An electromagnetic wave has an electric field with peak value 250 N/C . What is the average intensity of the wave ?

Options

A. `0.66 W//m^(2)`
B. `83 W//m^(2)`
C. `0.89 W//m^(2)`
D. `120 W//m^(2)`

Answer

To find the average intensity of an electromagnetic wave given the peak electric field, we can use the formula for intensity \( I \) in terms of the peak electric field \( E_0 \):

\[ I = \frac{\epsilon_0 c}{2} E_0^2 \]

Where:
- \( I \) is the average intensity,
- \( \epsilon_0 \) is the permittivity of free space (\( 8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2 \)),
- \( c \) is the speed of light in vacuum (\( 3 \times 10^8 \, \text{m/s} \)),
- \( E_0 \) is the peak electric field (in this case, \( 250 \, \text{N/C} \)).

### Step-by-step Solution:

1. **Identify the given values**:
- Peak electric field \( E_0 = 250 \, \text{N/C} \)
- Permittivity of free space \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2 \)
- Speed of light \( c = 3 \times 10^8 \, \text{m/s} \)

2. **Substitute the values into the intensity formula**:
\[ I = \frac{(8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2)(3 \times 10^8 \, \text{m/s})}{2} (250 \, \text{N/C})^2 \]

3. **Calculate \( (250 \, \text{N/C})^2 \)**:
\[ (250)^2 = 62500 \, \text{N}^2/\text{C}^2 \]

4. **Now substitute \( 62500 \, \text{N}^2/\text{C}^2 \) back into the equation**:
\[ I = \frac{(8.85 \times 10^{-12})(3 \times 10^8)}{2} \times 62500 \]

5. **Calculate \( \epsilon_0 c \)**:
\[ \epsilon_0 c = (8.85 \times 10^{-12})(3 \times 10^8) = 2.655 \times 10^{-3} \, \text{W/m} \]

6. **Now divide by 2**:
\[ \frac{2.655 \times 10^{-3}}{2} = 1.3275 \times 10^{-3} \, \text{W/m} \]

7. **Finally, multiply by \( 62500 \)**:
\[ I = (1.3275 \times 10^{-3}) \times 62500 \approx 83.0 \, \text{W/m}^2 \]

### Final Answer:
The average intensity of the wave is approximately \( 83 \, \text{W/m}^2 \).

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