can i get help ? A Geiger tube consists of two elements, a long metal cylindrica

Question

can i get help ?

A Geiger tube consists of two elements, a long metal cylindrical shell and a long straight metal wire running down its central axis. Model the tube as if both the wire and cylinder are infinitely long. The central wire is positively charged and the outer cylinder is negatively charged. The potential difference between the wire and the cylinder is 0.98 kV. Suppose the cylinder in the Geiger tube has an inside diameter of 4.04 cm and the wire has a diameter of 0.462 mm. The cylinder is grounded so its potential is equal to zero. What is the radius of the equipotential surface that has a potential equal to 51.5 V Is this surface closer to the wire or to the cylinder The equipotential surface is closer to the How far apart are the equipotential surfaces that have potentials of 185 and 250 V Compare your result in Part (b) to the distance between the two surfaces that have potentials of 710 and 730 V. respectively. What does this comparison tell you about the electric field strength as a function of the distance from the central wire

Explanation / Answer

a) distance betweeen surface of the wire and cyllinder = (4.04/2) - (0.0462/2)

= 1.9969 cm

let at x distance from center of wire potential is 515 volts.

so, at x = 515*1.9969/980

= 1.049 cm

This surface is closer to cyllinder.

b) E = 980/1.9969

= 490.76 V/cm

delta_x = deta_V/E

= (250 - 185)/490.76

= 0.132 cm

= 1.32 mm

c) delta_x = deta_V/E

= (730 - 710)/490.76

= 0.0407 cm

= 0.407 mm

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