In preparation for an experiment that you will do in your in

In preparation for an experiment that you will do in your introductory nuclear physics lab, you are shown the inside of a Geiger tube. You measure the radius and the length of the central wire of the Geiger tube to be 2.0 Times 10^-4 m and 1.2 x 10^-1 m, respectively. The outer surface of the tube is a conducting cylindrical shell that has an inner radius of 1.5 Times 10^-2 m. The shell is coaxial with the wire and has the same length (0.12 m). Calculate the capacitance of your tube, assuming that the gas in the tube has a dielectric constant of 1.00. Calculate the value of the linear charge density on the wire when the potential difference between the wire and shell is of 1.20 kV. A 1.9 pF capacitor is charged to a potential difference of 11.0 V. The wires connecting the capacitor to the battery are then disconnected from the battery and connected across a second, initially uncharged, capacitor. The potential difference across the 1.9 mu F capacitor then drops to 3 V. What is the capacitance of the second capacitor? The positively charged plate of a parallel-plate capacitor has a charge equal to Q. When the space between the plates is evacuated of air, the electric field strength between the plates is 2.8 Times 10^5 V/m. When the space is filled with a certain dielectric material, the field strength between the plates is reduced to 7.0 Times 10^4 V/m. What is the dielectric constant of the material? If Q = 13 nC, what is the area of the plates? What is the total induced bound charge on either face of the dielectric material?

Solution

Capacitance for cylinderical capacitor:

C = 2 pi e0 L / ln(Ro / Ri)

C= (2 x pi x 8.854 x 10^-12 x 0.12) / ln[(1.5 x 10^-2)/(2 x 10^-4)]

C = 1.546 x 10^-12 F ..........Ans (a)

(B) Q = C V = (1.546 x 10^-12 ) (1.20 x 10^3) = 1.855 x 10^-9 C

linear charge density = Q/L = 1.546 x 10^-8 C/m