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JEE Mains · Physics · STD 12 - Electromagnetic waves
  • JEE Mains 2014
  • Medium
  • MCQ

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:

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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 \(u_{E}=\frac{1}{4} \varepsilon_{0} E^{2}\)

\(\Rightarrow u_{E}=\frac{1}{4} \times 8.8 \times 10^{-12} \times(4)^{2}\)

\(=35.2 \times 10^{-12} \,\mathrm{J} / \mathrm{m}^{3}\)

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