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Question

Using $\mathrm{U}_{\mathrm{av}}=\frac{1}{2} \varepsilon_{0} \mathrm{E}^{2}$

But $U_{a v}=\frac{P}{4 \pi r^{2} 	imes c}$

$\frac{P}{4 \pi r^{2}}=

Vedclass · Accepted Answer

$2.45$

Vedclass · Answer

Using $\mathrm{U}_{\mathrm{av}}=\frac{1}{2} \varepsilon_{0} \mathrm{E}^{2}$

But $U_{a v}=\frac{P}{4 \pi r^{2} 	imes c}$

$\frac{P}{4 \pi r^{2}}=\frac{1}{2} \varepsilon_{0} E^{2} 	imes c$

$\mathrm{E}_{0}^{2}=\frac{2 \mathrm{P}}{4 \pi \mathrm{r}^{2} \varepsilon_{0} \mathrm{c}}=\frac{2 	imes 0.1 	imes 9 	imes 10^{9}}{1 	imes 3 	imes 10^{8}}$

$\mathrm{E}_{0}=\sqrt{6}=2.45 \,\mathrm{V} / \mathrm{m}$

A red $LED$ emits light at $0.1$ watt uniformly around it. The amplitude of the electric field of the light at a distance of $1\ m$ from the diode is....$ Vm^{-1}$

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