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    

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   

(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 first part of this problem (i.e. the first three answers)is simple. If you got the second answer, you should be able to getthe third. Just calculate the linear charge density of the cylinderusing the geometry and the given volume charge density. . in other words...    charge perlength = charge per volume * cross section area = 1 C/m3 * (R2 -r2) = .                         = (2.72 - 0.62) =    21.7713 uC/ meter .     Then the E field is just... .     E = 2k / r  = 2 * 8.99 x 109 * 21.7713 x 10-6 /5.4 =    72490N/C    (third answer...) . The next part, where the hole is off center... same idea but alittle trickier. You actually have to consider three "cylinders" ofcharge:    the hole, with radius0.6;     the full cylinder (imagine the holewasn't there), of radius 2.7;    and the actualcylinder, with the actual given charge (i.e. the big cylinder minusthe hole). . Then in each case you simply have to realizethat... .      E field to to actual chargedistribution = E field from full cylinder ofcharge - E field from "hole" of charge .    Since the full cylinder and the hole are symmetryon the y axis, you can calculate the E field from each and taketheir difference to get the E field from the actual chargedistribution. . This is really the same thing you did in the firstpart, but the symmetry was a bit simpler.

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