Tipler6 24. P.062. The radius and the length of the central wire in a Geiger tub

Question

Tipler6 24. P.062. The radius and the length of the central wire in a Geiger tube are 0.29 mm and 11 cm, respectively. The outer surface of the tube is a conducting cylindrical shell that has an inner radius of 1.5 cm. The shell is coaxial with the wire and has the same length (11 cm). The tube is filled with a gas that has a dielectric constant of 1.08 and a dielectric strength of 3 x 10 V/m. "What is the maximum potential difference that can be maintained between the wire and shell? kV Somit 3.432 You currently have 2 submissions for this question. Only 10 submission are allowed. You can make 8 more submissions for this question. Your submissions: 3.432 Computed value: 3.432 Submitted: Thursday, April 6 at 1:45 AM 2 What is the maximum charge per unit length on the wire? nCIm Submit You currently have 3 submissions for this question. Only 10 submission are allowed. You can make 7 more submissions for this question. Standard Exercise Tipler6 24. P.062. Standard Exercise Tipler 6 24. P.067. Standard Exercise Tipler6 24.P,088. Standard Exercise Tipler 6 25.P.004. Standard Exercise Tipler 6 25.P.034.

Explanation / Answer

Given Data

r = 0.29 mm = 0.29*10^-3 m

R = 1.5 cm = 1.5*10^-2 m

K = 1/4pi*eo = 1/4p*(8.85*10^-12) = 9*10^9

k = dielectric constant =1.08

E = 3 x 106 V/m.

L = 11 cm = 0.11 m

Solution:-

b)

lambda = charge per unit length

lambda =Q/L = E*kr / 2K

                    = [ (3*10^6)*1.08*(0.29*10^-3) ] / (2*9*10^9)

lambda = 5.22*10^-8 C/m or 52.2 nC/m

a) Electric field E is

E = 2K*lambda / k*r

=>(2K*lambda/k) = Er

=>(2K*lambda/k) = (3*10^6) *(0.29*10^-3) = 870

Maximum Potential Difference is

Vmax = [2K*lamba / k] * ln(R/r)

Vmax = [870] *ln(1.5 *10^-2 /0.29*10^-3)

Vmax = 3433 V or 3.433 KV

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