A long cylindrical charge distribution of radius 2.0 cm has a surface charge density σ = 18.0 [nC/m2] What is the approximate strength of the E-field a distance R = 12.0 [cm] from the central axis of the cylinder? Question 5 options:
|E|= 2.7 x 10+10 [N/C]
|E|= 27 [N/C]
|E|= 1.8 x 10+12 [N/C]
|E|= 2700 [N/C]
|E|= 340 [N/C]
|E| = 340 N/c
let r = 2 cm
= 0.02 m
R = 12 cm
= 0.12 m
Image a cyllinder with length L and radius around the actual
cyllinder.
use Gauss'a law
integral E.dS = Qin/epsilon
E*2*pi*R*L = sigma*(2*pi*r*L)/epsilon
E = sigma*r/(epsilon*R)
= 18*10^-9*0.02/(8.854*10^-12*0.12)
= 339 N/c (approximately 340 N/c)
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