Chapter 22, Problem 20P

To determine

The energy is transported a across area per hour by an EM wave.

Answer to Problem 20P

  $\text{1}{\text{.42 × 10}}^{\text{-6}}\text{J/h}\text{.}$

Explanation of Solution

Given info:

Electric field, $\text{38}\text{.6}\text{mV/m}$

Formula used:

  ${\text{S = cε}}_{\text{0}}{\text{E}}^{\text{2}}{}_{\text{rms}}$

  $\text{=}\frac{\text{ΔU}}{\text{AΔt}}$

Where, $\text{ΔU}$ is the energy, $\text{A}$ is the area, $\text{Δt}$ is the time, $\text{c}$ is the speed of light, and ${\text{ε}}_{\text{0}}$ is the constant.

  $\frac{\text{ΔU}}{\text{Δt}}{\text{=cε}}_{\text{0}}{\text{E}}^{\text{2}}{}_{\text{rms}}\text{A}$

Calculation:

Substituting the given values

  $\frac{\text{ΔU}}{\text{Δt}}{\text{=cε}}_{\text{0}}{\text{E}}^{\text{2}}{}_{\text{rms}}\text{A}$

  = ${\text{(3 × 10}}^{\text{8}}\text{m/s) (8}{\text{.85 × 10}}^{\text{-12}}{\text{C}}^{\text{2}}{\text{/N · m}}^{\text{2}}\text{)(0}{\text{.0386V /m)}}^{\text{2}}{\text{(1 × 10}}^{\text{-4}}{\text{m}}^{\text{2}}\text{)(3600s/h)}$

  $\text{=1}{\text{.42 × 10}}^{\text{-6}}\text{J/h}\text{.}$

Conclusion:

The amount of energy transported $\text{=1}{\text{.42 × 10}}^{\text{-6}}\text{J/h}\text{.}$