In free space the electric‑field amplitude of a plane electromagnetic wave is $E_0=5.0\times 10^4{\text{V m}}^{-1}$. Find the time‑averaged magnitude of the Poynting vector (i.e., the intensity $I$) of the wave.

Step-by-step solution

1. The time‑averaged intensity of an EM wave in vacuum is $I=\frac{1}{2}{\epsilon }_{0}c{E}_0^2$, where ${\epsilon }_0=8.85\times 10^{-12}{\text{F m}}^{-1}$ and $c=3.00\times 10^8{\text{m s}}^{-1}$.
2. Compute $E_0^2=(5.0\times 10^4{)}^2=2.5\times 10^9{\text{V}}^2{\text{m}}^{-2}$.
3. Now ${\epsilon }_{0}c=8.85\times 10^{-12}\times 3.00\times 10^8=2.655\times 10^{-3}{\text{A s V}}^{-1}{\text{m}}^{-1}$ (units simplify to ${\text{C}}^2{\text{N}}^{-1}{\text{m}}^{-2}$, giving W/m² after multiplication).
4. Plug into the formula: $I=\frac{1}{2}\times 2.655\times 10^{-3}\times 2.5\times 10^9$.
5. Calculate: $\frac{1}{2}\times 2.655\times 10^{-3}=1.3275\times 10^{-3}$; then $1.3275\times 10^{-3}\times 2.5\times 10^9=3.31875\times 10^6$? Wait, re‑evaluate: $2.5\times 10^9\times 1.3275\times 10^{-3}=2.5\times 1.3275\times 10^6=3.31875\times 10^6$ W/m². This is too large because a factor of $10^{-3}$ was missed. Correct calculation: ${\epsilon }_{0}c=2.655\times 10^{-3}$, multiply by $E_0^2=2.5\times 10^9$ gives $6.6375\times 10^6$, then half gives $3.3188\times 10^6$ W/m². However the standard result for $E_0=5\times 10^4$ V/m is $I=\frac{E_0^2}{2Z_0}$ where $Z_0=377\Omega$. Using that: $I=\frac{(5\times 10^4{)}^2}{2\times 377}=\frac{2.5\times 10^9}{754}=3.31\times 10^6{\text{W m}}^{-2}$. The earlier expression with ${\epsilon }_{0}c$ yields the same value. So the correct intensity is $I=3.33\times 10^6{\text{W m}}^{-2}$.
6. Round to three significant figures: $I=3.33\times 10^6{\text{W m}}^{-2}$.

Concept tested: intensity and Poynting vector of EM wave

Published 20 September 2026 · How this solution was prepared and checked