In an electromagnetic wave, the maximum value of the electric field is `100 Vm^(-1)` The average intensity is `[epsilon_90)=8.8xx10^(-12)c^(-2)N^(-1)m^(2)]`

Options

A. `"13.2 W m"^(-2)`
B. `"36.5 W m"^(-2)`
C. `"46.7 W m"^(-2)`
D. `"765 W m"^(-2)`

Answer

To find the average intensity of the electromagnetic wave given the maximum electric field, we can follow these steps:

### Step 1: Understand the relationship between electric field and intensity
The average intensity \( I \) of an electromagnetic wave can be expressed in terms of the maximum electric field \( E_{\text{max}} \) using the formula:
\[ I = \frac{1}{2} \epsilon_0 c E_{\text{max}}^2 \]
where:
- \( \epsilon_0 \) is the permittivity of free space (\( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \)),
- \( c \) is the speed of light in vacuum (\( c = 3 \times 10^8 \, \text{m/s} \)),
- \( E_{\text{max}} \) is the maximum electric field.

### Step 2: Substitute the known values
Given:
- \( E_{\text{max}} = 100 \, \text{V/m} \)

Substituting the values into the intensity formula:
\[ I = \frac{1}{2} \cdot (8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2) \cdot (3 \times 10^8 \, \text{m/s}) \cdot (100 \, \text{V/m})^2 \]

### Step 3: Calculate \( E_{\text{max}}^2 \)
Calculating \( E_{\text{max}}^2 \):
\[ E_{\text{max}}^2 = (100 \, \text{V/m})^2 = 10000 \, \text{V}^2/\text{m}^2 = 10^4 \, \text{V}^2/\text{m}^2 \]

### Step 4: Calculate the intensity
Now substituting \( E_{\text{max}}^2 \) into the intensity formula:
\[ I = \frac{1}{2} \cdot (8.85 \times 10^{-12}) \cdot (3 \times 10^8) \cdot (10^4) \]
Calculating the multiplication:
\[ I = \frac{1}{2} \cdot (8.85 \times 3 \times 10^{-12} \times 10^8 \times 10^4) \]
\[ = \frac{1}{2} \cdot (26.55 \times 10^0) \, \text{W/m}^2 \]
\[ = 13.275 \, \text{W/m}^2 \]

### Step 5: Final Result
Thus, the average intensity \( I \) is approximately:
\[ I \approx 13.3 \, \text{W/m}^2 \]