Answer:
0.09221 V/m
Explanation:
[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi \times 10^{-7}\ H/m[/tex]
P = Power of transmitter = 250 mW
r = Distance to source = 42 m
c = Speed of light = [tex]3\times 10^8\ m/s[/tex]
[tex]E_0[/tex] = Electric field amplitude
Intensity of a wave is given by
[tex]I=\frac{P}{4\pi r^2}\\\Rightarrow I=\frac{250\times 10^{-3}}{4\pi \times 42^2}\\\Rightarrow I=1.12779\times 10^{-5}\ W/m^2[/tex]
Intensity is also given by
[tex]I=\frac{E_0^2}{2\mu_0 c}\\\Rightarrow E_0=\sqrt{I2\mu_0 c}\\\Rightarrow E_0=\sqrt{1.12779\times 10^{-5}\times 2\times 4\pi \times 10^{-7}\times 3\times 10^8}\\\Rightarrow E_0=0.09221\ V/m[/tex]
The electric field amplitude at the receiver when it first fails is 0.09221 V/m