HW-1Solutions-01-08-08

HW-1Solutions-01-08-08 - Physics 21 Spring 2008 HW-1...

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Physics 21 Spring 2008 January 7, 2008 H W - 1 S o l u t i o n s 1-1. (HRW 21-3) A particle of charge is 12.0 cm distant from a second particle of charge . Calculate the magnitude of the electrostatic force between the particles. Solution: The magnitude of the mutual force of attraction at r = 0.120 m is () 66 12 92 2 22 3.00 10 C 1.50 10 C 8.99 10 N m C 2.81N. (0.120 m) qq Fk r −− ×× == × = 1-2. (HRW 21-10) Three particles are fixed on an x axis. Particle 1 of charge is at , and particle 2 of charge is at . If their net electrostatic force on particle 3 of charge is to be zero, what must be the ratio when particle 3 is at (a) and (b) ? Solution: (a) The individual force magnitudes (acting on Q ) are, by Eq. 21-1, 00 11 44 /2 qQ aa πε = which leads to | q 1 | = 9.0 | q 2 |. Since Q is located between q 1 and q 2 , we conclude q 1 and q 2 are like-sign. Consequently, q 1 / q 2 = 9.0. (b) Now we have 3/2 = which yields | q 1 | = 25 | q 2 |. Now, Q is not located between q 1 and q 2 , one of them must push and the other must pull. Thus, they are unlike-sign, so q 1 / q 2 = –25.
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This note was uploaded on 02/20/2008 for the course PHYSIC 2 taught by Professor Kim during the Spring '08 term at Lehigh University .

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HW-1Solutions-01-08-08 - Physics 21 Spring 2008 HW-1...

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