Question

What is the intensity of an electromagnetic wave with a peak electric field strength of 125V/m?

Short answer

The intensity is$20.74\mathrm{W}/{\mathrm{m}}^2$.

Step-by-step solution

Define electromagnetic waves

Electromagnetic waves are waves that are produced by changes in the electric and magnetic fields.

Evaluating the intensity of an electromagnetic wave

If one knows that the energy of every wave is proportional to the square of its amplitude, one may calculate the intensity as follows,

${\mathrm{I}}_{\mathrm{avg}}=\frac{c{\epsilon }_0{E}_0^2}{2}$

Where ${\mathrm{I}}_{\mathrm{avg}}$is average intensity, is light speed,${\mathrm{\epsilon }}_0$is the permittivity of free space, and ${\mathrm{E}}_0$is maximum field strength.

Substitute the values in the above equation,

$$\begin{gathered} {\mathrm{I}}_{\mathrm{avg}}=\frac{\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)\left(8.85\times {10}^{-12}\mathrm{F}/\mathrm{m}\right){\left(125\mathrm{V}/\mathrm{m}\right)}^2}{2} \\ =20.74\mathrm{W}/{\mathrm{m}}^2 \\ \mathrm{Therefore},\mathrm{the}\mathrm{intensity}\mathrm{is}20.74\mathrm{W}/{\mathrm{m}}^2. \end{gathered}$$