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(C) The power radiated by the bulb is $P = 100 	imes \frac{2.5}{100} = 2.5 \, W$.The intensity $I$ at a distance $r = 3 \, m$ is given by $I = \frac{

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$4.08$

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(C) The power radiated by the bulb is $P = 100 	imes \frac{2.5}{100} = 2.5 \, W$.The intensity $I$ at a distance $r = 3 \, m$ is given by $I = \frac{P}{4 \pi r^2} = \frac{2.5}{4 \pi (3)^2} = \frac{2.5}{36 \pi} \, W/m^2$.The relationship between intensity $I$ and the peak electric field $E_0$ is $I = \frac{1}{2} \varepsilon_0 E_0^2 c$.Rearranging for $E_0$,we get $E_0 = \sqrt{\frac{2I}{\varepsilon_0 c}}$.Substituting the values $\varepsilon_0 = \frac{1}{36 \pi 	imes 10^9} \, F/m$ and $c = 3 	imes 10^8 \, m/s$:$E_0 = \sqrt{\frac{2 	imes (2.5 / 36 \pi)}{(1 / 36 \pi 	imes 10^9) 	imes 3 	imes 10^8}} = \sqrt{\frac{2 	imes 2.5 	imes 36 \pi 	imes 10^9}{36 \pi 	imes 3 	imes 10^8}} = \sqrt{\frac{5 	imes 10}{3}} = \sqrt{16.67} \approx 4.08 \, V/m$.

Assume a bulb of efficiency $2.5\%$ as a point source. The peak value of the electric field produced by the radiation coming from a $100\, W$ bulb at a distance of $3\, m$ is approximately.....$V\,m^{-1}$.

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