In a plane EM wave, the electric field oscillates sinusoidally at a frequency of 5 × 10¹⁰ Hz and an amplitude of 50 V m⁻¹. The total average energy density of the electromagnetic field of the wave is: [Use ε₀ = 8.85 × 10⁻¹² C²/N m² ] (a) 4.425 × 10⁻⁸ J m⁻³ (b) 2.212 × 10⁻¹⁰ J m⁻³ (c) 2.212 × 10⁻⁸ J m⁻³ (d) 1.106 × 10⁻⁸ J m⁻³

Answer: (d). Total average energy density u = 1/2ε₀ E₀² (frequency is not needed). u = 1/2 × 8.85 × 10⁻¹² × 50² = 1.106 × 10⁻⁸ J m ⁻³ So the answer is (d) 1.106 × 10⁻⁸ J m ⁻³.

In a plane EM wave, the electric field oscillates sinusoidally at a… | JEE Main 2024 (31 Jan, Shift 1)

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Energy Density & Intensity

In a plane EM wave, the electric field oscillates sinusoidally at a frequency of $5 \times 10^{10}\text{ Hz}$ and an amplitude of $50\text{ V m}^{-1}$ . The total average energy density of the electromagnetic field of the wave is: [Use $\varepsilon_0 = 8.85 \times 10^{-12}\text{ C}^2/\text{N m}^2$ ]

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Answer: (d)

Total average energy density $u = \tfrac{1}{2}\varepsilon_0 E_0^2$ (frequency is not needed).
$u = \tfrac{1}{2} \times 8.85 \times 10^{-12} \times 50^2 = 1.106 \times 10^{-8}$ J m $^{-3}$
So the answer is (d) $1.106 \times 10^{-8}$ J m $^{-3}$ .

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