Radiation detector - cylindrical ionising chamber

[i_hate_math]

Homework Statement

A cylindrical detector has a central anode and a cathode with a radius of 36.8 μm and 30 mm respectively. At what applied voltage will an electron which is at 6 mm from the centre get enough energy to ionise helium gas (Ie » 23 eV) when traveling a mean free path length of 1 mm.

Relevant Equations

E(r) = Q/(2πϵLr) for a line of charge where Q, L are the total charge and length enclosed by the chosen gaussian surface
C = Q/V
Energy = q*V

Let r = position of the electron = 6mm - 36.8μm; λ = mean free path traversed.

Integrate E(r) = Q/(2πϵLr) between the two shells gives:
V = [Q/(2πϵL)]*log(r/(r-λ))

I know that the question is asking for the voltage at which the electron energy will get to 23eV, but i am unsure how to get rid of L or Q in the expression for V ?

Thanks

 

[BvU]

You haven't made use of the wire radius yet. Where is the charge and what does it mean for V on the outside of the wire ?

Edit: nor of the 30 mm... where V = ...

 

[i_hate_math]

You haven't made use of the wire radius yet. Where is the charge and what does it mean for V on the outside of the wire ?

Edit: nor of the 30 mm... where V = ...

The potential at the surface should be zero outside the surface since the total enclosed charge would be zero. Between the cathode and anode (outside the wire, inside the shell), the potential difference is:
V = [Q/(2πϵL)]*log(b/a)
where b=30mm and a=36.8μm

=> V/log(b/a) = [Q/(2πϵL)]

So perhaps i could substitute this into
ΔΦ = [Q/(2πϵL)]*log(r/(r-λ))
to get ΔΦ = V* [log(r/(r-λ)) / log(b/a)]
and solve for e*ΔΦ = 23 eV ?

 

[BvU]

Good plan ... !

 

[i_hate_math]

Good plan ... !

Thanks heaps!