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Step-by-Step Solution
1. Determine the Actual Radiated Power
The bulb is rated at 1000 W, but it is only 1.25% efficient in emitting this power as electromagnetic radiation. Thus, the effective radiated power
P_{\text{rad}} is:
P_{\text{rad}}
= \frac{1.25}{100} \times 1000 \text{ W}
= 12.5 \text{ W}.
2. Compute the Intensity at Distance 2 m
The point P is at a distance of 2 m from the source. Assuming the radiated power spreads uniformly over a spherical surface, the intensity
I at distance r = 2\,\text{m} is:
I = \frac{P_{\text{rad}}}{4\pi r^2}
= \frac{12.5}{4\pi \times (2)^2}
= \frac{12.5}{16\pi}
\approx 0.2487 \,\text{W\,m}^{-2}.
3. Relate the Intensity to the Peak Electric Field
For an electromagnetic wave in free space, the time-averaged intensity
I_{\text{avg}} is given by:
I_{\text{avg}} = \frac{1}{2} \,\varepsilon_0 \, c \, E_{0}^{2},
where
\varepsilon_0 = 8.85 \times 10^{-12}\,\text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}
is the permittivity of free space, and
c = 3 \times 10^{8}\,\text{m\,s}^{-1}
is the speed of electromagnetic waves in free space.
Here, E_{0} is the peak (maximum) electric field amplitude.
4. Solve for the Peak Electric Field E_{0}
We set the computed intensity I (0.2487 W/m2) equal to the expression for
I_{\text{avg}} :
0.2487
= \frac{1}{2} \times (8.85 \times 10^{-12})
\times (3 \times 10^{8}) \times E_{0}^{2}.
Therefore,
E_{0}^{2}
= \frac{2 \times 0.2487}{(8.85 \times 10^{-12}) \times (3 \times 10^{8})}
\approx 187.
Taking the square root:
E_{0} = \sqrt{187} \;\approx\; 13.7 \,\text{V\,m}^{-1}.
5. Express the Result in the Required Form
The problem states the peak electric field has the form
x \times 10^{-1}\,\text{V\,m}^{-1} . Since
13.7 \,\text{V\,m}^{-1} = 137 \times 10^{-1}\,\text{V\,m}^{-1} ,
it follows that
x = 137 . Rounding to the nearest integer still yields 137.
Final Answer
The value of x is 137.
