Question

Calculate the average total energy density of an electromagnetic wave coming from the sun that has the $RMS$ value of electric field as $720 \, N \, C^{- 1}$ .

• A. $3.3\times 10^{- 3} \, J \, m^{- 3}$
• B. $4.58\times 10^{- 6} \, J \, m^{- 3}$
• C. $6.37\times 10^{- 9} \, J \, m^{- 3}$
• D. $81.35\times 10^{- 12} \, J \, m^{- 3}$ .

Answer 1

Answer: (B)

Total average energy density of electromagnetic wave is
$ \begin{array}{l} U_{ av }=\frac{1}{2} \varepsilon_0 E_{ rms }^2+\frac{1}{2 \mu_0} B_{ rms }^2 \\ =\frac{1}{2} \varepsilon_0 E_{ rms }^2+\frac{1}{2 \mu_0}\left(\frac{E_{ rms }^2}{c^2}\right) \left(\because B_{ rms }=\frac{E_{ rms }}{c}\right) \\ =\frac{1}{2} \varepsilon_0 E_{ rms }^2+\frac{1}{2 \mu_0} E_{ rms }^2 \varepsilon_0 \mu_0 \left(\because c=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}\right) \\ =\frac{1}{2} \varepsilon_0 E_{ rms }^2+\frac{1}{2} \varepsilon_0 E_{ rms }^2=\varepsilon_0 E_{ rms }^2 \left(\because 0^{-12} \times(720)^2=4.58 \times 10^{-6} J m \right. \\ =8.85 \times 10^{-3} \end{array} $