The electric field in an electromagnetic wave is given by E = (50 NC ⁻¹) sin ω (t - x/c) The energy contained in a cylinder of volume V is 5.5 × 10 - 12 J. The value of V is _____________ cm3. (given ε 0 = 8.8 × 10 - 12C2N -1 m- 2) Numerical answer type. Work it out, then open the answer.

Answer: 500. Average energy density u = 1/2ε₀ E₀² = 1/2 × 8.8 × 10⁻¹² × 50² = 1.1 × 10⁻⁸ J/m ³ V = U/u = 5.5 × 10⁻¹²1.1 × 10⁻⁸ = 5 × 10⁻⁴ m ³ V = 500 cm ³

The electric field in an electromagnetic wave is given by E = (50 NC… | JEE Main 2021 (31 Aug, Shift 1)

Q20JEE Main 2021 · 31 AUG · SHIFT 1Medium

Energy Density & Intensity

The electric field in an electromagnetic wave is given by E = (50 NC ${}^{-1}$ ) sin $\omega$ (t $-$ x/c) The energy contained in a cylinder of volume V is 5.5 $\times$ 10 $-$ 12 J. The value of V is _____________ cm3. (given $\varepsilon$ 0 = 8.8 $\times$ 10 $-$ 12C2N $-1\,\text{m}-$ 2)

Numerical answer type. Work it out, then open the answer.

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Answer: 500

Average energy density $u = \tfrac{1}{2}\varepsilon_0 E_0^2 = \tfrac{1}{2} \times 8.8 \times 10^{-12} \times 50^2 = 1.1 \times 10^{-8}$ J/m $^3$
$V = \dfrac{U}{u} = \dfrac{5.5 \times 10^{-12}}{1.1 \times 10^{-8}} = 5 \times 10^{-4}$ m $^3$
$V = 500$ cm $^3$

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