solutions_25 - PHY2061 R. D. Field Chapter 25 Solutions...

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PHY2061 R. D. Field Solutions Chapter 25 Page 1 of 6 +Q d x q F 1 F 2 θ r Chapter 25 Solutions Problem 1: A +15 microC charge is located 40 cm from a +3.0 microC charge. The magnitude of the electrostatic force on the larger charge and on the smaller charge ( in N ) is, respectively, Answer: 2.5, 2.5 Solution: The magnitiude of the electrostatic for is given by, N cm C C C Nm r Q KQ F 53 . 2 ) 40 ( ) 3 )( 15 )( / 10 99 . 8 ( 2 2 2 9 2 2 1 = × = = µ . Remember the force on 1 due to 2 is equal and opposite to the force on 2 due to 1. Problem 2: Two point particles have charges q 1 and q 2 and are separated by a distance d . Particle q 2 experiences an electrostatic force of 12 milliN due to particle q 1 . If the charges of both particles are doubled and if the distance between them is doubled , what is the magnitude of the electrostatic force between them ( in milliN )? Answer: 12 Solution: The magnitude of the initial electristatic force is mN d q Kq F i 12 2 2 1 = = . The magnitude of the final electrostatic force is mN F d q q K F i f 12 ) 2 ( ) 2 )( 2 ( 2 2 1 = = = . Problem 3: Two identical point charges +Q are located on the y-axis at y=+d/2 and y=-d/2 , as shown in the Figure . A third charge q is placed on the x-axis. At what distance from the origin is the net force on q a maximum ?
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PHY2061 R. D. Field Solutions Chapter 25 Page 2 of 6 Answer: ) 2 2 /( d Solution: The net force on q is the superposition of the forces from each of the two charges as follows: () x d x KQqx x r KQq F F F y x r KQq F y x r KQq F & ) ) 2 / ( ( 2 & cos 2 & sin & cos & sin & cos 2 / 3 2 2 2 2 1 2 2 2 1 + = = + = + = = θ r r r r r where I used r 2 = x 2 + (d/2) 2 and cos θ = x/r. To find the maximum we take the derivative of F
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This note was uploaded on 05/31/2011 for the course PHY 2061 taught by Professor Fry during the Spring '08 term at University of Florida.

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solutions_25 - PHY2061 R. D. Field Chapter 25 Solutions...

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