JEE Mains · Physics · STD 12 - 8. Electromagnetic waves
An electromagnetic wave of frequency \(1\times10^{14}\, hertz\) is propagating along \(z-\) axis. The amplitude of electric field is \(4\, V/m\) . lf \({\varepsilon_0}=\, 8.8\times10^{-12}\, C^2/Nm^2\) , then average energy density of electric field will be:
- A \(35 .2\times10^{-10}\, J/m^3\)
- B \(35 .2\times10^{-11}\, J/m^3\)
- C \(35 .2\times10^{-12}\, J/m^3\)
- D \(35 .2\times10^{-13}\, J/m^3\)
Answer & Solution
Correct Answer
(C) \(35 .2\times10^{-12}\, J/m^3\)
Step-by-step Solution
Detailed explanation
Given: Amplitude of electric field, \(E_{0}=4\,\mathrm{v} / \mathrm{m}\) Absolute per mitivity, \(\varepsilon_{0}=8.8 \times 10^{-12}\, \mathrm{c}^{2}\, / \mathrm{N}-\mathrm{m}^{2}\) Average energy density \(u_{E}=?\) Applying formula, Average energy density…
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