Topic: Electronics Engineering
The electric field measured in the far field of an antenna at a distance of 50 m is $1~V/m$. The average power density at a distance of 500 m from the antenna is
- A. $26.6\mu W/m^{2}$
- B. $0.1\mu V/m^{2}$
- C. $10~\mu W/m^{2}$
- D. $13.3~\mu W/m^{2}$
Answer: A. $26.6\mu W/m^{2}$
Explanation: $E=\frac{V}{d}$. $V=1\times50=50~V$. At 500 m, $E=\frac{V}{d}=\frac{50}{500}=0.1~V/m$. $P=\frac{E^{2}}{120\pi}=\frac{(0.1)^{2}}{120\pi}=26.5~\mu W/m^{2}$.