chp_15_solutions

# chp_15_solutions - Problem Solutions 15.1 Since the charges...

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Unformatted text preview: Problem Solutions 15.1 Since the charges have opposite signs, the force is one of attraction . Its magnitude is ( )( ) ( ) 9 9 2 1 2 9 8 2 2 2 4.5 10 C 2.8 10 C N m 8.99 10 1.1 10 N C 3.2 m e k q q F r--- × × ⋅ = = × = × 15.10 The forces are as shown in the sketch at the right. ( )( ) ( ) 6 6 2 9 1 2 1 2 2 2-2 12 6.00 10 C 1.50 10 C N m 8.99 10 89.9 N C 3.00 10 m e k q q F r-- × × ⋅ = = × = × ( )( ) ( ) 6 6 2 1 3 9 2 2 2 2-2 13 6.00 10 C 2.00 10 C N m 8.99 10 43.2 N C 5.00 10 m e k q q F r-- × × ⋅ = = × = × ( )( ) ( ) 6 6 2 2 3 9 3 2 2 2-2 23 1.50 10 C 2.00 10 C N m 8.99 10 67.4 N C 2.00 10 m e k q q F r-- × × ⋅ = = × = × The net force on the 6 C μ charge is 6 1 2 46.7 N (to the left) F F F =- = The net force on the 1.5 C μ charge is 1.5 1 3 157 N (to the right) F F F = + = The net force on the 2 C μ- charge is 2 2 3 111 N (to the left) F F F- = + = 1 2 &#1; &#1; &#1; &#1; &#1; &#1; CHAPTER 15 15.11 In the sketch at the right, R F is the resultant of the forces 6 3 and F F that are exerted on the charge at the origin by the 6.00 nC and the –3.00 nC charges respectively. ( )( ) ( ) ( )( ) ( ) 9 9 2 9 6 2 2 6 9 9 2 9 5 3 2 2 6.00 10 C 5.00 10 C N m 8.99 10 C 0.300 m 3.00 10 N 3.00 10 C 5.00 10 C N m 8.99 10 1.35 10 N C 0.100 m F F------ × × ⋅ = × = × × × ⋅ = × = × The resultant is ( ) ( ) 2 2 5 6 3 1.38 10 N R F F F- = + = × at 1 3 6 tan 77.5 F F θ- = = ° or 5 1.38 10 N at 77.5 below axis R x- = × °- F 15.13 The forces on the 7.00 μ C charge are shown at the right....
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chp_15_solutions - Problem Solutions 15.1 Since the charges...

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