A 7.5 × 10¹⁴ Hz laser emits a 7.7-μs pulse,5.0 mm in diameter,with a beam energy density of 0.51 J/m³.What is the amplitude of the electric field of the emitted waves? (c = 3.00 × 10⁸ m/s,μ₀ = 4π × 10^-7 T ∙ m/A,ε₀ = 8.85 × 10^-12 C²/N ∙ m²) A)340 kV/m B)480 kV/m C)240 kV/m D)150 kV/m E)120 kV/m

Question

Quizplus · Accepted Answer

The amplitude of the electric field (E) can be calculated using the formula for energy density (u) in terms of electric field: u = (1/2)ε₀E². Solving for E gives E = sqrt(2u/ε₀). Substituting the given values, E = sqrt(2*0.51 J/m³ / 8.85 × 10⁻¹² C²/N∙m²) ≈ 340,000 V/m or 340 kV/m.

A 75 × 1014 Hz Laser Emits a 7

A 75 × 10¹⁴ Hz Laser Emits a 7