A plane electromagnetic wave in air has an electric field amplitude of $4.0\times 10^5,{\text{V m}}^{-1}$. What is its time‑averaged intensity in (\text{W m}^{-2})?

Step-by-step solution

1. Step 1: Write the formula for intensity of a plane EM wave: (I = \frac{1}{2}${\epsilon }_{0}c{E}_0^2$).
2. Step 2: Substitute the known values: (${\epsilon }_0$=8.854 $imes10^{-12}$,\text{F m}^{-1}), (c=3.00 $imes10^8$,\text{m s}^{-1}), and ($E_0$=4.0 $imes10^5$,\text{V m}^{-1}).
3. Step 3: Compute the product ($varepsilo{n}_{0}c$ = (8.854 $imes10^{-12}$)(3.00 $imes10^8$)=2.656 $imes10^{-3}$).
4. Step 4: Square the field amplitude: ((4.0 $imes10^5$)^{2}=1.60 $imes10^{11}$).
5. Step 5: Multiply (${\epsilon }_{0}c$) with ($E_0^2$): (2.656 $imes10^{-3}$\times1.60 $imes10^{11}$=4.25 $imes10^8$).
6. Step 6: Divide by 2 to get the intensity: (\frac{4.25 $imes10^8$}{2}=2.12 $imes10^8$).

Concept tested: Intensity of plane EM wave

Published 14 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked