A traveling sinusoidal electromagnetic wave in vacuum has an electric field amplitude of 83.7 V/m. Find the intensity of this wave and calculate the energy flowing during 15.5 s through an area of 0.0225 m2 that is perpendicular to the wave\'s direction of propagation.

Respuesta :

Answer with Explanation:

We are given that

Electric field,E=83.7V/m

Time,t=15.5 s

Area,A=[tex]0.0225m^2[/tex]

We have to find the intensity of the wave and energy .

Intensity,I=[tex]\frac{1}{2}c\epsilon_0E^2[/tex]

Where [tex]c=3\times 10^8m/s[/tex]

[tex]\epsilon_0=8.85\times 10^{-12}[/tex]

Substitute the values

[tex]I=\frac{1}{2}(3\times 10^8\times 8.85\times 10^{-12}\times (83.7)^2)=9.3W/m^2[/tex]

Energy,[tex]E=IAt[/tex]

Substitute the values

[tex]E=9.3\times 0.0225\times 15.5=3.24 J[/tex]