# hw4 - Homework 4 Solutions Spring 2010 Required Problems...

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Homework 4 Solutions Spring 2010 Required Problems Chapter 21 8. Use the charge per electron and the mass per electron. ( 29 ( 29 6 14 14 19 31 14 16 1 electron 46 10 C 2.871 10 2.9 10 electrons 1.602 10 C 9.109 10 kg 2.871 10 e 2.6 10 kg 1 e - - - - - - - × = × × - × × × = × 12. Let the right be the positive direction on the line of charges. Use the fact that like charges repel and unlike charges attract to determine the direction of the forces. In the following expressions, 9 2 2 8.988 10 N m C k = × . ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 75 2 2 48 2 2 85 2 2 75 C 48 C 75 C 85 C ˆ 147.2 N 150 N 0.35m 0.70m 75 C 48 C 48 C 85 C ˆ ˆ ˆ ˆ 563.5N 560 N 0.35m 0.35m 85 C 75 C 85 C 48 C ˆ 416.3N 420 N 0.35m ˆ ˆ ˆ ˆ ˆ ˆ k k k k k k μ + + - = - + = - ≈ - = + = = - - = - ≈ - F i F i i i i F i i i i i i i r r r 14. ( a ) If the force is repulsive, both charges must be positive since the total charge is positive. Call the total charge Q . ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 2 2 1 1 1 2 1 2 1 1 2 2 2 2 2 2 1 2 2 6 6 1 2 9 2 2 0 4 4 2 2 12.0N 1.16m 90.0 10 C 90.0 10 C 4 8.988 10 N m C kQ Q Q kQ Q Fd Q Q Q F Q QQ d d k Fd Fd Q Q Q Q k k Q - - - + = = = - + = ± - ± - = = = × ± × - ×

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6 6 60.1 10 C , 29.9 10 C - - = × × ( b ) If the force is attractive, then the charges are of opposite sign. The value used for F must then be negative. Other than that, the solution method is the same as for part ( a ). ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 2 2 1 1 1 2 1 2 1 1 2 2 2 2 2 2 1 2 2 6 6 1 2 9 2 2 6 6 0 4 4 2 2 12.0N 1.16m 90.0 10 C 90.0 10 C 4 8.988 10 N m C 106.8 10 C , 16.8 10 C kQ Q Q kQ Q Fd Q Q Q F Q QQ d d k Fd Fd Q Q Q Q k k Q - - - - - + = = = - + = ± - ± - = = - = × ± × - × = × - × 20. If all of the angles to the vertical (in both cases) are assumed to be small, then the spheres only have horizontal displacement, and so the electric force of repulsion is always horizontal. Likewise, the small angle condition leads to tan sin θ for all small angles. See the free-body diagram for each sphere, showing the three forces of gravity, tension, and the electrostatic force. Take to the right to be the positive horizontal direction, and up to be the positive vertical direction. Since the spheres are in equilibrium, the net force in each direction is zero. ( a ) 1 T1 1 E1 E1 T1 1 sin 0 sin x F F F F F = - = = 1 1 1 T1 1 1 T1 E1 1 1 1 1 1 1 1 cos sin tan cos cos y m g m g F F m g F F m g m g = - = = = = A completely parallel analysis would give E2 2 2 F m g = . Since the electric forces are a Newton’s third law pair, they can be set equal to each other in magnitude. E1 E2 1 1 2 2 1 2 2 1 1 F F m g m g m m θ θ = = = = ( b ) The same analysis can be done for this case. E1 E2 1 1 2 2 1 2 1 1 2 F F m g m g m m = = = = ( c ) The horizontal distance from one sphere to the other is s by the small angle approximation. See the diagram. Use the relationship derived above that E F mg = to solve for the distance.
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hw4 - Homework 4 Solutions Spring 2010 Required Problems...

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