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11_Chapter 15 HomeworkCH15 Electric Forces and Electric Fields

# 11_Chapter 15 HomeworkCH15 Electric Forces and Electric Fields

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Unformatted text preview: 11 Electric Forces and Electric Fields 15.11 ur 5.00 nC 6.00 nC F6 q ur 0.100 0.300 m m ur FR F3 3.00 nC In the sketch at the right, FR is the resultant of the forces F6 and F3 that are exerted on the charge at the origin by the 6.00 nC and the –3.00 nC charges respectively. 2 6.00 × 10 −9 C ) ( 5.00 × 10 −9 C ) 9 N⋅m ( F6 = 8.99 × 10 2 C2 ( 0.300 m ) = 3.00 × 10 −6 N −9 −9 N ⋅ m 2 ( 3.00 × 10 C ) ( 5.00 × 10 C ) = 1.35 × 10 −5 N F3 = 8.99 × 109 2 C2 0.100 m ) ( The resultant is FR = F 2 + ( F3 ) = 1.38 × 10 −5 N at θ = tan −1 3 F6 = 77.5° FR = 1.38 × 10 −5 N at 77.5° below − x axis Consider the arrangement of charges shown in the sketch at the right. The distance r is r= 2 2 ( 0.500 m ) + ( 0.500 m ) = 0.707 m The forces exerted on the 6.00 nC charge are N ⋅ m ( 6.00 × 10 C ) ( 2 .00 × 10 F2 = 8.99 × 109 2 C2 ( 0.707 m ) 2 −9 −9 C) 3.00 nC 0.500 m 15.12 2 0.500 m or ( F6 ) u r F2 r 45.0° 0.500 m r 45.0° 6.00 nC 2.00 nC = 2 .16 × 10 −7 N and 2 6.00 × 10 −9 C ) ( 3.00 × 10 −9 C ) 9 N⋅m ( F3 = 8.99 × 10 = 3.24 × 10 −7 N 2 C2 ( 0.707 m ) u r F3 ...
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