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Energy density or energy in a volume from the electric field amplitude

This type of question, asked 4 times: 2021 Paper 22, 11 Apr 2023, 31 Jan 2024, 28 Jan 2025. All 4 questions below, oldest first.

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  1. Question 1

    2021 Paper 22

    28
    The electric field in an electromagnetic wave is given by E=(50NC−1)sin⁡ω(t−x/c)E =\left( 50NC^{- 1} \right)\sin\omega(t - x/c) The energy contained in a cylinder of volume VV is 5.5×10−12 J5.5 \times10^{- 12}\text{ }J. The value of VV is ___ cm3cm^{3}. (given ∈0=8.8×10−12C2 N−1 m−2\in_{0}= 8.8 \times10^{- 12}C^{2}{\text{ }N}^{- 1}{\text{ }m}^{- 2} )

  2. Question 2

    11 Apr 2023

    1
    The electric field in an electromagnetic wave is given as E→=20sin⁡ω(t−xc)j→NC−1\overrightarrow{E}= 20\sin\omega\left( t -\frac{x}{c} \right)\overrightarrow{j}NC^{- 1} Where ω\omega and cc are angular frequency and velocity of electromagnetic wave respectively. The energy contained in a volume of 5×10−4 m35 \times10^{- 4}{\text{ }m}^{3} will be (Given ε0=8.85×10−12C2/Nm2\varepsilon_{0}= 8.85 \times10^{- 12}C^{2}/Nm^{2} )

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  3. Question 3

    31 Jan 2024

    10
    In a plane EM wave, the electric field oscillates sinusoidally at a frequency of 5×1010 Hz5 \times10^{10}\text{ }Hz and an amplitude of 50Vm−150Vm^{- 1}. The total average energy density of the electromagnetic field of the wave is : [Use ε0=8.85×10−12C2/Nm2\varepsilon_{0}= 8.85 \times10^{- 12}C^{2}/Nm^{2} ]

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  4. Question 4

    28 Jan 2025

    16
    Due to presence of an em-wave whose electric component is given by E=100sin⁡(ωt−kx)NC−1E = 100\sin(\omega t - kx)NC^{- 1}, a cylinder of length 200 cm holds certain amount of em-energy inside it. If another cylinder of same length but half diameter than previous one holds same amount of em-energy, the magnitude of the electric field of the corresponding em-wave should be modified as

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