Assume a bulb of efficiency 2.5% as a point s | vedclass

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Question

Here intensity, $I$ $=\frac{	ext { power }}{	ext { area }}$

$=\frac{100 	imes 2.5}{4 \pi(3)^{2} 	imes 100}=\frac{2.5}{36 \pi} \mathrm{\,W} \mat

Vedclass · Accepted Answer

$4.08$

Vedclass · Answer

Here intensity, $I$ $=\frac{	ext { power }}{	ext { area }}$

$=\frac{100 	imes 2.5}{4 \pi(3)^{2} 	imes 100}=\frac{2.5}{36 \pi} \mathrm{\,W} \mathrm{m}^{-2}$

We know, $I=\frac{1}{2} \varepsilon_{0} E_{0}^{2} C$

or $\quad \mathrm{E}_{0}=\sqrt{\frac{2 \mathrm{I}}{\varepsilon_{0} \mathrm{c}}}=\sqrt{\frac{2 	imes \frac{2.5}{36 \pi}}{\frac{1}{4 \pi 	imes 9 	imes 10^{9}} 	imes 3 	imes 10^{2}}}=4.08 \mathrm{\,Vm}^{-1}$

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

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