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(A) The average energy density $U_{av}$ of an electromagnetic wave is given by $U_{av} = \frac{1}{2} \varepsilon_{0} E_{0}^{2}$,where $E_{0}$ is the a

Vedclass · Accepted Answer

$2.45$

Vedclass · Answer

(A) The average energy density $U_{av}$ of an electromagnetic wave is given by $U_{av} = \frac{1}{2} \varepsilon_{0} E_{0}^{2}$,where $E_{0}$ is the amplitude of the electric field.The intensity $I$ at a distance $r$ from a point source is $I = \frac{P}{4 \pi r^{2}}$.Also,intensity is related to energy density by $I = U_{av} 	imes c$,where $c$ is the speed of light.Equating the two expressions for intensity: $\frac{P}{4 \pi r^{2}} = \frac{1}{2} \varepsilon_{0} E_{0}^{2} c$.Rearranging for $E_{0}^{2}$: $E_{0}^{2} = \frac{2P}{4 \pi r^{2} \varepsilon_{0} c} = \frac{2P}{r^{2}} 	imes \frac{1}{4 \pi \varepsilon_{0}} 	imes \frac{1}{c}$.Given $P = 0.1 \ W$,$r = 1 \ m$,$\frac{1}{4 \pi \varepsilon_{0}} = 9 	imes 10^{9} \ N \ m^{2} \ C^{-2}$,and $c = 3 	imes 10^{8} \ m/s$.$E_{0}^{2} = \frac{2 	imes 0.1}{1^{2}} 	imes (9 	imes 10^{9}) 	imes \frac{1}{3 	imes 10^{8}} = 0.2 	imes 3 	imes 10 = 6$.$E_{0} = \sqrt{6} \approx 2.45 \ V \ m^{-1}$.

$A$ red $LED$ emits light at $0.1 \ W$ uniformly in all directions. The amplitude of the electric field of the light at a distance of $1 \ m$ from the diode is.... $V \ m^{-1}$

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