Field intensity: 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:$$u = \frac{1}{2}ε_0E^2$$where:* ε_0 is the permittivity of free space* E is the electric field amplitudeSubstituting the given values into the equation, we get:$$u = \frac{1}{2}(8.85 × 10^{-12} C^2/N \cdot m^2)(0.90 V/m)^2 = 3.6 pJ/m^3$$Therefore, the average electromagnetic energy density at that point is 3.6 pJ/m^3.

Field Intensity: an 800-KHz Radio Signal Is Detected at a Point