logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

An electromagnetic field wave of frequency $\;1 \times 10^{14} \;$ hertz is propagating along z - axis . The amplitude of electric field is 4 V/m .If $\;\in_{0}=8.8 \times 10^{-12}\;C^{2}/N-m^{2}\;$ , then average energy density of electric field will be :

$(a)\;35.2 \times 10^{-10}\;J/m^{3} \qquad(b)\;35.2 \times 10^{-11}\;J/m^{3} \qquad(c)\;35.2 \times 10^{-12}\;J/m^{3} \qquad(d)\;35.2 \times 10^{-13}\;J/m^{3} $

Can you answer this question?
 
 

Please log in or register to answer this question.

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...