Calculating Min. Beam Diameter to Avoid Breakdown

[Dassinia]

Hello,

Homework Statement

The laser Nd:YAG can product a light impulse of λ=1064nm that lasts for 10ns and encloses an energy of 0.1 J and propagates as a cylindrical beam of uniform section in air (n=1)
The maximum electric field that can support air before atoms' ionization is ≈30 MV/m otherwise there's breakdown
What is the minimum diameter of the beam to avoid breakdown

Homework Equations

The Attempt at a Solution

I don't know how I can link all these datas and which formulas will help me to

Thanks

 

[GregoryS]

An average flux density of EM wave can be calculated with
$$
I = \frac{1}{2} \sqrt{\frac{ε_0} {μ_0}} E^2,
$$
where $E = 30 MV/m.$

On the other side an energy ($W = 0.1 J$) is
$$
W = I S τ
$$
where S is a beam area and $τ = 10^{-8} s$.

So you can calculate a beam area S and then find a diameter of the beam.

I don't know why data include a wave lenght.

 

[Dassinia]

S is the area of the beam or of a section because if it is the total area it will depend on h as it is a cylinder and we don't have it. Thanks

 

[berkeman]

An average flux density of EM wave can be calculated with
$$
I = \frac{1}{2} \sqrt{\frac{ε_0} {μ_0}} E^2,
$$
where $E = 30 MV/m.$

On the other side an energy ($W = 0.1 J$) is
$$
W = I S τ
$$
where S is a beam area and $τ = 10^{-8} s$.

So you can calculate a beam area S and then find a diameter of the beam.

I don't know why data include a wave lenght.

Gregory -- please be sure to check your PMs. You received a message about how we handle Homework Help here on the PF. Thank you.

 

[berkeman]

S is the area of the beam or of a section because if it is the total area it will depend on h as it is a cylinder and we don't have it. Thanks

You *do* have the height of the cylinder. You are given the duration of the pulse...

 

[Dassinia]

Is it just L=c*t ?

 

[berkeman]

Is it just L=c*t ?

Yep!

So can you show us the complete solution now?

 

[Dassinia]

With
μ0 = 4 π 10^-7 Hm^-1
ε0= 8,85* 10^-12 Fm⁻¹
ε=(ε0/μO)^1/2

W=0.5*ε*E²*t*2π*r*c*t
r=W/(ε*E²*t²*2π*c)

t=10^-8 s
E=30*10^6 V/m
W=0.1 J