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:
Pick an option to check your answer.
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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