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JEE Main 2014 (11 Apr, Shift 1) · Energy Density & Intensity

Energy Density & Intensity question from JEE Main 2014 (11 Apr, Shift 1)

Q34JEE Main 2014 · 11 APR · SHIFT 1Easy

Energy Density & Intensity

An electromagnetic wave of frequency 1 ×\times 1014 hertz is propagating along z - axis. The amplitude of electric field is 4 V/m. If ε\varepsilon0 = 8.8 ×\times 10−-12 C2/N-m2, then average energy density of electric field will be:

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Answer: (c)

  1. Average electric energy density ⟨uE⟩=14ε0E02\langle u_E \rangle = \tfrac{1}{4}\varepsilon_0 E_0^2 (frequency is not needed).
  2. ⟨uE⟩=14×8.8×10−12×16=35.2×10−12\langle u_E \rangle = \tfrac{1}{4} \times 8.8 \times 10^{-12} \times 16 = 35.2 \times 10^{-12} J/m3^3
  3. So the answer is (c) 35.2×10−1235.2 \times 10^{-12} J/m3^3.
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