Question

The maximum electric field strength in air is $3.0\mathrm{MV}/\mathrm{m}$. Stronger electric fields ionize the air and create a spark. What is the maximum power that can be delivered by a $1.0-\mathrm{cm}$-diameter laser beam propagating through air$?$

Short answer

Maximum power is$3.75\times {10}^6\mathrm{W}$.

Step-by-step solution

Calculation for intensity

Given info:

Electric field is$3\times {10}^6\frac{\mathrm{V}}{\mathrm{m}}$.

Distance is$0.5\times {10}^{-2}\mathrm{m}$

${\text{c=3×10}}^8\frac{\mathrm{m}}{\mathrm{s}}$

Intensity:

$\mathrm{E}={\left(\frac{2\times {\mathrm{I}}_{\mathrm{avg}}}{\mathrm{c}\times {\mathrm{\epsilon }}_{\mathrm{o}}}\right)}^{1/2}$

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

${\mathrm{I}}_{\mathrm{avg}}=\frac{3\times {10}^8{\displaystyle \frac{\mathrm{m}}{\mathrm{s}}}\times 8.85\times {10}^{-12}{\displaystyle \frac{{\mathrm{C}}^2}{{\mathrm{Nm}}^2}}\times {\left(3\times {10}^6\right)}^2{\displaystyle \frac{\mathrm{V}}{\mathrm{m}}}}{2}$

${\mathrm{I}}_{\mathrm{avg}}=119.47\times {10}^8\frac{\mathrm{W}}{{\mathrm{m}}^2}$

Calculation of power

Power:

${\mathrm{I}}_{\mathrm{avg}}=\frac{\mathrm{P}}{\mathrm{A}}$

$\mathrm{P}={\mathrm{I}}_{\mathrm{avg}}\times \mathrm{A}$

$\mathrm{P}=4\mathrm{\pi }{\left(0.5\times {10}^{-2}\right)}^2{\mathrm{m}}^2\times 119.47\times {10}^8\frac{\mathrm{W}}{{\mathrm{m}}^2}$

$\mathrm{P}=3.75\times {10}^6\mathrm{W}$