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 230 mV/m.Assume that the signal power is radiated uniformly in all directions and that radio waves incident upon the ground are completely absorbed.The average electromagnetic energy density at that point is closest to: A)2.3 × 10-13 J/m³ B)3.3 × 10-13 J/m³ C)4.7 × 10-13 J/m³ D)6.6 × 10-13 J/m³ E)9.4 × 10-13 J/m³

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Quizplus · Accepted Answer

The average electromagnetic energy density is given by:$$u=\frac{1}{2}\epsilon_0E_0^2$$where:* $\epsilon_0$ is the permittivity of free space* $E_0$ is the electric field amplitudeSubstituting the given values into the equation, we get:$$u=\frac{1}{2}(8.85 imes10^{-12} ext{ F/m})(0.23 ext{ V/m})^2=2.3 imes10^{-13} ext{ J/m}^3$$Therefore, the average electromagnetic energy density at that point is closest to 2.3 × 10-13 J/m³.

An 800 KHz Radio Signal Is Detected at a Point