T2a5s - Physics 102Assignment 5 Solutions(Spring Term{1 15.8 The magnitude of the repulsive force between electrons must equal the weight of an

# T2a5s - Physics 102Assignment 5 Solutions(Spring Term{1...

• 3

This preview shows page 1 - 2 out of 3 pages.

Physics 102Assignment 5 Solutions (Spring Term) {1} 15.8 The magnitude of the repulsive force between electrons must equal the weight of an electron, Thus, 2 2 e e k e r m g = or ( )( ) ( ) ( ) 2 9 2 2 19 2 31 2 8.99 10 N m C 1.60 10 C 9.11 10 kg 9.80 m s e e k e r m g × × = = = × 5.08 m {2} 15.15 Consider the free-body diagram of one of the spheres given at the right. Here, T is the tension in the string and e F is the repulsive electrical force exerted by the other sphere. 0 cos5.0 y F T mg = ° = Σ , or cos5.0 mg T = ° 0 sin5.0 tan5.0 x e F F T mg = = ° = ° Σ At equilibrium, the distance separating the two spheres is 2 sin5.0 r L = ° . Thus, tan5.0 e F mg = ° becomes ( ) 2 2 tan5.0 2 sin5.0 e k q mg L = ° ° and yields ( ) tan5.0 2 sin5.0 e mg q L k ° = ° ( ) ( ) ( ) 3 2 9 2 2 0.20 10 kg 9.80 m s tan5.0 2 0.300 m sin5.0 8.99 10 N m C × ° = ° = × 7.2 nC {3} 15.16 The required position is shown in the sketch at the right. Note that this places q closer to the smaller charge, which will allow the two forces to cancel. Requiring that 6 3 F F = gives ( ) ( ) ( ) 2 2 6.00 nC 3.00 nC 0.600 m e e k q k q x x = + , or ( ) 2
• • • 