11_Ch 15 College Physics ProblemCH15 Electric Forces and Electric Fields

# 11_Ch 15 College Physics ProblemCH15 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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## This note was uploaded on 12/09/2011 for the course PHYS 2020 taught by Professor Staff during the Fall '10 term at FIU.

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