AP EAMCET · PHYSICS · Electromagnetic Waves
An electromagnetic wave of frequency \(\mathrm{I} \times 10^{14} \mathrm{~Hz}\) is propagating along \(\mathrm{z}\)-axis. The amplitude of electric field is \(4 \mathrm{Vm}^{-1}\), then energy density of the electric field will be (Permittivity of free space \(=8.8 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}\) )
- A \(35.2 \times 10^{-13} \mathrm{Jm}^{-3}\)
- B \(70.4 \times 10^{-13} \mathrm{Jm}^{-3}\)
- C \(70.4 \times 10^{-12} \mathrm{Jm}^{-3}\)
- D \(352 \times 10^{-12} \mathrm{Jm}^{-3}\)
Answer & Solution
Correct Answer
(C) \(70.4 \times 10^{-12} \mathrm{Jm}^{-3}\)
Step-by-step Solution
Detailed explanation
Given, electromagnetic wave frequency, \(f_m=1.0 \times 10^{14} \mathrm{~Hz}\) amplitude of the electric field, \(E_0=4 \mathrm{Vm}^{-1}\) permittivity of free space, \(\varepsilon_0=8.8 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}\) The value of energy density…
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