Optics Question 314
Question: A radio receiver antenna that is 2 m long is oriented along the direction of the electromagnetic wave and receives a signal of intensity $ 5\times {{10}^{-16}}W/m^{2} $ . The maximum instantaneous potential difference across the two ends of the antenna is
Options:
A) 1.23 mV
B) 1.23 mV
C) 1.23 V
D) 12.3 mV
Show Answer
Answer:
Correct Answer: A
Solution:
$ I=\frac{1}{2}{\varepsilon _{0}}CE _{0}^{2} $
Therefore $ E _{0}=\sqrt{\frac{2I}{{\varepsilon _{0}}C}}=\sqrt{\frac{2\times 5\times {{10}^{-16}}}{8.85}}=0.61\times {{10}^{-6}}\frac{V}{m} $ Also $ E _{0}=\frac{V _{0}}{d} $
Therefore $ V _{0}=E _{0}d=0.61\times {{10}^{-6}}\times 2=1.23\mu V $
