Chapter 22, Problem 50GP

To determine

The distance where we can able to hear the transmission.

Answer to Problem 50GP

The answer is $61km$ .

Explanation of Solution

Given data:

Electric field strength, $P=25\times {10}^3\text{W}$

Speed of the light, $c=3\times {10}^8\text{m/s}$

Permittivity of vacuum, ${\epsilon }_0=8.85\times {10}^{-12}{\text{C}}^{\text{2}}\text{/N}{\text{.m}}^{\text{2}}$

The electric field amplitude, ${\text{E}}_0=\text{0}\text{.020}\text{V/m}$

Formula used:

It is known that: $\overline{I}=\left(\frac{1}{2}\right){\epsilon }_{0}c{E}_0^2$

Where,

  $c$ is speed of the light.

  $\overline{I}$ is average intensity.

  $E_0$ is the electric field amplitude.

  ${\epsilon }_0$ is the permittivity of vacuum.

Calculation:

Substitute the given data into the formula $\overline{I}=\left(\frac{1}{2}\right){\epsilon }_{0}c{E}_0^2$

  $\overline{I}=\left(\frac{1}{2}\right){\epsilon }_{0}c{E}_0^2=\frac{1}{2}\left(8.85\times {10}^{-12}{\text{C}}^{\text{2}}\text{/N}{\text{.m}}^{\text{2}}\right)\times 3\times {10}^8\text{m/s}\times {\left(\text{0}\text{.020}\text{V/m}\right)}^2=5.31\times {10}^{-7}W/m^2$

  $\therefore r=\sqrt{\frac{5.31\times {10}^{-7}W/m^2}{4\pi \times 25\times {10}^3\text{W}}}=61km$

Conclusion: The receiver should be within the radius of $61km$ to receive the transmission.