Question

A radio station on the surface of the earth radiates $50kW$. If transmitter radiates equally in all directions above the surface of earth find the amplitude of electric field detected $100km$ away.

• A. $2.45{Vm}^{-1}$
• B. $2.45\times {10}^{-1}{Vm}^{-1}$
• C. $2.45\times {10}^{-2}{Vm}^{-1}$
• D. $2.45\times {10}^{-3}{Vm}^{-1}$

Answer 1

Answer: (D)

$2.45\times {10}^{-3}{Vm}^{-1}$

Power $P\propto$ Intensity, $I\propto$Amplitude,$A^2$
Intensity $I\propto \frac{1}{r^2}$,

Hence, $A\propto \frac{1}{r}$
Hence, $A=\frac{\sqrt{P}}{r}=\frac{\sqrt{50\times {10}^3}}{100\times {10}^3}=2.45\times {10}^{-3}V{m}^{-1}$