Question:
The peak electric field produced by the radiation coming from the $8 \mathrm{~W}$ bulb at a distance of $10 \mathrm{~m}$ is $\frac{x}{10} \sqrt{\frac{\mu_{0} c}{\pi}} \frac{V}{m}$. The efficiency of the bulb is
$10 \%$ and it is a point source. The value of $x$ is
Solution:
(2)
$I=\frac{1}{2} c \in_{0} E_{0}^{2}$
$\frac{8}{4 \pi \times 10^{2}}=\frac{1}{2} \times c \times \frac{1}{\mu_{0} c^{2}} \times E_{0}^{2}$
$E_{0}=\frac{2}{10} \sqrt{\frac{\mu_{0} c}{\pi}}$
$\Rightarrow x=2$
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