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JEE Main
Exam
JEE Main
Subject
Physics
Chapter
Electromagnetic Waves

JEE Main · Physics · Electromagnetic Waves

An electromagnetic wave of frequency 1 × 10 ^ 14 Hz is propagating...

An electromagnetic wave of frequency 1×1014Hz1 \times {10}^{14} \text{Hz} is propagating along zz -axis. The amplitude of the electric field is 4V/m4\text{V/m}. If ε0=8.8×10−12C2/N−m2{ε}_{0} = 8\text{.}8 \times 1{0}^{- 12}C^{2} / {N-m}^{2}, then the average energy density of electric field will be

Options

  1. A35.2×10−12J/m335.2 \times {10}^{- 12} {\text{J/m}}^{3}
  2. B35.2×10−10J/m335.2 \times {10}^{- 10} {\text{J/m}}^{3}
  3. C35.2×10−11J/m335.2 \times {10}^{- 11} {\text{J/m}}^{3}
  4. D35.2×10−13J/m335.2 \times {10}^{- 13} {\text{J/m}}^{3}

Correct answer

A. 35.2×10−12J/m335.2 \times {10}^{- 12} {\text{J/m}}^{3}

Step-by-step solution

f=1014Hzf = 1 {0}^{1 4} \text{Hz}
E0=4V/m{E}_{0} = 4\text{V/m}
∈0=8.8×10−12C2N−m2{∈}_{0} = \text{8.8} \times 1 {0}^{- 1 2} \dfrac{C^{2}}{\text{N} - m^{2}}
Energy density of electric field
= 12\dfrac{1}{2} (Total energy density)
=12⋅12∈0E2= \dfrac{1}{2} \cdot \dfrac{1}{2} {∈}_{0} E^{2}
=12.12{8.8×10−12×42}= \dfrac{1}{2} \text{.} \dfrac{1}{2} \left\{\text{8.8} \times 1 {0}^{- 1 2} \times {4}^{2}\right\}
=12⋅12×16×8.8×10−12J/m3= \dfrac{1}{2} \cdot \dfrac{1}{2} \times 16 \times 8.8 \times {10}^{- 12} {\text{J/m}}^{3}
=35.2×10−12J/m3= 35.2 \times {10}^{- 12} {\text{J/m}}^{3}
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