A hole of radius R 1 = 0.6 m is drilled alongthe symmetry axis of an infinite, n

Question

A hole of radius R1 = 0.6 m is drilled alongthe symmetry axis of an infinite, non-conducting solid cylinder ofradius R2 = 2.7 m. The resulting annularcylinder is given a uniform volume charge density r = +1 µC/m3.

(a) Calculate the magnitude of the electric fieldE at the following values of r, wherer is the radial distance measured perpendicularly from theaxis of symmetry (z-axis).

At r = 0.3 m: E = N/C    *
0   OK

At r = 0.9 m: E = N/C    *
2.84e4   OK

At r = 5.4 m: E = N/C     *
72490   OK

HELP:  Rather than approaching thisproblem as a "standard" Gauss's Law problem with a "thick"cylindrical shell of charge, it is helpful to view it as asuperposition of two cylindrical charge distributions. Do you seehow? (The benefits will become clear in the second part of theproblem.)

Suppose instead the hole was parallel to the z-axisalong the line (x, y) = (0, 0.9) m.

(b) Calculate the resulting x- andy-components of the electric field at the point (x, y, z)= (0, 5.4, 0) m.

Ex = N/C    *
0   OK

Ey = N/C   
0   NO

(c) Calculate also the x- and y-components ofthe electric field at the point (x, y, z) = (0, 0.9, 0) m.

Ex = N/C    *
0   OK

Ey = N/C    

Explanation / Answer

The electric fields in the second part is the superposition offields due to 2 cylindrical charge distributions, one is the solidcylinder with radius R2 = 2.7 m carrying charge density = 1 C/m3, another is also a solid cylinderwith the center at (0, 0.9, 0) and radius R1 = 0.6 mcarrying charge density ' = - = -1C/m3, b) At P(0, 5.4, 0), y = 5.4 m the field due to R2 is, (using Gauss Law, I thinkyou know how to get it form the first part) E1 =R22/(2oy), direction:+y the field due to R1 is E2 ='R12/(2oy'),(direction: +y) where y' = 5.4 - 0.9 = 4.5 m so the net field at P isR22/(2oy) +'R12/(2oy') =R22/(2oy) -R12/(2oy') =3.95*103 N/C, direction: +y this is the y-component c) At Q(0, 0.9, 0), y = 0.9 m, this point is the center of the"hole" By symmetry, now E2 = 0, E1 = y/(2o) so E = E1 = y/(2o) =5.08*104 N/C direction: +y this is the y-component

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