Chapter 22, Problem 51GP

To determine

To Show: The peak electric field in the wave is equal to the $\sqrt{\frac{{\mu }_{0}c{P}_0}{2\pi r^2}}$ .

Explanation of Solution

Given data:

Average power rate $=P_0$

Frequency $=f$

Formula used:

Intensity of wave:

  $S=\frac{P_0}{A}$

Where,

  $P_o=$ Power

  $A=$ Area

Calculation:

Now, intensity of wave:

  $\begin{array}{l}S=\frac{P_0}{A}\\ S=\frac{P_0}{4\pi r^2}\end{array}$

Now, intensity of wave in terms of electric field

  $\begin{array}{l}\frac{1}{2}c{\epsilon }_0{E}_0^2=\frac{P_0}{4\pi r^2}\\ \frac{1}{2}c\left(\frac{1}{c^2{\mu }_0}\right)E_0{}^2=\frac{P_0}{4\pi r^2}\\ E_0{}^2=\frac{2c^2{\mu }_0{P}_0}{4\pi r^{2}c}\\ E_0{}^2=\frac{c{\mu }_0{P}_0}{2\pi r^2}\\ E_0=\sqrt{\frac{c{\mu }_0{P}_0}{2\pi r^2}}\end{array}$

Conclusion: Hence, the peak electric field is equal to the $\sqrt{\frac{{\mu }_{0}c{P}_0}{2\pi r^2}}$ .