An 800-kHz radio signal is detected at a point 8.5 km distant from a transmitter tower.The electric field amplitude of the signal at that point is 0.90 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)3.6 pJ/m³ B)5.1 pJ/m³ C)7.2 pJ/m³ D)10 pJ/m³ E)14 pJ/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:* u is the average electromagnetic energy density (in J/m^3)* ε0 is the permittivity of free space (in F/m)* E0 is the electric field amplitude (in V/m)Plugging in the given values, we get:$$u=\frac{1}{2}(8.85	imes10^{-12}	ext{ C}^2/	ext{N}\cdot	ext{m}^2)(0.90	ext{ V/m})^2=\boxed{3.6	ext{ pJ/m}^3}$$

An 800-KHz Radio Signal Is Detected at a Point 8