Average power density of a plane wave in free space
BlinkNBuild practice problem · GATE standard, authored and verified in-house
A uniform plane wave propagates in free space with a peak electric field amplitude of $E_0 = 10\ \text{V/m}$. Take the intrinsic impedance of free space as $\eta_0 = 377\ \Omega$.
The time-average power density, in W/m$^2$, is ________.
Show the step-by-step derivation
Answer
0.1326 W/m$^2$ (accept 0.13 to 0.135) W/m$^2$ (accept 0.13 to 0.135)
Step-by-step derivation
- For a uniform plane wave in a lossless medium, $\vec{E}$ and $\vec{H}$ are in phase and perpendicular, related by the intrinsic impedance: $$H_0 = \frac{E_0}{\eta_0}.$$
- The time-average Poynting vector magnitude is $$S_{avg} = \tfrac{1}{2}\,E_0 H_0 = \frac{E_0^2}{2\eta_0},$$ where the factor of $\tfrac{1}{2}$ comes from averaging $\cos^2$ over a cycle.
- Substitute the given values: $$S_{avg} = \frac{10^2}{2 \times 377} = \frac{100}{754}.$$
- Evaluate: $$S_{avg} = \mathbf{0.1326\ W/m^2}.$$
- Cross-check via RMS. The RMS field is $E_{rms} = 10/\sqrt{2} = 7.071$ V/m, and $E_{rms}^2/\eta_0 = 50/377 = 0.1326$ W/m$^2$. Same answer - the $\tfrac{1}{2}$ and the $\sqrt{2}$ are the same fact stated twice. ✓
The trap this question is built around
Omitting the factor of $\tfrac{1}{2}$ and answering $0.265$ W/m$^2$. That is the peak power density, not the average. If the field is given as RMS, the half disappears - so always check which the question specifies.