An electric bulb of power `30piW` has an efficiency of `10%` and it can be assumed to behave like a point source of light. At a distance of 3 m from the bulb, the peak value of the electric field in the light produced by the bulb is `["Take "epsilon_(0)~~9xx10^(-12)C^(2)N^(-1)m^(-2)]`

Options

A. `(100)/(9)" V m"^(-1)`
B. `(10)/(9)"V m"^(-1)`
C. `(10)/(9sqrt2)"V m"^(-1)`
D. `(100)/(9sqrt2)"V m"^(-1)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the Effective Power of the Bulb
Given that the bulb has a power of \( P = 30\pi \, W \) and an efficiency of \( 10\% \), we can find the effective power that contributes to the light intensity.

\[ P_{\text{effective}} = \text{Efficiency} \times P = 0.10 \times 30\pi = 3\pi \, W \]

### Step 2: Calculate the Intensity at a Distance of 3 m
The intensity \( I \) at a distance \( r \) from a point source is given by:

\[ I = \frac{P_{\text{effective}}}{A} \]

Where \( A \) is the area of a sphere with radius \( r \):

\[ A = 4\pi r^2 \]

Substituting \( r = 3 \, m \):

\[ A = 4\pi (3^2) = 36\pi \, m^2 \]

Now substituting back to find the intensity:

\[ I = \frac{3\pi}{36\pi} = \frac{3}{36} = \frac{1}{12} \, W/m^2 \]

### Step 3: Relate Intensity to Electric Field
The intensity \( I \) is related to the electric field \( E \) by the formula:

\[ I = \frac{1}{2} \epsilon_0 c E_{\text{rms}}^2 \]

Where:
- \( \epsilon_0 = 9 \times 10^{-12} \, C^2/N \cdot m^2 \)
- \( c = 3 \times 10^8 \, m/s \)

### Step 4: Solve for \( E_{\text{rms}} \)
Rearranging the formula gives:

\[ E_{\text{rms}}^2 = \frac{2I}{\epsilon_0 c} \]

Substituting the values:

\[ E_{\text{rms}}^2 = \frac{2 \times \frac{1}{12}}{(9 \times 10^{-12})(3 \times 10^8)} \]

Calculating the denominator:

\[ (9 \times 10^{-12})(3 \times 10^8) = 27 \times 10^{-4} = 2.7 \times 10^{-3} \]

Now substituting back to find \( E_{\text{rms}}^2 \):

\[ E_{\text{rms}}^2 = \frac{2 \times \frac{1}{12}}{2.7 \times 10^{-3}} = \frac{\frac{1}{6}}{2.7 \times 10^{-3}} = \frac{1}{6 \times 2.7 \times 10^{-3}} = \frac{1}{0.0162} \approx 61.73 \, (V/m)^2 \]

### Step 5: Calculate the Peak Electric Field \( E_0 \)
The peak electric field \( E_0 \) is related to the rms value by:

\[ E_0 = \sqrt{2} E_{\text{rms}} \]

Calculating \( E_0 \):

\[ E_0 = \sqrt{2} \times \sqrt{61.73} \approx 1.414 \times 7.85 \approx 11.09 \, V/m \]

### Final Answer
The peak value of the electric field \( E_0 \) at a distance of 3 m from the bulb is approximately:

\[ E_0 \approx 11.09 \, V/m \]

---