{1}{\sqrt{\mu \epsilon}} \]
where \( \mu \) is the permeability and \( \epsilon \) is the permittivity of the medium.

2. **Express permeability and permittivity in terms of relative values**:
The permeability (\( \mu \)) can be expressed as:
\[ \mu = \mu_r \mu_0 \]
and the permittivity (\( \epsilon \)) as:
\[ \epsilon = \epsilon_r \epsilon_0 \]
where \( \mu_0 \) and \( \epsilon_0 \) are the permeability and permittivity of free space, respectively.

3. **Substituting into the speed formula**:
Substitute the expressions for \( \mu \) and \( \epsilon \) into the speed formula:
\[ V = \frac{1}{\sqrt{\mu_r \mu_0 \epsilon_r \epsilon_0}} \]

4. **Square both sides**:
Squaring both sides gives:
\[ V^2 = \frac{1}{\mu_r \mu_0 \epsilon_r \epsilon_0} \]

5. **Rearranging the equation to find \( \epsilon_r \)**:
Rearranging the equation to solve for \( \epsilon_r \):
\[ \epsilon_r = \frac{1}{\mu_r \mu_0} \cdot \frac{1}{V^2} \]

6. **Substituting known values**:
Since \( \mu_r = 1 \) and \( V = 2 \times 10^8 \, \text{m/s} \), we can substitute these values:
\[ \epsilon_r = \frac{1}{1 \cdot \mu_0} \cdot \frac{1}{(2 \times 10^8)^2} \]

7. **Using the speed of light**:
We know that the speed of light in vacuum \( c \) is related to \( \mu_0 \) and \( \epsilon_0 \) as:
\[ c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} \]
Therefore, \( c^2 = \frac{1}{\mu_0 \epsilon_0} \).

8. **Finding \( \epsilon_r \)**:
We can express \( \epsilon_r \) in terms of \( c \) and \( V \):
\[ \epsilon_r = \frac{c^2}{V^2} \]
Substituting \( c = 3 \times 10^8 \, \text{m/s} \):
\[ \epsilon_r = \frac{(3 \times 10^8)^2}{(2 \times 10^8)^2} \]

9. **Calculating the values**:
\[ \epsilon_r = \frac{9 \times 10^{16}}{4 \times 10^{16}} = \frac{9}{4} = 2.25 \]

10. **Conclusion**:
The relative permittivity \( \epsilon_r \) of the medium is \( 2.25 \).