QB_PH_064: Poynting vector

In free space, a plane wave has electric field amplitude $E_0 = 30\text{ V/m}$. What is the time-averaged magnitude of the Poynting vector in $\text{W/m}^2$ (use $c = 3 \times 10^8\text{ m/s}, \epsilon_0 = 8.854 \times 10^{-12}\text{ F/m}$)?

Correct Answer

### Solution: $\langle S \rangle = \frac{1}{2} c \epsilon_0 E_0^2 = 0.5 \times 3 \times 10^8 \times 8.854 \times 10^{-12} \times 900 \approx 1.195\text{ W/m}^2$.