An 800-kHz radio signal is detected at a point 9.5 km distant from a transmitter tower.The $\vec { E }$ 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 are completely absorbed.What is the average electromagnetic energy density at that point? (c = 3.0 × 10⁸ m/s,μ₀ = 4π × 10^-7 T ∙ m/A,ε₀ = 8.85 × 10^-12 C²/N ∙ m²) 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:$$u=\frac{1}{2}\epsilon_0 E^2$$where:* $\epsilon_0$ is the permittivity of free space* $E$ is the electric field amplitudeSubstituting the given values into the equation, we get:$$u=\frac{1}{2}(8.85\times10^{-12}\text{ C}^2/\text{N}\cdot\text{m}^2)(0.23\text{ V/m})^2$$$$u=2.3\times10^{-13}\text{ J/m}^3$$Therefore, the correct answer is A.

An 800-KHz Radio Signal Is Detected at a Point 9 E⃗\vec { E }E

An 800-KHz Radio Signal Is Detected at a Point 9 $\vec { E }$