The peak value of the electric field component of an electromagnetic wave is E. At a particular instant the intensity of the wave is of 0.020 `W//m^(2)` . If the electric field were increased to 5E what would be the intensity of the wave ?

Options

A. `0.020 W//m^(2)`
B. `0.25 W//m^(2)`
C. `0.10 W//m^(2)`
D. `0.50 W//m^(2)`

Answer

To solve the problem, we need to understand the relationship between the electric field (E) and the intensity (I) of an electromagnetic wave. The intensity of an electromagnetic wave is given by the formula:

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

Where:
- \( I \) is the intensity,
- \( \epsilon_0 \) is the permittivity of free space,
- \( c \) is the speed of light in vacuum,
- \( E \) is the peak value of the electric field.

Given that the initial intensity \( I \) is \( 0.020 \, \text{W/m}^2 \) and the peak electric field is \( E \), we can express the intensity in terms of \( E \):

1. **Initial Intensity**:
\[ I = \frac{1}{2} \epsilon_0 c E^2 = 0.020 \, \text{W/m}^2 \]

2. **New Electric Field**:
If the electric field is increased to \( 5E \), we can find the new intensity \( I' \):
\[ I' = \frac{1}{2} \epsilon_0 c (5E)^2 \]

3. **Calculating New Intensity**:
Expanding the equation for \( I' \):
\[ I' = \frac{1}{2} \epsilon_0 c (25E^2) = 25 \left(\frac{1}{2} \epsilon_0 c E^2\right) \]
Since \( \frac{1}{2} \epsilon_0 c E^2 = I \), we can substitute:
\[ I' = 25 I \]

4. **Substituting the Value of I**:
Now substituting the value of \( I \):
\[ I' = 25 \times 0.020 \, \text{W/m}^2 = 0.50 \, \text{W/m}^2 \]

Thus, the new intensity of the wave when the electric field is increased to \( 5E \) is:

\[ \boxed{0.50 \, \text{W/m}^2} \]