A plane electromagnetic wave traveling in vacuum has an electric‑field amplitude $E_0=2.0\times 10^5{\text{V m}}^{-1}$. Calculate the time‑averaged intensity $I$ of the wave.

Step-by-step solution

1. The intensity of an electromagnetic wave is given by $I=\frac{1}{2}c{\epsilon }_0{E}_0^2$, where $c=3.0\times 10^8{\text{m s}}^{-1}$ and ${\epsilon }_0=8.854\times 10^{-12}{\text{F m}}^{-1}$.
2. Calculate $E_0^2=(2.0\times 10^5{)}^2=4.0\times 10^{10}{\text{V}}^2{\text{m}}^{-2}$.
3. Compute ${\epsilon }_0{E}_0^2=8.854\times 10^{-12}\times 4.0\times 10^{10}=0.354{\text{J m}}^{-3}$.
4. Multiply by $c$: $c{\epsilon }_0{E}_0^2=3.0\times 10^8\times 0.354=1.062\times 10^8{\text{W m}}^{-2}$.
5. Finally, $I=\frac{1}{2}\times 1.062\times 10^8=5.31\times 10^7{\text{W m}}^{-2}$, which is written as $5.3\times 10^7{\text{W m}}^{-2}$.

Concept tested: Intensity of an EM wave

Published 17 September 2026 · How this solution was prepared and checked