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Unformatted text preview: Solution for Homework 3 Beginning Electric Fields Solution to Homework Problem 3.1(Electric Force Between a Van de Graaff and a Pith Ball) Problem: The Van de Graaff develops a charge of  1 10 5 C and a pith ball has charge of 1 10 9 C . Which of the following describes the relationship between the size of the force the Van de Graaff exerts on the pith ball and the size of the force the pith ball exerts on the Van de Graaff? Select One of the Following: (a) The Van de Graaff exerts a much larger force on the pith ball. (b) The pith ball exerts a much larger force on the Van de Graaff (cAnswer) The forces have the same magnitude. Solution Newtons Third Law states that the forces are equal in magnitude and opposite in direction. Total Points for Problem: 3 Points Solution to Homework Problem 3.2(Electric Force is Electric Charge Multiplied by Field  Magnitude Only) Problem: An electric charge with charge q is placed in an electric field with field magnitude E . Due to the field, the charge experiences a force F . If the charge is doubled and the field tripled, how much force is then exerted on the charge? Select One of the Following: (a) F (b) 2 F (c) 3 F (dAnswer) 6 F (e) 9 F Solution The magnitude of the force F on a particle with charge q in an electric field with magnitude E is equal to F = qE . Thus, the new force will be 6 F . Total Points for Problem: 3 Points Solution to Homework Problem 3.3(If Two Fields Add, There Cannot Be a Point of Zero Field) 1 Problem: A positive + q charge and a negative 2 q charge are placed along the x axis as drawn to the right. Is there a point in the region between the dashed lines (except at infinity) where the magnitude of the electric field is zero? Select One of the Following: (a) Yes, there is a point where the magnitude of the electric field is zero between the dashed lines. (bAnswer) No, there is no point where the mag nitude of the electric field is zero between the dashed lines. x y +q2q Solution The field cannot be zero between the particles, because both fields point in the same direction at all points in the region. Total Points for Problem: 3 Points Solution to Homework Problem 3.4(Basic Electrostatics) Problem: A point charge with charge 3nC is at +4cmy . Calculate the acceleration (as a vector) of a point charge with charge 2nC and mass 1g = 1 10 3 kg placed at point P at P = 3cmx 2cmy . Select One of the Following: (a) vectora = 5 . 36 10 2 m s 2 x + 1 . 07 10 2 m s 2 y (b) vectora = +7 . 36 10 3 m s 2 x + 3 . 07 10 2 m s 2 y (c) vectora = 3 . 26 10 3 m s 2 x + 1 . 07 10 3 m s 2 y (d) vectora = +1 . 11 10 1 m s 2 x + 2 . 07 10 4 m s 2 y (eAnswer) vectora = +5 . 36 10 3 m s 2 x + 1 . 07 10 2 m s 2 y Solution (a) Let vector r 1 = 4cmy = (0 , 4cm) the position of the point charge. Let vector r 2 = 3cmx 2cmy = ( 3cm , 2cm) the position of the point P . So the displacement vector is....
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This note was uploaded on 03/29/2010 for the course PHYS 2469 taught by Professor Stewat during the Spring '10 term at University of Arkansas Community College at Batesville.
 Spring '10
 Stewat
 Charge, Electric Fields, Force, Work

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