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Unformatted text preview: Solution for Homework 9 Computing Potential Difference Solution to Homework Problem 9.1(Which Direction Potential Increases?) Problem: The potential increases in an electric field in which of the following directions? Select One of the Following: (a) Perpendicular to a field line (b) In the same direction as a field line (c-Answer) Opposite the direction of a field line (d) At the angle 42 . 35 ◦ Solution The potential increases in an electric field in the opposite direction of a field line. Total Points for Problem: 2 Points Solution to Homework Problem 9.2(Electric Field and Potential Between Two Charged Plates) Problem: The two conducting plates have net charges of equal magnitude and opposite charge distributed evenly across their sur- faces. Which of the following is true? Select One of the Following: (a) At point 1 the electric potential is lower than it is at point 2 . (b) The electric potential is the same at points 1 and 2 . (c) The magnitude of the electric field is the same at points 1 and 2 . (d) A positive point charge, when released from rest at point 2 , will move at a constant velocity in the direction of point 1 . (e-Answer) Both (a) and (c) are correct. + + + + + _ _ _ _ _ 1 2 Solution (a) The electric field of equal and opposite planes of charge is uniform between the charges and points from the positive plate to the negative plate, so answer (c) is correct. (b) The electric field points to lower potential, so point 1 has lower potential than point 2 . Therefore, answer (a) is also correct. The correct answer to the question is then (e). Total Points for Problem: 3 Points 1 Solution to Homework Problem 9.3(Finite Line Charge along y-Axis) Problem: The figure to the right shows a finite line charge with uniform linear charge density λ . The line charge occupies the y axis from to L . Write the integral that you would evaluate to compute the electric potential at a point P a distance R from the line charge along the x axis. Let the potential at ∞ be zero. Select One of the Following: (a) integraltext L kλdy √ y (b) integraltext L R kλdy √ y 2 (c) integraltext L kλdy √ y 2 (d-Answer) integraltext L kλdy √ R 2 + y 2 (e) integraltext L − L kλdy √ R 2 + y 2 L P x y R Solution (a) Chop the system up into bits.: Divide the charge into a set of segments. The length of the i th segment is Δ y and therefore the charge of the i th segment is q i = λ Δ y . The center of the i th segment is (0 ,y i , 0) . The distance of this point to the point P located at ( R, , 0) is d i = radicalBig R 2 + y 2 i L P x y R y i...
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This note was uploaded on 05/04/2008 for the course PHYS 2074 taught by Professor Stewart during the Winter '08 term at Arkansas.
- Winter '08