A 1000 W bulb has optical efficiency `1.2%`. Find the amplitude (V/m) of electric field at distance `2 m` from bulb?

Answer

To solve the problem, we need to find the amplitude of the electric field (E₀) at a distance of 2 meters from a 1000 W bulb with an optical efficiency of 1.2%.

### Step-by-Step Solution:

1. **Calculate the Effective Power Output**:
The optical efficiency of the bulb is given as 1.2%. This means that only 1.2% of the total power is converted into light.
\[ P_{\text{effective}} = 1000 \, \text{W} \times \frac{1.2}{100} = 12 \, \text{W} \]

2. **Use the Intensity Formula**:
The intensity (I) of the light at a distance (r) from a point source can be calculated using the formula:
\[ I = \frac{P_{\text{effective}}}{4 \pi r^2} \]
Substituting the values:
\[ I = \frac{12 \, \text{W}}{4 \pi (2 \, \text{m})^2} = \frac{12}{16 \pi} = \frac{3}{4 \pi} \, \text{W/m}^2 \]

3. **Relate Intensity to Electric Field Amplitude**:
The intensity can also be expressed in terms of the electric field amplitude (E₀):
\[ I = \frac{1}{2} \epsilon_0 c E_0^2 \]
Where:
- \(\epsilon_0\) (the permittivity of free space) = \(8.85 \times 10^{-12} \, \text{F/m}\)
- \(c\) (the speed of light) = \(3 \times 10^8 \, \text{m/s}\)

4. **Set the Two Intensity Equations Equal**:
Equating the two expressions for intensity:
\[ \frac{3}{4 \pi} = \frac{1}{2} \epsilon_0 c E_0^2 \]

5. **Rearranging for E₀**:
Rearranging the equation to solve for \(E_0^2\):
\[ E_0^2 = \frac{3}{4 \pi} \cdot \frac{2}{\epsilon_0 c} \]
Substituting the values of \(\epsilon_0\) and \(c\):
\[ E_0^2 = \frac{3}{4 \pi} \cdot \frac{2}{(8.85 \times 10^{-12})(3 \times 10^8)} \]

6. **Calculating E₀**:
Calculate the right-hand side:
\[ E_0^2 = \frac{3 \cdot 2}{4 \pi \cdot 8.85 \times 10^{-12} \cdot 3 \times 10^8} \]
\[ E_0^2 = \frac{6}{4 \pi \cdot 2.655 \times 10^{-3}} \approx \frac{6}{3.32 \times 10^{-3}} \approx 1800 \]
Taking the square root:
\[ E_0 \approx \sqrt{1800} \approx 42.43 \, \text{V/m} \]

### Final Answer:
The amplitude of the electric field at a distance of 2 m from the bulb is approximately **42.43 V/m**.