In preparation for an experiment that you will do in your introductory nuclear p
ID: 1837880 • Letter: I
Question
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?Explanation / Answer
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