Power of a Source producing electromagnetic wave is 1500 W. Maximum value of electric field at a point 3m from the source is found to be ` E_0` measureed in `V//m , E_0 // 20 = V//m`

Explanation

To solve the problem, we need to find the maximum value of the electric field \( E_0 \) at a point 3 meters from a source producing electromagnetic waves with a power of 1500 W. We will also find \( \frac{E_0}{20} \).

### Step-by-Step Solution:

1. **Understand the relationship between power, intensity, and electric field:**
The intensity \( I \) of an electromagnetic wave can be expressed in two ways:
\[ I = \frac{P}{4 \pi r^2} \]
and
\[ I = \frac{1}{2} \epsilon_0 E_0^2 c \]
where:
- \( P \) is the power of the source,
- \( r \) is the distance from the source,
- \( \epsilon_0 \) 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.

2. **Set the two expressions for intensity equal to each other:**
\[ \frac{P}{4 \pi r^2} = \frac{1}{2} \epsilon_0 E_0^2 c \]

3. **Substitute the known values:**
- Power \( P = 1500 \, \text{W} \)
- Distance \( r = 3 \, \text{m} \)
- \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \)
- \( c = 3 \times 10^8 \, \text{m/s} \)

Plugging in these values:
\[ \frac{1500}{4 \pi (3^2)} = \frac{1}{2} (8.85 \times 10^{-12}) E_0^2 (3 \times 10^8) \]

4. **Calculate the left side:**
\[ 4 \pi (3^2) = 4 \pi \times 9 = 36 \pi \approx 113.097 \]
\[ \frac{1500}{113.097} \approx 13.27 \, \text{W/m}^2 \]

5. **Set up the equation:**
\[ 13.27 = \frac{1}{2} (8.85 \times 10^{-12}) E_0^2 (3 \times 10^8) \]

6. **Simplify the right side:**
\[ \frac{1}{2} \times 8.85 \times 10^{-12} \times 3 \times 10^8 \approx 1.3275 \times 10^{-3} E_0^2 \]

7. **Equate and solve for \( E_0^2 \):**
\[ 13.27 = 1.3275 \times 10^{-3} E_0^2 \]
\[ E_0^2 = \frac{13.27}{1.3275 \times 10^{-3}} \approx 10000 \]
\[ E_0 = \sqrt{10000} = 100 \, \text{V/m} \]

8. **Find \( \frac{E_0}{20} \):**
\[ \frac{E_0}{20} = \frac{100}{20} = 5 \, \text{V/m} \]

### Final Answer:
The value of \( \frac{E_0}{20} \) is \( 5 \, \text{V/m} \).