It was shown in Example 21.11 (Section 21.5) in the textbook that the electric f

Question

It was shown in Example 21.11 (Section 21.5) in the textbook that the electric field due to an infinite line of charge is perpendicular to the line and has magnitude E=/20r. Consider an imaginary cylinder with a radius of r = 0.145 m and a length of l = 0.455 m that has an infinite line of positive charge running along its axis. The charge per unit lengthWhat is the flux through the cylinder if its radius is increased to r= 0.540 m ?on the line is = 4.00 C/m . What is the flux through the cylinder if its length is increased to l= 0.760 m ?

Explanation / Answer

Field through flat circle part of cylinder is zero.

field passing through curved surface = (lambda) / (2 pi e0 r)

surface area of curved surafce = (2 pi r L )

and at every point on curved surface, field is perpendiculr to the surface hence parallel
to area vector.
flux = B.A = B A

flux = (lambda x 2pi r L) / (2pi e0 r) = (lambda L) / e0

so flux is independent of radius of cylinder.

so when L = 0.455 m

flux = (4 x 10^-6 x 0.455) / (8.854 x 10^-12) = 205556.81 V m

when L = 0.760 m

flux = (4 x 10^-6 x 0.760) / (8.854 x 10^-12) = 343347.64 V m

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