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 x 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

Question

Quizplus · Accepted Answer

The average electromagnetic energy density (u) is given by u = (ε₀E²)/2. Substituting the given values, u = (8.85 × 10⁻¹² C²/N∙m² × (0.90 V/m)²)/2 = 3.6 × 10⁻¹² J/m³ = 3.6 pJ/m³.

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