Question

23.2 energy transport and the poynting vector
a sinusoidal electromagnetic wave has an electric field whose rms value is $100.0\frac{v}{m}$. if we assume that this wave is coming from an isotropic source $1.00\\ m$ away, what is the source power $p_s$?
$\bigcirc 333\\ w/m^2$
$\bigcirc 333\\ w$
$\bigcirc 26.5\\ w$
$\bigcirc 26.5\\ w/m^2$

Explanation:

Step1: Calculate the intensity \(I\)

The formula for the intensity of an electromagnetic wave in terms of the rms electric field \(E_{rms}\) is \(I=\frac{E_{rms}^{2}}{c\mu_{0}}\), where \(c = 3\times10^{8}\space m/s\) (speed of light) and \(\mu_{0}=4\pi\times 10^{-7}\space T\cdot m/A\).
Substitute \(E_{rms} = 100.0\space V/m\), \(c = 3\times10^{8}\space m/s\) and \(\mu_{0}=4\pi\times 10^{-7}\space T\cdot m/A\) into the formula:

$$
LATEXBLOCK0
$$

Step2: Calculate the source power \(P_s\)

For an isotropic source, the intensity \(I=\frac{P_s}{4\pi r^{2}}\), where \(r = 1.00\space m\).
Rearrange the formula to solve for \(P_s\): \(P_s=I\times4\pi r^{2}\)
Substitute \(I = 26.5\space W/m^{2}\) and \(r = 1.00\space m\) into the formula:

$$
LATEXBLOCK1
$$

Answer:

333 W