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JEE Main 202311 AprilMorning Shift
Exam
JEE Main
Year
2023
Date
11 April
Shift
Morning Shift
Subject
Physics
Chapter
Capacitance

JEE Main 2023 · 11 April · Morning Shift · Physics · Capacitance

The electric field in an electromagnetic wave is given as → E =...

The electric field in an electromagnetic wave is given as
E→=20sinω(t−xc)j→NC−1\overset{\to}{E} = 20sin\omega\left(t - \dfrac{x}{c}\right)\overset{\to}{j}N{C}^{- 1}, where ω\omega and cc are angular frequency and velocity of electromagnetic wave respectively. The energy contained in a volume of 5×10−4m35 \times {10}^{- 4}{m}^{3} will be
(Given ε0=8.85×10−12C2N−1m−2{ε}_{0} = 8 . 85 \times {10}^{- 12}{C}^{2}{N}^{- 1}{m}^{- 2})

Options

  1. A88.5×10−13J88 . 5 \times {10}^{- 13}J
  2. B17.7×10−13J17 . 7 \times {10}^{- 13}J
  3. C28.5×10−13J28 . 5 \times {10}^{- 13}J
  4. D8.85×10−13J8 . 85 \times {10}^{- 13}J

Correct answer

D. 8.85×10−13J8 . 85 \times {10}^{- 13}J

Step-by-step solution

Given that, E→=20sinω(t−xc)j→NC−1\overset{\to}{E} = 20sin\omega\left(t - \dfrac{x}{c}\right)\overset{\to}{j}N{C}^{- 1}

The average energy density of an electromagnetic wave is 12ε0E20\dfrac{1}{2}{ε}_{0}{E}^{2}0.

The total energy stored is

Etotal=12ε0E02×volume{E}_{\text{total}} = \dfrac{1}{2}{ε}_{0}{E}_{0}^{2} \times volume

⇒Etotal=12×8.85×10−12×(20)2×5×10−4=8.85×10−13J⇒ {E}_{total} = \dfrac{1}{2} \times 8 . 85 \times {10}^{- 12} \times ( 20{)}^{2} \times 5 \times {10}^{- 4} = 8 . 85 \times {10}^{- 13}J

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