HW-2Solutions-01-08-08

# HW-2Solutions-01-08-08 - Physics 21 Spring 2008 HW-2...

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Physics 21 Spring 2008 January 8, 2008 HW-2 Solutions 2-1. (HRW 21-40) A particle of charge Q is fixed at the origin of an xy coordinate system. At a particle ( ) is located on the x axis at , moving with a speed of in the positive y direction. For what value of Q will the moving particle execute circular motion? (Neglect the gravitational force on the particle.) Solution: For the Coulomb force to be sufficient for circular motion at that distance (where r = 0.200 m and the acceleration needed for circular motion is a = v 2 / r) the following equality is required: 2 2 0 4 Qq mv r r πε = − . With q = 4.00 × 10 6 C, m = 0.000800 kg, v = 50.0 m/s, this leads to 2 4 2 5 0 9 2 2 6 4 (0.200 m)(8.00 10 kg)(50.0 m/s) 1.11 10 C (8.99 10 N m C )(4.00 10 C) rmv Q q πε × = − = − = − × × × . 2-2. (HRW 21-42) A charge of is to be split into two parts that are then separated by . What is the maximum possible magnitude of the electrostatic force between those two parts? Solution: Let q 1 be the charge of one part and q 2 that of the other part; thus, q 1 + q 2 = Q = 6.0 μ C.

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• Spring '08
• Kim
• Charge, Electric charge, HRW

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