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  • In an electromagnetic wave, the amplitude of electric field is <img alt="1 \mathrm{~V} / \mathrm{m}" src="https://entrancecorner.oncodecogs.com/gif.latex?1%20%5Cmathrm%7B%7EV%7D%20/%20%5Cmathrm%7Bm

In an electromagnetic wave, the amplitude of electric field is 1 \mathrm{~V} / \mathrm{m}. The frequency of wave is 5 \times 10^{-14} \mathrm{~Hz}. The wave is propagating along \mathrm{z}-axis. The average energy density of electric field, in \text{Joule}/ \mathrm{m}^3 , will be:
 

Option: 1

1.1 \times 10^{-11}

 


Option: 2

2.2 \times 10^{-12}
 


Option: 3

3.3 \times 10^{-13}
 


Option: 4

4.4 \times 10^{-14}


Answers (1)

best_answer

Average energy density is given by

\mathrm{U}_{\mathrm{E}}=\frac{1}{2} \varepsilon_0 \mathrm{E}^2=\frac{1}{2} \varepsilon_0\left(\frac{\mathrm{E}_0}{\sqrt{2}}\right)^2=\frac{1}{4} \varepsilon_0 \mathrm{E}_0^2

\mathrm{ =\frac{1}{4} \times 0.85 \times 10^{-12} \times(1)^2 }

\mathrm{ =2.2 \times 10^{-12} \mathrm{~J} / \mathrm{m}^3 }

Hence option 2 is correct.





 

Posted by

Riya

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