Chapter 22, Problem 36P

To determine

The produce voltage over the length of the antenna and the rate of energy transport.

Answer to Problem 36P

The required answers are $7.14\times {10}^{-4}V/m$ and $\text{1}{\text{.35×10}}^{\text{-9}}{\text{W/m}}^{\text{2}}$ .

Explanation of Solution

Given data:

Power, $V_{rms}=1.0\times {10}^{-3}V$

Distance, $d=1.60m$

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

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

Formula used:

It is known that: ${\text{E}}_{\text{rms}}\text{=}\frac{{\text{V}}_{\text{rms}}}{\text{d}}$ and $S=c{\epsilon }_0{E}_{rms}{}^2$

Where,

  $\text{S}$ is the rate of energy transport.

  $c$ is speed of the light.

  $E_{rms}$ is root square means value of the electric field.

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

Calculation:

Substitute the given data into the formula ${\text{E}}_{\text{rms}}\text{=}\frac{{\text{V}}_{\text{rms}}}{\text{d}}$ ,

  ${\text{E}}_{\text{rms}}\text{=}\frac{{\text{V}}_{\text{rms}}}{\text{d}}=\frac{1.0\times {10}^{-3}V}{1.60m}=7.14\times {10}^{-4}V/m$

Now, we calculate the rate of energy transport,

  $S=c{\epsilon }_0{E}_{rms}{}^2=\left(3\times {10}^8\text{m/s}\times 8.85\times {10}^{-12}{\text{C}}^{\text{2}}\text{/N}{\text{.m}}^{\text{2}}\right){\left(7.14\times {10}^{-4}V/m\right)}^2$

Simplify further,

  $\text{S}=\text{1}{\text{.35×10}}^{\text{-9}}{\text{W/m}}^{\text{2}}$

Conclusion: The produce voltage over the length of the antenna and the rate of energy transport are $7.14\times {10}^{-4}V/m$ and $\text{1}{\text{.35×10}}^{\text{-9}}{\text{W/m}}^{\text{2}}$ respectively.