The velocity of an electromagnetic wave in a medium is `2xx10^(8) mS^(-1)` . If the relative permeability is 1 the relative permittivity of the medium is `(C_(0)=3xx10^(8) mS^(-1))`

Options

A. `2.25`
B. `1.5`
C. `4//9`
D. `2//3`

Explanation

To solve the problem, we need to find the relative permittivity (ε_r) of a medium given the velocity of an electromagnetic wave in that medium (v = 2 x 10^8 m/s), the speed of light in vacuum (c = 3 x 10^8 m/s), and the relative permeability (μ_r = 1).

### Step-by-Step Solution:

1. **Understand the relationship between speed, permeability, and permittivity**:
The speed of electromagnetic waves in a medium 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 (μ) and permittivity (ε) can be expressed in terms of their relative values:
\[ \mu = \mu_r \cdot \mu_0 \]
\[ \epsilon = \epsilon_r \cdot \epsilon_0 \]
where \( \mu_0 \) and \( \epsilon_0 \) are the permeability and permittivity of free space, respectively.

3. **Substituting into the speed formula**:
Substituting the expressions for μ and ε into the speed formula gives:
\[ v = \frac{1}{\sqrt{\mu_r \cdot \mu_0 \cdot \epsilon_r \cdot \epsilon_0}} \]
Given that \( \mu_r = 1 \), this simplifies to:
\[ v = \frac{1}{\sqrt{\mu_0 \cdot \epsilon_r \cdot \epsilon_0}} \]

4. **Using the speed of light in vacuum**:
The speed of light in vacuum is given by:
\[ c = \frac{1}{\sqrt{\mu_0 \cdot \epsilon_0}} \]
Therefore, we can express \( \mu_0 \cdot \epsilon_0 \) in terms of c:
\[ \mu_0 \cdot \epsilon_0 = \frac{1}{c^2} \]

5. **Substituting back into the speed formula**:
Now substituting this back into our equation for v:
\[ v = \frac{1}{\sqrt{1 \cdot \epsilon_r \cdot \frac{1}{c^2}}} = c \cdot \frac{1}{\sqrt{\epsilon_r}} \]
Rearranging gives:
\[ \sqrt{\epsilon_r} = \frac{c}{v} \]

6. **Calculating ε_r**:
Plugging in the values:
\[ \sqrt{\epsilon_r} = \frac{3 \times 10^8 \, \text{m/s}}{2 \times 10^8 \, \text{m/s}} = 1.5 \]
Squaring both sides gives:
\[ \epsilon_r = (1.5)^2 = 2.25 \]

### Final Answer:
The relative permittivity (ε_r) of the medium is **2.25**.