A point source of electromagnetic radiation has an average power output of1500W. The maximum value of electric field at a distance 3m from this source in `Vm^-1` is

Options

A. `500`
B. `100`
C. `500/3`
D. `250/3`

Explanation

To find the maximum value of the electric field \( E_0 \) at a distance of 3 meters from a point source of electromagnetic radiation with an average power output of 1500 W, we can follow these steps:

### Step 1: Understand the relationship between power, intensity, and electric field
The intensity \( I \) of electromagnetic waves is related to the electric field \( E_0 \) by the equation:
\[ I = \frac{1}{2} \epsilon_0 c E_0^2 \]
where:
- \( I \) is the intensity (W/m²),
- \( \epsilon_0 \) is the permittivity of free space (\( 8.85 \times 10^{-12} \, \text{F/m} \)),
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)),
- \( E_0 \) is the maximum electric field (V/m).

### Step 2: Calculate the area over which the power is distributed
For a point source radiating power uniformly in all directions, the area \( A \) at a distance \( r \) is given by:
\[ A = 4 \pi r^2 \]
For \( r = 3 \, \text{m} \):
\[ A = 4 \pi (3)^2 = 36 \pi \, \text{m}^2 \]

### Step 3: Calculate the intensity
The intensity \( I \) can also be defined as the power \( P \) divided by the area \( A \):
\[ I = \frac{P}{A} \]
Substituting the values:
\[ I = \frac{1500 \, \text{W}}{36 \pi} \approx \frac{1500}{113.097} \approx 13.27 \, \text{W/m}^2 \]

### Step 4: Relate intensity to electric field
From the intensity equation:
\[ I = \frac{1}{2} \epsilon_0 c E_0^2 \]
We can rearrange this to solve for \( E_0 \):
\[ E_0 = \sqrt{\frac{2I}{\epsilon_0 c}} \]

### Step 5: Substitute the values
Substituting \( I \), \( \epsilon_0 \), and \( c \):
\[ E_0 = \sqrt{\frac{2 \times 13.27}{8.85 \times 10^{-12} \times 3 \times 10^8}} \]

Calculating the denominator:
\[ \epsilon_0 c = 8.85 \times 10^{-12} \times 3 \times 10^8 \approx 2.655 \times 10^{-3} \]

Now substituting back into the equation for \( E_0 \):
\[ E_0 = \sqrt{\frac{26.54}{2.655 \times 10^{-3}}} \approx \sqrt{10000} = 100 \, \text{V/m} \]

### Final Answer
The maximum value of the electric field \( E_0 \) at a distance of 3 m from the source is:
\[ E_0 \approx 100 \, \text{V/m} \]

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