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 =

At r = 0.9 m: E =

At r = 5.4 m: E =

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 =

Ey =

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

Ex =

Ey =

Explanation / Answer

R1 = 0.6 m, R2 = 2.7 m. = +1 µC/m3. a) use Gauss law, At r = 0.30 m E*2rL = 0, so E = 0 At r = 0.90 m, E*2rL = *(r2 -R12)L/o E = *(r2 -R12)/(2or) =2.82*104 N/C At r = 5.4 m E*2rL = *(R22 -R12)L/o E = *(R22 -R12)/(2or) =7.25*104 N/C b) 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, 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') =7.17*104 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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