It was shown in Example 21.11 (Section 21.5) in the textbook that the electric f
ID: 1527560 • Letter: I
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 = 0.110 m and a length of l = 0.440 m that has an infinite line of positive charge running along its axis. The charge per unit length on the line is = 4.65 C/m .
Part A
What is the electric flux through the cylinder due to this infinite line of charge?
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Part B
What is the flux through the cylinder if its radius is increased to r= 0.530 m ?
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Part C
What is the flux through the cylinder if its length is increased to l= 0.940 m ?
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 = 0.110 m and a length of l = 0.440 m that has an infinite line of positive charge running along its axis. The charge per unit length on the line is = 4.65 C/m .
Part A
What is the electric flux through the cylinder due to this infinite line of charge?
= Nm2/CSubmitMy AnswersGive Up
Part B
What is the flux through the cylinder if its radius is increased to r= 0.530 m ?
= Nm2/CSubmitMy AnswersGive Up
Part C
What is the flux through the cylinder if its length is increased to l= 0.940 m ?
Explanation / Answer
(A) From Gauss law,
total flux = Qin / e0
= ( lambda L ) / e0
= (4.65 x 10^-6 x 0.440) / (8.854 x 10^-12)
= 2.31 x 10^5 N m^2 / C
(B) By changing radius, Qin does not change hence flux will remain same.
flux = 2.31 x 10^5 N m^2 / C
(C) now Qin will change.
Qin = 0.940 x 4.65 x 10^-6 = 4.94 x 10^5 N m^2 / C
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