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 and the permittivity of free space, we can use the formula for average intensity \( I \):

\[ I = \frac{1}{2} \epsilon_0 E_{max}^2 c \]

where:
- \( I \) is the average intensity,
- \( \epsilon_0 \) is the permittivity of free space,
- \( E_{max} \) is the maximum electric field,
- \( c \) is the speed of light in vacuum.

### Step 1: Identify the given values
- Maximum electric field \( E_{max} = 100 \, \text{V/m} \)
- Permittivity of free space \( \epsilon_0 = 8.8 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \)
- Speed of light \( c = 3 \times 10^8 \, \text{m/s} \)

### Step 2: Substitute the values into the formula
Substituting the given values into the intensity formula:

\[ I = \frac{1}{2} \times (8.8 \times 10^{-12}) \times (100)^2 \times (3 \times 10^8) \]

### Step 3: Calculate \( E_{max}^2 \)
Calculate \( E_{max}^2 \):

\[ E_{max}^2 = (100)^2 = 10000 \, \text{V}^2/\text{m}^2 \]

### Step 4: Substitute \( E_{max}^2 \) back into the formula
Now substituting \( E_{max}^2 \) back into the equation:

\[ I = \frac{1}{2} \times (8.8 \times 10^{-12}) \times (10000) \times (3 \times 10^8) \]

### Step 5: Simplify the expression
Calculating the multiplication step-by-step:

1. Calculate \( \frac{1}{2} \times 8.8 \times 10^{-12} = 4.4 \times 10^{-12} \)
2. Now multiply \( 4.4 \times 10^{-12} \times 10000 = 4.4 \times 10^{-8} \)
3. Finally, multiply by \( 3 \times 10^8 \):

\[ I = 4.4 \times 10^{-8} \times 3 \times 10^8 = 13.2 \, \text{W/m}^2 \]

### Step 6: Final Result
Thus, the average intensity \( I \) is:

\[ I = 13.2 \, \text{W/m}^2 \]