Answer :
Explanation:
It is given that,
Power of EM waves, P = 1800 W
We need to find the intensity at a distance of 5 m. Also, the rms value of the electric field.
Intensity,
[tex]I=\dfrac{P}{4\pi r^2}\\\\I=\dfrac{1800}{4\pi\times (5)^2}\\\\I=5.72\ W/m^2[/tex]
The formula that is used to find the rms value of the electric field is as follows :
[tex]I=\epsilon_o cE^2_{rms}[/tex]
c is speed of light and [tex]\epsilon_o[/tex] is permittivity of free space
So,
[tex]E_{rms}=\sqrt{\dfrac{I}{\epsilon_o c}}\\\\E_{rms}=\sqrt{\dfrac{5.72}{8.85\times 10^{-12}\times 3\times 10^8}}\\\\E_{rms}=46.41\ V/m[/tex]
Hence, this is the required solution.