Energy Transport and the Poynting Vector — Question 1

Question 1

Peak field and rms field give the same time-averaged energy flux

A sinusoidal vacuum plane wave has electric amplitude $E_0=150$ V/m. Use $\epsilon_0=8.854\times10^{-12}$ F/m and $c=3.00\times10^8$ m/s. A flat detector of area $A=2.0\times10^{-3}$ square meters has its normal 60 degrees from propagation.

1. Find rms electric field, intensity, and mean total energy density.

2. Find the average power crossing the detector and explain which area enters the flux.

Question 1 - Solution

Peak field and rms field give the same time-averaged energy flux

(a) Method and calculation

$$E_{\mathrm{rms}}=E_0/\sqrt2=106.066\,\mathrm{V/m},\quad I=\epsilon_0cE_{\mathrm{rms}}^2=\boxed{29.8823\,\mathrm{W/m^2}}.$$$$\boxed{\langle u\rangle=I/c=9.96075\times10^{-8}\,\mathrm{J/m^3}}.$$

(b) Interpretation and further calculation

$$P=\langle\mathbf S\rangle\cdot\hat n A=IA\cos60^\circ.$$$$\boxed{P=2.98823\times10^{-2}\,\mathrm W};\quad A_\perp=A\cos60^\circ.$$

Physical check

The intensity is defined per area perpendicular to propagation, while the physical detector area is tilted.