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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)?

Question

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Solution

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 I of an electromagnetic wave is given by:

I=12ϵ0cE2 I = \frac{1}{2} \epsilon_0 c E^2

where:

  • ϵ0 \epsilon_0 is the permittivity of free space (ϵ08.85×1012F/m \epsilon_0 \approx 8.85 \times 10^{-12} \, \text{F/m} ),
  • c c is the speed of light in a vacuum (c3×108m/s c \approx 3 \times 10^8 \, \text{m/s} ),
  • E E is the electric field strength.

Given that the electric field strength required for dielectric breakdown of air is E=3MV/m=3×106V/m 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:

E2=(3×106V/m)2=9×1012V2/m2 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=12×8.85×1012F/m×3×108m/s×9×1012V2/m2 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=12×8.85×1012×3×108×9×1012 I = \frac{1}{2} \times 8.85 \times 10^{-12} \times 3 \times 10^8 \times 9 \times 10^{12}

I=12×8.85×3×9×108 I = \frac{1}{2} \times 8.85 \times 3 \times 9 \times 10^8

I=12×239.85×108 I = \frac{1}{2} \times 239.85 \times 10^8

I=119.925×108 I = 119.925 \times 10^8

I=1.19925×1010W/m2 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 1.2×1010W/m2 1.2 \times 10^{10} \, \text{W/m}^2 .

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