PHYS-102hw2soln

# PHYS-102hw2soln - 2 0.600 2 = 2.50 N F 2 = 8.99 10 9 10.0...

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PHYS-102 Homework 2 Solutions Spring 06/07 ____________________________________________________________ Chapter 19 Electric Forces and Electric Fields P19.31 ! E = EA cos " A = ! r 2 = 0.200 ( ) 2 = 0.126 m 2 5.20 ! 10 5 = E 0.126 ( ) cos0 ° E = 4.14 ! 10 6 N C = 4.14 MN C P19.36 mg = qE = q 2 " 0 # \$ % ( = q Q A 2 " 0 # \$ % ( Q A = 2 ! 0 mg q = 2 8.85 " 10 # 12 ( ) 0.01 ( ) 9.8 ( ) # 0.7 " 10 # 6 = # 2.48 μ C m 2 P19.55 F = k e q 1 q 2 r 2 : tan = 15.0 60.0 = 14.0 ° F 1 = 8.99 ! 10 9 ( ) 10.0 ! 10 " 6 ( ) 2 0.150 ( ) 2 = 40.0 N F 3 = 8.99 ! 10 9 ( ) 10.0 ! 10 " 6 ( )
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Unformatted text preview: 2 0.600 ( ) 2 = 2.50 N F 2 = 8.99 ! 10 9 ( ) 10.0 ! 10 " 6 ( ) 2 0.619 ( ) 2 = 2.35 N F x = " F 3 " F 2 cos14.0 ° = " 2.50 " 2.35cos14.0 ° = " 4.78 N F y = " F 1 " F 2 sin14.0 ° = " 40.0 " 2.35sin14.0 ° = " 40.6 N F net = F x 2 + F y 2 = 4.78 ( ) 2 + 40.6 ( ) 2 = 40.9 N tan = F y F x = " 40.6 " 4.78 = 263 ° FIG.P19.55...
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## This note was uploaded on 06/02/2008 for the course PHYS 102 taught by Professor N/a during the Spring '08 term at Drexel.

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