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(B) The intensity of the electromagnetic wave at a distance $r$ is given by $I = \frac{P}{4\pi r^2}$.Also,the intensity is related to the electric fie

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$100$

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(B) The intensity of the electromagnetic wave at a distance $r$ is given by $I = \frac{P}{4\pi r^2}$.Also,the intensity is related to the electric field by $I = \frac{1}{2} c \epsilon_0 E_0^2$,where $E_0$ is the maximum (peak) electric field.Equating the two expressions: $\frac{1}{2} c \epsilon_0 E_0^2 = \frac{P}{4\pi r^2}$.Solving for $E_0$: $E_0 = \sqrt{\frac{2P}{4\pi r^2 c \epsilon_0}} = \sqrt{\frac{P}{2\pi r^2 c \epsilon_0}}$.Given $P = 1500 \, W$,$r = 3 \, m$,$c = 3 	imes 10^8 \, m/s$,and $\epsilon_0 = 8.854 	imes 10^{-12} \, C^2 N^{-1} m^{-2}$.$E_0 = \sqrt{\frac{1500}{2 	imes 3.14 	imes (3)^2 	imes 3 	imes 10^8 	imes 8.854 	imes 10^{-12}}}$.$E_0 = \sqrt{\frac{1500}{50.24 	imes 10^{-4}}} \approx \sqrt{298567} \approx 546 \, V/m$. However,using the approximation $c \epsilon_0 = \frac{1}{120\pi}$:$E_0 = \sqrt{\frac{2P}{4\pi r^2} \cdot 120\pi} = \sqrt{\frac{2P \cdot 30}{r^2}} = \frac{1}{r} \sqrt{60P} = \frac{1}{3} \sqrt{60 	imes 1500} = \frac{1}{3} \sqrt{90000} = \frac{300}{3} = 100 \, V/m$.

Electromagnetic waves are emitted from a point source. The output power of the source is $1500 \, W$. The maximum value of the electric field at a point $3 \, m$ away from the source is ........ $V \, m^{-1}$.

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