What would be the average intensity of a laser beam so strong that its electric field produced dielectric breakdown of air (which requires E, = 3 MV/m)?

To find the average intensity of a laser beam that produces dielectric breakdown of air, we need to use the relationship between the electric field and the intensity of an electromagnetic wave. The intensity $I$ of an electromagnetic wave is given by:

$I = \frac{1}{2} \epsilon_0 c E^2$

where:

• $\epsilon_0$ is the permittivity of free space ($\epsilon_0 \approx 8.85 \times 10^{-12} \, \text{F/m}$),
• $c$ is the speed of light in a vacuum ($c \approx 3 \times 10^8 \, \text{m/s}$),
• $E$ is the electric field strength.

Given that the electric field strength required for dielectric breakdown of air is $E = 3 \, \text{MV/m} = 3 \times 10^6 \, \text{V/m}$, we can substitute these values into the formula.

First, square the electric field strength:

$E^2 = (3 \times 10^6 \, \text{V/m})^2 = 9 \times 10^{12} \, \text{V}^2/\text{m}^2$

Next, multiply by the permittivity of free space and the speed of light:

$I = \frac{1}{2} \times 8.85 \times 10^{-12} \, \text{F/m} \times 3 \times 10^8 \, \text{m/s} \times 9 \times 10^{12} \, \text{V}^2/\text{m}^2$

Now, perform the multiplication:

$I = \frac{1}{2} \times 8.85 \times 10^{-12} \times 3 \times 10^8 \times 9 \times 10^{12}$

$I = \frac{1}{2} \times 8.85 \times 3 \times 9 \times 10^8$

$I = \frac{1}{2} \times 239.85 \times 10^8$

$I = 119.925 \times 10^8$

$I = 1.19925 \times 10^{10} \, \text{W/m}^2$

Therefore, the average intensity of the laser beam required to produce dielectric breakdown of air is approximately $1.2 \times 10^{10} \, \text{W/m}^2$.