Question

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]

Answer 1

|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)