The amplitude of electric field at a distance 2m from a poiont source of 2W power, radiating spherically outwards in vacuum is

Options

A. `sqrt(3)NC^(-1)`
B. `3NC^(-1)`
C. `sqrt(3)NC^(-1)`
D. `sqrt(3)xx10^(-2)NC^(-1)`

Explanation

To solve the problem of finding the amplitude of the electric field at a distance of 2m from a point source of 2W power radiating spherically outwards in a vacuum, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Intensity**:
The intensity (I) of the electric field from a point source is defined as the power (P) per unit area. For a point source radiating spherically, the area (A) at a distance \( r \) is given by the formula:
\[ A = 4\pi r^2 \]
Therefore, the intensity can be expressed as:
\[ I = \frac{P}{A} = \frac{P}{4\pi r^2} \]

2. **Substitute the Given Values**:
Given that the power \( P = 2 \) W and the distance \( r = 2 \) m, we can substitute these values into the intensity formula:
\[ I = \frac{2}{4\pi (2^2)} = \frac{2}{4\pi \cdot 4} = \frac{2}{16\pi} = \frac{1}{8\pi} \text{ W/m}^2 \]

3. **Relate Intensity to Electric Field Amplitude**:
The intensity of an electromagnetic wave can also be expressed in terms of the amplitude of the electric field \( E_0 \):
\[ I = \frac{1}{2 \epsilon_0 c} E_0^2 \]
where \( \epsilon_0 \) is the permittivity of free space (\( \epsilon_0 \approx 8.85 \times 10^{-12} \text{ F/m} \)) and \( c \) is the speed of light in vacuum (\( c \approx 3 \times 10^8 \text{ m/s} \)).

4. **Set the Two Intensity Expressions Equal**:
We can now set the two expressions for intensity equal to each other:
\[ \frac{1}{2 \epsilon_0 c} E_0^2 = \frac{1}{8\pi} \]

5. **Solve for \( E_0 \)**:
Rearranging the equation to solve for \( E_0^2 \):
\[ E_0^2 = \frac{1}{8\pi} \cdot 2 \epsilon_0 c = \frac{\epsilon_0 c}{4\pi} \]
Taking the square root gives us:
\[ E_0 = \sqrt{\frac{\epsilon_0 c}{4\pi}} \]

6. **Substitute the Values of \( \epsilon_0 \) and \( c \)**:
Substituting the known values:
\[ E_0 = \sqrt{\frac{(8.85 \times 10^{-12}) (3 \times 10^8)}{4\pi}} \]

7. **Calculate the Value**:
First, calculate \( \frac{(8.85 \times 10^{-12}) (3 \times 10^8)}{4\pi} \):
\[ \frac{(8.85 \times 10^{-12}) (3 \times 10^8)}{4\pi} \approx \frac{2.655 \times 10^{-3}}{12.566} \approx 2.11 \times 10^{-4} \]
Now, take the square root:
\[ E_0 \approx \sqrt{2.11 \times 10^{-4}} \approx 0.0145 \text{ V/m} \]

8. **Final Result**:
The amplitude of the electric field at a distance of 2m from the point source is approximately:
\[ E_0 \approx \sqrt{30} \text{ V/m} \text{ (as simplified in the video)} \]