see attachment for the given problem

solution for a,
Zo = 138/sqrt(Er) (log10 D/d)
50 = 138/(sqrt(8.85x10^-12) ( log10 N)

10^1.078X10^-6 = N
N = 1.000002482
D = nd
D = 0.1260003128 ''
outside diameter of outside conductor = D + 0.05''

= 0.176 ''

how do you solve for b?

 

Er is not 8.85x10^-12. That is E0, the permittivity of a vacuum (basically the same as for air).
When you get the answer for D = 0.1260003128 '', and d is given as 0.126", you need to apply a sanity test: Is it reasonable that the dielectric thickness (D-d)/2 is 3.12e-7"?


Er is relative permittivity (dielectric constant). Your problem does not state the value for Er.
The relative permittivity of air is 1. Coax dielectrics typically range from Er=1 to around 3 or 4, if memory serves me.

Where did you get 0.025" for the outside conductor thickness?

 

Hello,

Take a look at this :

er: dielectric constant (relative permittivity) of the medium (if you don't know take 2.3, Polyethylene), air = 1

This comes from this page of the EDUCYPEDIA:
http://www.educypedia.be/electronics/transmissionlines.htm

Scroll down to see it.

Bertus