Q. 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
A) $1.23\ \mu V$
B) $1.23\ mV$
C) $1.23\ V$
D) $12.3\ mV$
Explanation
(a) $I = \frac{1}{2}\varepsilon_{0}CE_{0}^{2}$
$\Rightarrow E_{0} = \sqrt{\frac{2I}{\varepsilon_{0}c}} = \sqrt{\frac{2 \times 5 \times 10^{- 16}}{8.85 \times 10^{- 12} \times 3 \times 10^{8}}} = 0.61 \times 10^{- 6}\frac{V}{m}$
Also $E_{0} = \frac{V_{0}}{d} \Rightarrow V_{0} = E_{0}d = 0.61 \times 10^{- 6} \times 2 = 1.23\mu V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.