20 Physics II Online Spring17 Activities and Due Dates I HW: Electric Fields and
ID: 1600846 • Letter: 2
Question
20 Physics II Online Spring17 Activities and Due Dates I HW: Electric Fields and Gauss's Law O 3/24/2017 11:55 PM CA 74.9/100 2/28/2017 05:52 PM Gradebook Print Calculator Periodic Table Question 29 of 32 Map A EA Sapling Learning learning A long, conductive cylinder of radius R1 3.45 cm and uniform charge per unit length 604 pC/m is coaxial with a long, cylindrical, non-conducting shell of inner and outer radii R 12.1 cm and R3 13.8 cm, respectively. If the cylindrical shell carries a uniform charge density of p 53.6 pC/m3, find the magnitude of the electric field at the following radial distances from the central axis: Number D 1.24 cm Number R, R NIC 8.45 cm Number NIC 12.2 cm Number NIC 23.8 cm. A Previous Give Up & View solution Check Answer Next Exit HintExplanation / Answer
electric field due to conducting cylinder E = 2*k*lambda/r r > R1
electric field due to non-conducting cylindricla shell
E = (rho/(2*e0))*(r^2-R2^2)/r for R2 < r < R3
E = (rho/(2*e0))*(R3^2-R2^2)/r for r >/= R3
for r = 1.24 cm
r < R1 , R2
E = 0 N/C <<<<---------answer
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for r = 8.45 cm
r > R1 and r < R2
E = 2*k*lambda/r
E = 2*9*10^9*604*10^-12/0.0845
E = 128.66 N/C <<<-------------answer
================
r = 12.2 cm
r > R and R2 <r < R3
E = 2*k*lambda/r + (rho/(2*e0))*(r^2-R2^2)/r
E = (2*9*10^9*604*10^-12/0.122) + ((53.6*10^-12/(2*8.85*10^-12))*((0.122^2-0.121^2)/0.122))
E = 89.12 N/C <<<<--------answer
=====================
r = 23.8 cm
r > R1 and r > R3
E = 2*k*lambda/r + (rho/(2*e0))*(R3^2-R2^2)/r
E = (2*9*10^9*604*10^-12/0.238) + ((53.6*10^-12/(2*8.85*10^-12))*((0.138^2-0.121^2)/0.238))
E = 45.74 N/C <<<----answer
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from gauss law
totla flux = Qinside/e0
-533 = (-(19*10^-9) + (32*10^-9) + q3)/(8.85*10^-12)
q3 = -17.7 nC <<<------answer
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