Question

The peak electric field produced by the radiation coming from $8\text{W}$ bulb at a distance of $10\text{m}$ is $\frac{x}{10}\sqrt{\frac{{\mu }_{0}c}{\pi }}\text{V/m}$. The efficiency of the bulb is $10\%$ and it is a point source. The value of $x$ is_________.

Answer 1

Intensity $I=\frac{1}{2}c{ϵ}_0(E_0{)}^2$
$\text{Intensity}=\text{Power/Area}=8/(4\pi \times {10}^2)$
$\frac{8}{4\pi \times {10}^2}\times \frac{1}{2}=\frac{1}{4}\times c\times \frac{1}{{\mu }_0{c}^2}\times E_0^2$
$[\text{Multiplying by}1/2\text{and}{ϵ}_0=1/{\mu }_0{c}^2]$
$E_0=\frac{2}{10}\sqrt{\frac{{\mu }_{0}c}{\pi }}$
$\Rightarrow x=2$