Answer:
a) intensity [tex]= 2.19 \times 10^{-6} W/m^2[/tex]
b) E = 0.0406 V/m
Explanation:
given data:
Power = 30 kW
R = 33×10^3 M
1) Intensity is given as
[tex] Intensity I = \frac{P}{4\times \pi \times r^2}[/tex]
[tex]= \frac{30 \times 10^3}{4\times 3.14 \times (33\times10^3)^2}[/tex]
[tex]= 2.19 \times 10^{-6} W/m^2[/tex]
2) electric field is given as
[tex]I = 1/2 \times \epsilon \times c \times E^2[/tex]
Solving for E
[tex]E = \sqrt{\frac{2I}{\epsilon c}}[/tex]
[tex]E =\sqrt{\frac{2\times 2.19 \times 10^{-6}}{8.85\times 10^{-12} 3\times 10^8}}[/tex]
E = 0.0406 V/m
