An electromagnetic plane wave in air has an electric field amplitude of $3.0\times 10^5{\text{V m}}^{-1}$. What is the magnitude of the time‑averaged Poynting vector of this wave?

Step-by-step solution

1. For a plane wave in free space, the time‑averaged Poynting vector magnitude is $S_{\text{avg}}=\frac{E_0^2}{2c}$.
2. Compute $E_0^2=(3.0\times 10^5{)}^2=9.0\times 10^{10}{\text{(V m}}^{-1}{\text{)}}^2$.
3. Divide by $2c$: $S_{\text{avg}}=\frac{9.0\times 10^{10}}{2\times 3.00\times 10^8}=1.5\times 10^2{\text{W m}}^{-2}$.

Concept tested: Time‑averaged Poynting vector in free space

Published 11 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked