The sunlight strikes the upper atmosphere of the Earth with intensity 1.38 k W/`m^(2)` . The peak value of electric field at that point will be ( in kilovolt /meter )

Options

A. 2.04
B. 4.08
C. 8.16
D. 1.02

Answer

To find the peak value of the electric field (E₀) from the intensity (I) of sunlight striking the upper atmosphere, we can use the relationship between intensity and the electric field in electromagnetic waves. The formula we will use is:

\[ I = \frac{1}{2} \epsilon_0 c E_0^2 \]

Where:
- \( I \) is the intensity (in W/m²),
- \( \epsilon_0 \) is the permittivity of free space (\( 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \)),
- \( c \) is the speed of light in vacuum (\( 3 \times 10^8 \, \text{m/s} \)),
- \( E_0 \) is the peak electric field (in V/m).

### Step-by-Step Solution:

1. **Convert Intensity to Standard Units**:
Given intensity \( I = 1.38 \, \text{kW/m}^2 = 1380 \, \text{W/m}^2 \).

2. **Rearrange the Formula**:
We need to express \( E_0 \) in terms of \( I \):
\[ I = \frac{1}{2} \epsilon_0 c E_0^2 \]
Rearranging gives:
\[ E_0 = \sqrt{\frac{2I}{\epsilon_0 c}} \]

3. **Substitute Known Values**:
Now, substitute the known values into the equation:
- \( I = 1380 \, \text{W/m}^2 \)
- \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \)
- \( c = 3 \times 10^8 \, \text{m/s} \)

Thus,
\[ E_0 = \sqrt{\frac{2 \times 1380}{(8.85 \times 10^{-12}) \times (3 \times 10^8)}} \]

4. **Calculate the Denominator**:
First, calculate \( \epsilon_0 \times c \):
\[ \epsilon_0 \times c = (8.85 \times 10^{-12}) \times (3 \times 10^8) \approx 2.655 \times 10^{-3} \]

5. **Calculate the Full Expression**:
Now substitute back into the equation:
\[ E_0 = \sqrt{\frac{2760}{2.655 \times 10^{-3}}} \]

6. **Final Calculation**:
\[ E_0 = \sqrt{1.040 \times 10^6} \approx 1020 \, \text{V/m} \]

7. **Convert to kV/m**:
Since we need the answer in kilovolts per meter:
\[ E_0 \approx 1.02 \, \text{kV/m} \]

### Final Answer:
The peak value of the electric field at that point is approximately **1.02 kV/m**.