An 800-kHz radio signal is detected at a point 9.5 km distant from a transmitter tower. The electric field amplitude of the signal at that p

Question

An 800-kHz radio signal is detected at a point 9.5 km distant from a transmitter tower. The electric field amplitude of the signal at that point is 0.23 V/m. Assume that the signal power is radiated uniformly in all directions and that radio waves incident upon the ground is completely absorbed. What is the average electromagnetic energy density at that point

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Thu Giang 3 years 2021-08-05T23:24:23+00:00 1 Answers 33 views 0

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  1. thanhha
    0
    2021-08-05T23:25:39+00:00
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    Answer:

    E_{avg}=2.34*10^{-13}J/m^3

    Explanation:

    From the question we are told that:

    Frequency F=800KHz =800*10^3Hz

    Distance 9.5km=9.5*10^3

    Electric field amplitude E=0.23V/m

    Generally the equation for Average electromagnetic energy density is mathematically given by

     E_{avg}=\frac{1}{2} \epsilon_0 E^2

     E_{avg}=\frac{1}{2} 8.85*10^{-12}*(0.23)^2

     E_{avg}=\frac{1}{2}* 8.85*10^{-12}*(0.23)^2

     E_{avg}=2.34*10^{-13}J/m^3

    Therefore the  average electromagnetic energy density at that point

     E_{avg}=2.34*10^{-13}J/m^3

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