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 step by step, we will follow the outlined approach based on the information provided in the question and the video transcript.

### Step 1: Calculate the Output Power
The efficiency of the bulb is given as 10%, and its input power is \(30\pi \, W\).

\[ \text{Output Power} = \text{Efficiency} \times \text{Input Power} = \frac{10}{100} \times 30\pi = 3\pi \, W \]

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

\[ I = \frac{P}{A} \]

where \(A\) is the area of the sphere with radius \(r\). The area \(A\) is given by:

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

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

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

Now, substituting the output power into the intensity formula:

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

### Step 3: Relate Intensity to Electric Field
The intensity \(I\) is also related to the electric field \(E_0\) by the equation:

\[ I = \frac{1}{2} \epsilon_0 E_0^2 c \]

where:
- \(\epsilon_0 = 9 \times 10^{-12} \, C^2/(N \cdot m^2)\) (permittivity of free space)
- \(c = 3 \times 10^8 \, m/s\) (speed of light)

Rearranging the equation to solve for \(E_0\):

\[ E_0 = \sqrt{\frac{2I}{\epsilon_0 c}} \]

### Step 4: Substitute Known Values
Substituting the values we have:

\[ E_0 = \sqrt{\frac{2 \times \frac{1}{12}}{(9 \times 10^{-12}) \times (3 \times 10^8)}} \]

Calculating the denominator:

\[ \epsilon_0 c = (9 \times 10^{-12}) \times (3 \times 10^8) = 27 \times 10^{-4} = 2.7 \times 10^{-3} \]

Now substituting back into the equation for \(E_0\):

\[ E_0 = \sqrt{\frac{2 \times \frac{1}{12}}{2.7 \times 10^{-3}}} \]

Calculating the numerator:

\[ 2 \times \frac{1}{12} = \frac{2}{12} = \frac{1}{6} \]

Thus:

\[ E_0 = \sqrt{\frac{1/6}{2.7 \times 10^{-3}}} = \sqrt{\frac{1}{6 \times 2.7 \times 10^{-3}}} \]

Calculating \(6 \times 2.7 = 16.2\):

\[ E_0 = \sqrt{\frac{1}{16.2 \times 10^{-3}}} = \sqrt{\frac{1000}{16.2}} \times 10^{1.5} \]

### Step 5: Final Calculation
Calculating \(E_0\):

\[ E_0 \approx \sqrt{61.73} \times 10^{1.5} \approx 7.85 \times 10^{0.75} \]

This gives us the peak value of the electric field.

### Final Answer
The peak value of the electric field \(E_0\) is approximately \( \frac{100}{9\sqrt{2}} \, N/C\).

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