Fall 2011 Homework 4 Solutions

# Fall 2011 Homework 4 Solutions - Physics 21 Fall 2011...

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Unformatted text preview: Physics 21 Fall, 2011 Solution to HW-4 22-7 The electric field due to an infinite line of charge is perpendicular to the line and has magnitude E = λ/ 2 π r . Consider an imaginary cylinder with a radius of r = 0 . 200 m and length l = 0 . 465 m that has an infinite line of positive charge running along its axis. The charge per unit length on the line is λ = 7 . 15 µ C / m. (a) What is the electric ﬂux through the cylinder due to this infinite line of charge? (b) What is the ﬂux through the cylinder if its radius is increased to r = 0 . 515 m? c) What is the ﬂux through the cylinder if its length is increased to l = 0 . 760 m? (a) Because the electric field lines are perpendicular to the line of charge, they will also be perpendicular to the surface of the curved or “barrel” part of the cylinder, and parallel to the ends of the cylinder. Because the line of charge is concentric with the cylinder, the field strength will also be constant along the barrel part of the cylinder. Then the electric ﬂux can just be calculated by: Φ = EA barrel = λ 2 π r 2 πrl = 375 , 700 Nm 2 / C Note that the r ’s cancel; the ﬂux is independent of the ra- dius of the cylinder. We can also use Gauss’ law to do the calculation and obtain the same expression. Φ = Q enc = λl = 375 , 700 Nm 2 / C (b) We already found in (a) that the ﬂux is independent of r ; Gauss’ law also tells us that the ﬂux through the cylinder only depends on the charge enclosed. Because increasing the radius does not change how much charge is enclosed, the...
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Fall 2011 Homework 4 Solutions - Physics 21 Fall 2011...

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